MATH 114 - QUANTITATIVE REASONING -
Factoring and Simplifying Algebraic Expressions
Question Bank
Question 1
Step-by-step solution: To factor the expression 6x2+ 11x−10, we need to
find two numbers that multiply to −60 (the product of 6 and −10) and add up
to 11 (the coefficient of the middle term).
The two numbers are 15 and −4, as 15 ×(−4) = −60 and 15 + (−4) = 11.
Now, we rewrite the middle term 11xusing the two new terms: 6x2+ 15x−
4x−10.
Factor by grouping: 3x(2x+ 5) −2(2x+ 5).
Factor out the common factor 2x+ 5: (3x−2)(2x+ 5).
Therefore, the factored form of 6x2+ 11x−10 is (3x−2)(2x+ 5).Question
1: Factor the expression 6x2+ 11x−10.
Step-by-step solution: To factor the expression 6x2+ 11x−10, we
need to find two numbers that multiply to −60 (the product of 6and
−10) and add up to 11 (the coefficient of the middle term).
The two numbers are 15 and −4, as 15×(−4) = −60 and 15+(−4) = 11.
Now, we rewrite the middle term 11xusing the two new terms:
6x2+ 15x−4x−10.
Factor by grouping: 3x(2x+ 5) −2(2x+ 5).
Factor out the common factor 2x+ 5:(3x−2)(2x+ 5).
Therefore, the factored form of 6x2+ 11x−10 is (3x−2)(2x+ 5).
Question 2
Step-by-Step Solution: To factor the given expression 4x2−25, we
will utilize the difference of squares formula: a2−b2= (a+b)(a−b).
First, we recognize that 4x2is the square of 2xand 25 is the square
of 5. So, we can rewrite the expression as: 4x2−25 = (2x)2−52.
Now, applying the difference of squares formula, we get: 4x2−25 =
(2x+ 5)(2x−5).
Therefore, the factored form of 4x2−25 is (2x+ 5)(2x−5).Question
2: Factor the following algebraic expression: 4x2−25
1
Step-by-Step Solution: To factor the given expression 4x2−25, we
will utilize the difference of squares formula: a2−b2= (a+b)(a−b).
First, we recognize that 4x2is the square of 2xand 25 is the square
of 5. So, we can rewrite the expression as: 4x2−25 = (2x)2−52.
Now, applying the difference of squares formula, we get: 4x2−25 =
(2x+ 5)(2x−5).
Therefore, the factored form of 4x2−25 is (2x+ 5)(2x−5).
Question 3
Factor the following algebraic expression: 4x2+ 12xy + 9y2
Solution:
To factor the given expression 4x2+ 12xy + 9y2, we will look for two
numbers that multiply to 4×9 = 36 (coefficient of x2and constant
term) and add up to the middle coefficient, which is 12 (coefficient of
xy).
The two numbers are 6and 6as 6×6 = 36 and 6 + 6 = 12.
Now we rewrite the middle term 12xy as 6xy + 6xy:
4x2+ 6xy + 6xy + 9y2
Now, we group the terms and factor by grouping:
(4x2+ 6xy) + (6xy + 9y2)
2x(2x+ 3y)+3y(2x+ 3y)
Finally, we factor out the common binomial factor 2x+ 3y:
(2x+ 3y)(2x+ 3y)
Thus, the factored form of 4x2+ 12xy + 9y2is (2x+ 3y)2.Question 3:
Factor the following algebraic expression: 4x2+ 12xy + 9y2
Solution:
To factor the given expression 4x2+ 12xy + 9y2, we will look for two
numbers that multiply to 4×9 = 36 (coefficient of x2and constant
term) and add up to the middle coefficient, which is 12 (coefficient of
xy).
The two numbers are 6and 6as 6×6 = 36 and 6 + 6 = 12.
Now we rewrite the middle term 12xy as 6xy + 6xy:
4x2+ 6xy + 6xy + 9y2
Now, we group the terms and factor by grouping:
(4x2+ 6xy) + (6xy + 9y2)
2x(2x+ 3y)+3y(2x+ 3y)
2
Finally, we factor out the common binomial factor 2x+ 3y:
(2x+ 3y)(2x+ 3y)
Thus, the factored form of 4x2+ 12xy + 9y2is (2x+ 3y)2.
Question 4
Factor the following algebraic expression: x2−2x−15
Solution:
To factor the expression x2−2x−15, we need to find two numbers
that multiply to −15 and add up to −2. The numbers that satisfy
these conditions are −5and 3.
Therefore, we can rewrite the expression by splitting the middle
term using −5and 3:
x2−2x−15 = x2−5x+ 3x−15
=x(x−5) + 3(x−5)
= (x+ 3)(x−5)
Thus, the factored form of x2−2x−15 is (x+ 3)(x−5).Question 4:
Factor the following algebraic expression: x2−2x−15
Solution:
To factor the expression x2−2x−15, we need to find two numbers
that multiply to −15 and add up to −2. The numbers that satisfy
these conditions are −5and 3.
Therefore, we can rewrite the expression by splitting the middle
term using −5and 3:
x2−2x−15 = x2−5x+ 3x−15
=x(x−5) + 3(x−5)
= (x+ 3)(x−5)
Thus, the factored form of x2−2x−15 is (x+ 3)(x−5).
Question 5
Factor the following algebraic expression:
3x2+ 12x
Solution:
3
To factor the given expression 3x2+12x, we first look for the great-
est common factor (GCF) of the terms 3x2and 12x.
The GCF of 3x2and 12xis 3x.
Next, we factor out the GCF 3xfrom the expression 3x2+ 12x:
3x2+ 12x= 3x(x+ 4)
Therefore, the factored form of the expression 3x2+ 12xis 3x(x+
4).Question 5:
Factor the following algebraic expression:
3x2+ 12x
Solution:
To factor the given expression 3x2+12x, we first look for the great-
est common factor (GCF) of the terms 3x2and 12x.
The GCF of 3x2and 12xis 3x.
Next, we factor out the GCF 3xfrom the expression 3x2+ 12x:
3x2+ 12x= 3x(x+ 4)
Therefore, the factored form of the expression 3x2+ 12xis 3x(x+4).
Question 6
Solution: To factor the given expression 6x2−9xy, we first look for
the greatest common factor (GCF) of the terms.
1. Identify the common factors of the terms: 6x2and 9xy.
The factors of 6x2are 1,2,3,6, x, x2,2x, 3x, 6x, 6x2.
The factors of 9xy are 1,3,9, x, y, xy, 3x, 3y, 9x, 9y, 9xy.
2. Determine the common factor of 6x2and 9xy.
The common factor between 6x2and 9xy is 3x.
3. Factor out the common factor.
6x2−9xy = 3x(2x−3y)
Therefore, the factored form of 6x2−9xy is 3x(2x−3y).Question 6:
Factor the following algebraic expression: 6x2−9xy
Solution: To factor the given expression 6x2−9xy, we first look for
the greatest common factor (GCF) of the terms.
1. Identify the common factors of the terms: 6x2and 9xy.
The factors of 6x2are 1,2,3,6, x, x2,2x, 3x, 6x, 6x2.
The factors of 9xy are 1,3,9, x, y, xy, 3x, 3y, 9x, 9y, 9xy.
2. Determine the common factor of 6x2and 9xy.
The common factor between 6x2and 9xy is 3x.
3. Factor out the common factor.
6x2−9xy = 3x(2x−3y)
Therefore, the factored form of 6x2−9xy is 3x(2x−3y).
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Question 7
Factor the following algebraic expression: 2x3−8x2+ 4x
Solution:
To factor the given algebraic expression 2x3−8x2+ 4x, we first look
for the greatest common factor (GCF) among the terms. In this case,
the GCF is 2x.
2x3−8x2+ 4x= 2x(x2−4x+ 2)
= 2x(x2−2x−2x+ 2)
= 2x[(x2−2x)−2(x−1)]
= 2x[x(x−2) −2(x−1)]
= 2x(x−2)(x−1)
Therefore, the factored form of the given algebraic expression 2x3−
8x2+ 4xis 2x(x−2)(x−1).Question 7:
Factor the following algebraic expression: 2x3−8x2+ 4x
Solution:
To factor the given algebraic expression 2x3−8x2+ 4x, we first look
for the greatest common factor (GCF) among the terms. In this case,
the GCF is 2x.
2x3−8x2+ 4x= 2x(x2−4x+ 2)
= 2x(x2−2x−2x+ 2)
= 2x[(x2−2x)−2(x−1)]
= 2x[x(x−2) −2(x−1)]
= 2x(x−2)(x−1)
Therefore, the factored form of the given algebraic expression 2x3−
8x2+ 4xis 2x(x−2)(x−1).
Question 8
Solution: To factor the given expression 4x2−16xy, we first look
for the greatest common factor of the two terms.
Step 1: Identify the common factor of the terms. The common
factor of 4x2and 16xy is 4x.
Step 2: Factor out the common factor.
4x2−16xy = 4x(x−4y)
5
Therefore, the factored form of 4x2−16xy is 4x(x−4y).Question 8:
Factor the following algebraic expression: 4x2−16xy
Solution: To factor the given expression 4x2−16xy, we first look
for the greatest common factor of the two terms.
Step 1: Identify the common factor of the terms. The common
factor of 4x2and 16xy is 4x.
Step 2: Factor out the common factor.
4x2−16xy = 4x(x−4y)
Therefore, the factored form of 4x2−16xy is 4x(x−4y).
Question 9
Solution: To factor the given expression 3x2−12xy + 9y2, we will
look for common factors in each term.
First, we notice that all terms are divisible by 3, so we factor out
3:
3x2−12xy + 9y2= 3(x2−4xy + 3y2)
Next, we observe that the expression x2−4xy + 3y2can be factored
as a perfect square trinomial:
x2−4xy + 3y2= (x−3y)(x−y)
Thus, the fully factored form of the given expression 3x2−12xy +9y2
is:
3x2−12xy + 9y2= 3(x−3y)(x−y)
Question 9: Factor the following expression: 3x2−12xy + 9y2
Solution: To factor the given expression 3x2−12xy + 9y2, we will
look for common factors in each term.
First, we notice that all terms are divisible by 3, so we factor out
3:
3x2−12xy + 9y2= 3(x2−4xy + 3y2)
Next, we observe that the expression x2−4xy + 3y2can be factored
as a perfect square trinomial:
x2−4xy + 3y2= (x−3y)(x−y)
Thus, the fully factored form of the given expression 3x2−12xy +9y2
is:
3x2−12xy + 9y2= 3(x−3y)(x−y)
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Question 10
Factor the following algebraic expression completely:
12x2y−18xy2
Solution:
To factor the given expression 12x2y−18xy2, we first look for the
greatest common factor (GCF) of the two terms.
The GCF of 12x2yand −18xy2is 6xy. So, we factor out 6xy from
the expression:
12x2y−18xy2= 6xy(2x−3y)
Hence, the completely factored form of the expression is 6xy(2x−
3y). Question 10:
Factor the following algebraic expression completely:
12x2y−18xy2
Solution:
To factor the given expression 12x2y−18xy2, we first look for the
greatest common factor (GCF) of the two terms.
The GCF of 12x2yand −18xy2is 6xy. So, we factor out 6xy from
the expression:
12x2y−18xy2= 6xy(2x−3y)
Hence, the completely factored form of the expression is 6xy(2x−
3y).
Question 11
Step-by-step solution: To factor the given algebraic expression
6x2+ 15x, we first identify the greatest common factor (GCF) of the
terms. 1. Identify the GCF of the terms 6x2and 15x: - The GCF of
6 and 15 is 3. - The GCF of x2and xis x. Hence, the GCF of the
terms is 3x.
2. Factor out the GCF from the expression:
6x2+ 15x= 3x(2x+ 5).
Therefore, the factored form of the expression 6x2+ 15xis 3x(2x+
5).Question 11: Factor the following algebraic expression: 6x2+ 15x.
Step-by-step solution: To factor the given algebraic expression
6x2+ 15x, we first identify the greatest common factor (GCF) of the
terms. 1. Identify the GCF of the terms 6x2and 15x: - The GCF of
6 and 15 is 3. - The GCF of x2and xis x. Hence, the GCF of the
terms is 3x.
2. Factor out the GCF from the expression:
6x2+ 15x= 3x(2x+ 5).
Therefore, the factored form of the expression 6x2+15xis 3x(2x+5).
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Question 12
Factor the following algebraic expression: 4x2−9
Solution:
To factor the expression 4x2−9, we will use the difference of squares
formula, which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−9
First, rewrite the expression as a difference of squares: 4x2−32
Now, apply the difference of squares formula:
4x2−32= (2x+ 3)(2x−3)
Therefore, the factored form of the expression 4x2−9is (2x+3)(2x−
3).Question 12:
Factor the following algebraic expression: 4x2−9
Solution:
To factor the expression 4x2−9, we will use the difference of squares
formula, which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−9
First, rewrite the expression as a difference of squares: 4x2−32
Now, apply the difference of squares formula:
4x2−32= (2x+ 3)(2x−3)
Therefore, the factored form of the expression 4x2−9is (2x+3)(2x−
3).
Question 13
Step-by-step solution: To factor the given expression, we can use
the difference of squares formula which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−25
Here, a= 2xand b= 5.
Applying the difference of squares formula: 4x2−25 = (2x+5)(2x−5)
Therefore, the factored form of the expression 4x2−25 is (2x+5)(2x−
5).Question 13: Factor the following algebraic expression: 4x2−25
Step-by-step solution: To factor the given expression, we can use
the difference of squares formula which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−25
Here, a= 2xand b= 5.
Applying the difference of squares formula: 4x2−25 = (2x+5)(2x−5)
Therefore, the factored form of the expression 4x2−25 is (2x+
5)(2x−5).
Question 14
Step-by-step solution: To factor the given expression x2−4y2, we
can recognize that it is a difference of squares because x2is a perfect
8
square and 4y2is also a perfect square.
The general form of a difference of squares can be factored as
(a2−b2) = (a+b)(a−b).
In this case, we have a=xand b= 2y. Therefore, we can factor
the expression as:
x2−4y2= (x+ 2y)(x−2y)
Thus, the factored form of the given expression x2−4y2is (x+2y)(x−
2y).Question 14: Factor the following algebraic expression: x2−4y2
Step-by-step solution: To factor the given expression x2−4y2, we
can recognize that it is a difference of squares because x2is a perfect
square and 4y2is also a perfect square.
The general form of a difference of squares can be factored as
(a2−b2)=(a+b)(a−b).
In this case, we have a=xand b= 2y. Therefore, we can factor
the expression as:
x2−4y2= (x+ 2y)(x−2y)
Thus, the factored form of the given expression x2−4y2is (x+
2y)(x−2y).
Question 15
Step-by-step solution: To factor the given expression, we first need
to find the greatest common factor (GCF) of the two terms. The GCF
of 15x2and 10xy is 5x. Therefore, we can rewrite the expression as:
5x(3x+ 2y).
Thus, the factored form of the expression 15x2+ 10xy is 5x(3x+
2y).Question 15: Factor the following algebraic expression: 15x2+10xy.
Step-by-step solution: To factor the given expression, we first need
to find the greatest common factor (GCF) of the two terms. The GCF
of 15x2and 10xy is 5x. Therefore, we can rewrite the expression as:
5x(3x+ 2y).
Thus, the factored form of the expression 15x2+ 10xy is 5x(3x+ 2y).
Question 16
Solution: To factor the given expression 6x2−3x−15, we look for
a common factor for all the terms. In this case, the greatest common
factor is 3. So, we factor out 3:
9
6x2−3x−15 = 3(2x2−x−5)
Next, we need to factor the quadratic expression inside the paren-
theses 2x2−x−5. To do this, we find two numbers that multiply to
the leading coefficient (2) times the constant term (-5), which is -10,
and add up to the coefficient of the linear term (-1).
These two numbers are -2 and 5 because −2×5 = −10 and −2+5 = 3.
So, we rewrite the expression as:
2x2−x−5=2x2+ 5x−6x−5
=x(2x+ 5) −1(2x+ 5)
= (x−1)(2x+ 5)
Therefore, the fully factored form of the expression 6x2−3x−15 is
3(x−1)(2x+ 5).Question 16: Factor the following algebraic expression:
6x2−3x−15.
Solution: To factor the given expression 6x2−3x−15, we look for
a common factor for all the terms. In this case, the greatest common
factor is 3. So, we factor out 3:
6x2−3x−15 = 3(2x2−x−5)
Next, we need to factor the quadratic expression inside the paren-
theses 2x2−x−5. To do this, we find two numbers that multiply to
the leading coefficient (2) times the constant term (-5), which is -10,
and add up to the coefficient of the linear term (-1).
These two numbers are -2 and 5 because −2×5 = −10 and −2+5 = 3.
So, we rewrite the expression as:
2x2−x−5=2x2+ 5x−6x−5
=x(2x+ 5) −1(2x+ 5)
= (x−1)(2x+ 5)
Therefore, the fully factored form of the expression 6x2−3x−15 is
3(x−1)(2x+ 5).
Question 17
Solution: To factor the given expression 4x2+ 12xy + 9y2, we look
for two binomials in the form of (ax +by)(cx +dy)where a,b,c, and d
are constants.
10
We want to find values of a,b,c, and dsuch that when we multiply
(ax +by)(cx +dy), we get the original expression 4x2+ 12xy + 9y2.
Let’s calculate the product of (ax +by)(cx +dy):
(ax +by)(cx +dy) = acx2+adxy +bcxy +bdy2=acx2+ (ad +bc)xy +bdy2
Now we compare the above expansion with the given expression
4x2+ 12xy + 9y2:
ac = 4
ad +bc = 12
bd = 9
By trying different values of a,b,c, and d, we can determine that
the factorization of 4x2+ 12xy + 9y2is (2x+ 3y)2.
Therefore, the factored form of the expression 4x2+ 12xy + 9y2
is (2x+ 3y)2.Question 17: Factor the following algebraic expression:
4x2+ 12xy + 9y2.
Solution: To factor the given expression 4x2+ 12xy + 9y2, we look
for two binomials in the form of (ax +by)(cx +dy)where a,b,c, and d
are constants.
We want to find values of a,b,c, and dsuch that when we multiply
(ax +by)(cx +dy), we get the original expression 4x2+ 12xy + 9y2.
Let’s calculate the product of (ax +by)(cx +dy):
(ax +by)(cx +dy) = acx2+adxy +bcxy +bdy2=acx2+ (ad +bc)xy +bdy2
Now we compare the above expansion with the given expression
4x2+ 12xy + 9y2:
ac = 4
ad +bc = 12
bd = 9
By trying different values of a,b,c, and d, we can determine that
the factorization of 4x2+ 12xy + 9y2is (2x+ 3y)2.
Therefore, the factored form of the expression 4x2+ 12xy + 9y2is
(2x+ 3y)2.
Question 18
Step-by-step Solution: To factor the expression 6a2b−15ab2+ 9ab,
we will first look for the greatest common factor (GCF) of the terms.
1. Identify the common factor among the terms: The common
factor among the terms 6a2b,−15ab2, and 9ab is 3ab.
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2. Rewrite the expression using the common factor: 3ab(2a−5b+3)
Therefore, the factored form of the expression 6a2b−15ab2+ 9ab is
3ab(2a−5b+ 3).Question 18: Factor the following algebraic expression:
6a2b−15ab2+ 9ab
Step-by-step Solution: To factor the expression 6a2b−15ab2+ 9ab,
we will first look for the greatest common factor (GCF) of the terms.
1. Identify the common factor among the terms: The common
factor among the terms 6a2b,−15ab2, and 9ab is 3ab.
2. Rewrite the expression using the common factor: 3ab(2a−5b+3)
Therefore, the factored form of the expression 6a2b−15ab2+ 9ab is
3ab(2a−5b+ 3).
Question 19
Step-by-step Solution: To factor the expression 6x2−17x−3, we
need to find two numbers that multiply to the coefficient of x2which
is 6 and add up to the coefficient of xwhich is -17. Let’s denote these
two numbers as m and n.
1. Multiply the coefficient of x2by the constant term: 6× −3 = −18
2. Find the factors of -18 that add up to -17 (coefficient of x): -18 =
-18 + 1 3. Rewrite the middle term -17x using the factors found in
step 2: 6x2−18x+ 1x−34. Group the terms: (6x2−18x) + (1x−3) 5.
Factor out the greatest common factor in each group: 6x(x−3)+1(x−3)
6. Factor out the common binomial factor: (6x+ 1)(x−3)
Therefore, the factored form of 6x2−17x−3is (6x+1)(x−3).Question
19: Factor the following algebraic expression: 6x2−17x−3
Step-by-step Solution: To factor the expression 6x2−17x−3, we
need to find two numbers that multiply to the coefficient of x2which
is 6 and add up to the coefficient of xwhich is -17. Let’s denote these
two numbers as m and n.
1. Multiply the coefficient of x2by the constant term: 6× −3 = −18
2. Find the factors of -18 that add up to -17 (coefficient of x): -18 =
-18 + 1 3. Rewrite the middle term -17x using the factors found in
step 2: 6x2−18x+ 1x−34. Group the terms: (6x2−18x) + (1x−3) 5.
Factor out the greatest common factor in each group: 6x(x−3)+1(x−3)
6. Factor out the common binomial factor: (6x+ 1)(x−3)
Therefore, the factored form of 6x2−17x−3is (6x+ 1)(x−3).
Question 20
Factor the following algebraic expression:
6x2+ 3x−15
Solution:
12
To factor the given expression 6x2+ 3x−15, we first look for a
common factor among the coefficients of the terms.
Step 1: Factor out the greatest common factor, which is 3.
3(2x2+x−5)
Step 2: To further factor the expression inside the parentheses,
we look for two numbers that multiply to the constant term (-5) and
add up to the coefficient of the linear term (1). The numbers are 5
and -1.
Step 3: Rewrite the expression with the new set of numbers:
3(2x2+ 5x−1x−5)
Step 4: Factor by grouping:
3(2x2+ 5x−1x−5)
= 3(2x(x+ 5) −1(x+ 5))
Step 5: Factor out the common factor in each pair of terms:
3((2x−1)(x+ 5))
Therefore, the factored form of 6x2+3x−15 is 3(2x−1)(x+5).Question
20:
Factor the following algebraic expression:
6x2+ 3x−15
Solution:
To factor the given expression 6x2+ 3x−15, we first look for a
common factor among the coefficients of the terms.
Step 1: Factor out the greatest common factor, which is 3.
3(2x2+x−5)
Step 2: To further factor the expression inside the parentheses,
we look for two numbers that multiply to the constant term (-5) and
add up to the coefficient of the linear term (1). The numbers are 5
and -1.
Step 3: Rewrite the expression with the new set of numbers:
3(2x2+ 5x−1x−5)
Step 4: Factor by grouping:
3(2x2+ 5x−1x−5)
= 3(2x(x+ 5) −1(x+ 5))
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Step-by-Step Solution: To factor the given expression 4x2−25, we
will utilize the difference of squares formula: a2−b2= (a+b)(a−b).
First, we recognize that 4x2is the square of 2xand 25 is the square
of 5. So, we can rewrite the expression as: 4x2−25 = (2x)2−52.
Now, applying the difference of squares formula, we get: 4x2−25 =
(2x+ 5)(2x−5).
Therefore, the factored form of 4x2−25 is (2x+ 5)(2x−5).
Question 3
Factor the following algebraic expression: 4x2+ 12xy + 9y2
Solution:
To factor the given expression 4x2+ 12xy + 9y2, we will look for two
numbers that multiply to 4×9 = 36 (coefficient of x2and constant
term) and add up to the middle coefficient, which is 12 (coefficient of
xy).
The two numbers are 6and 6as 6×6 = 36 and 6 + 6 = 12.
Now we rewrite the middle term 12xy as 6xy + 6xy:
4x2+ 6xy + 6xy + 9y2
Now, we group the terms and factor by grouping:
(4x2+ 6xy) + (6xy + 9y2)
2x(2x+ 3y)+3y(2x+ 3y)
Finally, we factor out the common binomial factor 2x+ 3y:
(2x+ 3y)(2x+ 3y)
Thus, the factored form of 4x2+ 12xy + 9y2is (2x+ 3y)2.Question 3:
Factor the following algebraic expression: 4x2+ 12xy + 9y2
Solution:
To factor the given expression 4x2+ 12xy + 9y2, we will look for two
numbers that multiply to 4×9 = 36 (coefficient of x2and constant
term) and add up to the middle coefficient, which is 12 (coefficient of
xy).
The two numbers are 6and 6as 6×6 = 36 and 6 + 6 = 12.
Now we rewrite the middle term 12xy as 6xy + 6xy:
4x2+ 6xy + 6xy + 9y2
Now, we group the terms and factor by grouping:
(4x2+ 6xy) + (6xy + 9y2)
2x(2x+ 3y)+3y(2x+ 3y)
2
Finally, we factor out the common binomial factor 2x+ 3y:
(2x+ 3y)(2x+ 3y)
Thus, the factored form of 4x2+ 12xy + 9y2is (2x+ 3y)2.
Question 4
Factor the following algebraic expression: x2−2x−15
Solution:
To factor the expression x2−2x−15, we need to find two numbers
that multiply to −15 and add up to −2. The numbers that satisfy
these conditions are −5and 3.
Therefore, we can rewrite the expression by splitting the middle
term using −5and 3:
x2−2x−15 = x2−5x+ 3x−15
=x(x−5) + 3(x−5)
= (x+ 3)(x−5)
Thus, the factored form of x2−2x−15 is (x+ 3)(x−5).Question 4:
Factor the following algebraic expression: x2−2x−15
Solution:
To factor the expression x2−2x−15, we need to find two numbers
that multiply to −15 and add up to −2. The numbers that satisfy
these conditions are −5and 3.
Therefore, we can rewrite the expression by splitting the middle
term using −5and 3:
x2−2x−15 = x2−5x+ 3x−15
=x(x−5) + 3(x−5)
= (x+ 3)(x−5)
Thus, the factored form of x2−2x−15 is (x+ 3)(x−5).
Question 5
Factor the following algebraic expression:
3x2+ 12x
Solution:
3
To factor the given expression 3x2+12x, we first look for the great-
est common factor (GCF) of the terms 3x2and 12x.
The GCF of 3x2and 12xis 3x.
Next, we factor out the GCF 3xfrom the expression 3x2+ 12x:
3x2+ 12x= 3x(x+ 4)
Therefore, the factored form of the expression 3x2+ 12xis 3x(x+
4).Question 5:
Factor the following algebraic expression:
3x2+ 12x
Solution:
To factor the given expression 3x2+12x, we first look for the great-
est common factor (GCF) of the terms 3x2and 12x.
The GCF of 3x2and 12xis 3x.
Next, we factor out the GCF 3xfrom the expression 3x2+ 12x:
3x2+ 12x= 3x(x+ 4)
Therefore, the factored form of the expression 3x2+ 12xis 3x(x+4).
Question 6
Solution: To factor the given expression 6x2−9xy, we first look for
the greatest common factor (GCF) of the terms.
1. Identify the common factors of the terms: 6x2and 9xy.
The factors of 6x2are 1,2,3,6, x, x2,2x, 3x, 6x, 6x2.
The factors of 9xy are 1,3,9, x, y, xy, 3x, 3y, 9x, 9y, 9xy.
2. Determine the common factor of 6x2and 9xy.
The common factor between 6x2and 9xy is 3x.
3. Factor out the common factor.
6x2−9xy = 3x(2x−3y)
Therefore, the factored form of 6x2−9xy is 3x(2x−3y).Question 6:
Factor the following algebraic expression: 6x2−9xy
Solution: To factor the given expression 6x2−9xy, we first look for
the greatest common factor (GCF) of the terms.
1. Identify the common factors of the terms: 6x2and 9xy.
The factors of 6x2are 1,2,3,6, x, x2,2x, 3x, 6x, 6x2.
The factors of 9xy are 1,3,9, x, y, xy, 3x, 3y, 9x, 9y, 9xy.
2. Determine the common factor of 6x2and 9xy.
The common factor between 6x2and 9xy is 3x.
3. Factor out the common factor.
6x2−9xy = 3x(2x−3y)
Therefore, the factored form of 6x2−9xy is 3x(2x−3y).
4
Question 7
Factor the following algebraic expression: 2x3−8x2+ 4x
Solution:
To factor the given algebraic expression 2x3−8x2+ 4x, we first look
for the greatest common factor (GCF) among the terms. In this case,
the GCF is 2x.
2x3−8x2+ 4x= 2x(x2−4x+ 2)
= 2x(x2−2x−2x+ 2)
= 2x[(x2−2x)−2(x−1)]
= 2x[x(x−2) −2(x−1)]
= 2x(x−2)(x−1)
Therefore, the factored form of the given algebraic expression 2x3−
8x2+ 4xis 2x(x−2)(x−1).Question 7:
Factor the following algebraic expression: 2x3−8x2+ 4x
Solution:
To factor the given algebraic expression 2x3−8x2+ 4x, we first look
for the greatest common factor (GCF) among the terms. In this case,
the GCF is 2x.
2x3−8x2+ 4x= 2x(x2−4x+ 2)
= 2x(x2−2x−2x+ 2)
= 2x[(x2−2x)−2(x−1)]
= 2x[x(x−2) −2(x−1)]
= 2x(x−2)(x−1)
Therefore, the factored form of the given algebraic expression 2x3−
8x2+ 4xis 2x(x−2)(x−1).
Question 8
Solution: To factor the given expression 4x2−16xy, we first look
for the greatest common factor of the two terms.
Step 1: Identify the common factor of the terms. The common
factor of 4x2and 16xy is 4x.
Step 2: Factor out the common factor.
4x2−16xy = 4x(x−4y)
5
Therefore, the factored form of 4x2−16xy is 4x(x−4y).Question 8:
Factor the following algebraic expression: 4x2−16xy
Solution: To factor the given expression 4x2−16xy, we first look
for the greatest common factor of the two terms.
Step 1: Identify the common factor of the terms. The common
factor of 4x2and 16xy is 4x.
Step 2: Factor out the common factor.
4x2−16xy = 4x(x−4y)
Therefore, the factored form of 4x2−16xy is 4x(x−4y).
Question 9
Solution: To factor the given expression 3x2−12xy + 9y2, we will
look for common factors in each term.
First, we notice that all terms are divisible by 3, so we factor out
3:
3x2−12xy + 9y2= 3(x2−4xy + 3y2)
Next, we observe that the expression x2−4xy + 3y2can be factored
as a perfect square trinomial:
x2−4xy + 3y2= (x−3y)(x−y)
Thus, the fully factored form of the given expression 3x2−12xy +9y2
is:
3x2−12xy + 9y2= 3(x−3y)(x−y)
Question 9: Factor the following expression: 3x2−12xy + 9y2
Solution: To factor the given expression 3x2−12xy + 9y2, we will
look for common factors in each term.
First, we notice that all terms are divisible by 3, so we factor out
3:
3x2−12xy + 9y2= 3(x2−4xy + 3y2)
Next, we observe that the expression x2−4xy + 3y2can be factored
as a perfect square trinomial:
x2−4xy + 3y2= (x−3y)(x−y)
Thus, the fully factored form of the given expression 3x2−12xy +9y2
is:
3x2−12xy + 9y2= 3(x−3y)(x−y)
6
Question 10
Factor the following algebraic expression completely:
12x2y−18xy2
Solution:
To factor the given expression 12x2y−18xy2, we first look for the
greatest common factor (GCF) of the two terms.
The GCF of 12x2yand −18xy2is 6xy. So, we factor out 6xy from
the expression:
12x2y−18xy2= 6xy(2x−3y)
Hence, the completely factored form of the expression is 6xy(2x−
3y). Question 10:
Factor the following algebraic expression completely:
12x2y−18xy2
Solution:
To factor the given expression 12x2y−18xy2, we first look for the
greatest common factor (GCF) of the two terms.
The GCF of 12x2yand −18xy2is 6xy. So, we factor out 6xy from
the expression:
12x2y−18xy2= 6xy(2x−3y)
Hence, the completely factored form of the expression is 6xy(2x−
3y).
Question 11
Step-by-step solution: To factor the given algebraic expression
6x2+ 15x, we first identify the greatest common factor (GCF) of the
terms. 1. Identify the GCF of the terms 6x2and 15x: - The GCF of
6 and 15 is 3. - The GCF of x2and xis x. Hence, the GCF of the
terms is 3x.
2. Factor out the GCF from the expression:
6x2+ 15x= 3x(2x+ 5).
Therefore, the factored form of the expression 6x2+ 15xis 3x(2x+
5).Question 11: Factor the following algebraic expression: 6x2+ 15x.
Step-by-step solution: To factor the given algebraic expression
6x2+ 15x, we first identify the greatest common factor (GCF) of the
terms. 1. Identify the GCF of the terms 6x2and 15x: - The GCF of
6 and 15 is 3. - The GCF of x2and xis x. Hence, the GCF of the
terms is 3x.
2. Factor out the GCF from the expression:
6x2+ 15x= 3x(2x+ 5).
Therefore, the factored form of the expression 6x2+15xis 3x(2x+5).
7
Question 12
Factor the following algebraic expression: 4x2−9
Solution:
To factor the expression 4x2−9, we will use the difference of squares
formula, which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−9
First, rewrite the expression as a difference of squares: 4x2−32
Now, apply the difference of squares formula:
4x2−32= (2x+ 3)(2x−3)
Therefore, the factored form of the expression 4x2−9is (2x+3)(2x−
3).Question 12:
Factor the following algebraic expression: 4x2−9
Solution:
To factor the expression 4x2−9, we will use the difference of squares
formula, which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−9
First, rewrite the expression as a difference of squares: 4x2−32
Now, apply the difference of squares formula:
4x2−32= (2x+ 3)(2x−3)
Therefore, the factored form of the expression 4x2−9is (2x+3)(2x−
3).
Question 13
Step-by-step solution: To factor the given expression, we can use
the difference of squares formula which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−25
Here, a= 2xand b= 5.
Applying the difference of squares formula: 4x2−25 = (2x+5)(2x−5)
Therefore, the factored form of the expression 4x2−25 is (2x+5)(2x−
5).Question 13: Factor the following algebraic expression: 4x2−25
Step-by-step solution: To factor the given expression, we can use
the difference of squares formula which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−25
Here, a= 2xand b= 5.
Applying the difference of squares formula: 4x2−25 = (2x+5)(2x−5)
Therefore, the factored form of the expression 4x2−25 is (2x+
5)(2x−5).
Question 14
Step-by-step solution: To factor the given expression x2−4y2, we
can recognize that it is a difference of squares because x2is a perfect
8
square and 4y2is also a perfect square.
The general form of a difference of squares can be factored as
(a2−b2) = (a+b)(a−b).
In this case, we have a=xand b= 2y. Therefore, we can factor
the expression as:
x2−4y2= (x+ 2y)(x−2y)
Thus, the factored form of the given expression x2−4y2is (x+2y)(x−
2y).Question 14: Factor the following algebraic expression: x2−4y2
Step-by-step solution: To factor the given expression x2−4y2, we
can recognize that it is a difference of squares because x2is a perfect
square and 4y2is also a perfect square.
The general form of a difference of squares can be factored as
(a2−b2)=(a+b)(a−b).
In this case, we have a=xand b= 2y. Therefore, we can factor
the expression as:
x2−4y2= (x+ 2y)(x−2y)
Thus, the factored form of the given expression x2−4y2is (x+
2y)(x−2y).
Question 15
Step-by-step solution: To factor the given expression, we first need
to find the greatest common factor (GCF) of the two terms. The GCF
of 15x2and 10xy is 5x. Therefore, we can rewrite the expression as:
5x(3x+ 2y).
Thus, the factored form of the expression 15x2+ 10xy is 5x(3x+
2y).Question 15: Factor the following algebraic expression: 15x2+10xy.
Step-by-step solution: To factor the given expression, we first need
to find the greatest common factor (GCF) of the two terms. The GCF
of 15x2and 10xy is 5x. Therefore, we can rewrite the expression as:
5x(3x+ 2y).
Thus, the factored form of the expression 15x2+ 10xy is 5x(3x+ 2y).
Question 16
Solution: To factor the given expression 6x2−3x−15, we look for
a common factor for all the terms. In this case, the greatest common
factor is 3. So, we factor out 3:
9
6x2−3x−15 = 3(2x2−x−5)
Next, we need to factor the quadratic expression inside the paren-
theses 2x2−x−5. To do this, we find two numbers that multiply to
the leading coefficient (2) times the constant term (-5), which is -10,
and add up to the coefficient of the linear term (-1).
These two numbers are -2 and 5 because −2×5 = −10 and −2+5 = 3.
So, we rewrite the expression as:
2x2−x−5=2x2+ 5x−6x−5
=x(2x+ 5) −1(2x+ 5)
= (x−1)(2x+ 5)
Therefore, the fully factored form of the expression 6x2−3x−15 is
3(x−1)(2x+ 5).Question 16: Factor the following algebraic expression:
6x2−3x−15.
Solution: To factor the given expression 6x2−3x−15, we look for
a common factor for all the terms. In this case, the greatest common
factor is 3. So, we factor out 3:
6x2−3x−15 = 3(2x2−x−5)
Next, we need to factor the quadratic expression inside the paren-
theses 2x2−x−5. To do this, we find two numbers that multiply to
the leading coefficient (2) times the constant term (-5), which is -10,
and add up to the coefficient of the linear term (-1).
These two numbers are -2 and 5 because −2×5 = −10 and −2+5 = 3.
So, we rewrite the expression as:
2x2−x−5=2x2+ 5x−6x−5
=x(2x+ 5) −1(2x+ 5)
= (x−1)(2x+ 5)
Therefore, the fully factored form of the expression 6x2−3x−15 is
3(x−1)(2x+ 5).
Question 17
Solution: To factor the given expression 4x2+ 12xy + 9y2, we look
for two binomials in the form of (ax +by)(cx +dy)where a,b,c, and d
are constants.
10
We want to find values of a,b,c, and dsuch that when we multiply
(ax +by)(cx +dy), we get the original expression 4x2+ 12xy + 9y2.
Let’s calculate the product of (ax +by)(cx +dy):
(ax +by)(cx +dy) = acx2+adxy +bcxy +bdy2=acx2+ (ad +bc)xy +bdy2
Now we compare the above expansion with the given expression
4x2+ 12xy + 9y2:
ac = 4
ad +bc = 12
bd = 9
By trying different values of a,b,c, and d, we can determine that
the factorization of 4x2+ 12xy + 9y2is (2x+ 3y)2.
Therefore, the factored form of the expression 4x2+ 12xy + 9y2
is (2x+ 3y)2.Question 17: Factor the following algebraic expression:
4x2+ 12xy + 9y2.
Solution: To factor the given expression 4x2+ 12xy + 9y2, we look
for two binomials in the form of (ax +by)(cx +dy)where a,b,c, and d
are constants.
We want to find values of a,b,c, and dsuch that when we multiply
(ax +by)(cx +dy), we get the original expression 4x2+ 12xy + 9y2.
Let’s calculate the product of (ax +by)(cx +dy):
(ax +by)(cx +dy) = acx2+adxy +bcxy +bdy2=acx2+ (ad +bc)xy +bdy2
Now we compare the above expansion with the given expression
4x2+ 12xy + 9y2:
ac = 4
ad +bc = 12
bd = 9
By trying different values of a,b,c, and d, we can determine that
the factorization of 4x2+ 12xy + 9y2is (2x+ 3y)2.
Therefore, the factored form of the expression 4x2+ 12xy + 9y2is
(2x+ 3y)2.
Question 18
Step-by-step Solution: To factor the expression 6a2b−15ab2+ 9ab,
we will first look for the greatest common factor (GCF) of the terms.
1. Identify the common factor among the terms: The common
factor among the terms 6a2b,−15ab2, and 9ab is 3ab.
11
2. Rewrite the expression using the common factor: 3ab(2a−5b+3)
Therefore, the factored form of the expression 6a2b−15ab2+ 9ab is
3ab(2a−5b+ 3).Question 18: Factor the following algebraic expression:
6a2b−15ab2+ 9ab
Step-by-step Solution: To factor the expression 6a2b−15ab2+ 9ab,
we will first look for the greatest common factor (GCF) of the terms.
1. Identify the common factor among the terms: The common
factor among the terms 6a2b,−15ab2, and 9ab is 3ab.
2. Rewrite the expression using the common factor: 3ab(2a−5b+3)
Therefore, the factored form of the expression 6a2b−15ab2+ 9ab is
3ab(2a−5b+ 3).
Question 19
Step-by-step Solution: To factor the expression 6x2−17x−3, we
need to find two numbers that multiply to the coefficient of x2which
is 6 and add up to the coefficient of xwhich is -17. Let’s denote these
two numbers as m and n.
1. Multiply the coefficient of x2by the constant term: 6× −3 = −18
2. Find the factors of -18 that add up to -17 (coefficient of x): -18 =
-18 + 1 3. Rewrite the middle term -17x using the factors found in
step 2: 6x2−18x+ 1x−34. Group the terms: (6x2−18x) + (1x−3) 5.
Factor out the greatest common factor in each group: 6x(x−3)+1(x−3)
6. Factor out the common binomial factor: (6x+ 1)(x−3)
Therefore, the factored form of 6x2−17x−3is (6x+1)(x−3).Question
19: Factor the following algebraic expression: 6x2−17x−3
Step-by-step Solution: To factor the expression 6x2−17x−3, we
need to find two numbers that multiply to the coefficient of x2which
is 6 and add up to the coefficient of xwhich is -17. Let’s denote these
two numbers as m and n.
1. Multiply the coefficient of x2by the constant term: 6× −3 = −18
2. Find the factors of -18 that add up to -17 (coefficient of x): -18 =
-18 + 1 3. Rewrite the middle term -17x using the factors found in
step 2: 6x2−18x+ 1x−34. Group the terms: (6x2−18x) + (1x−3) 5.
Factor out the greatest common factor in each group: 6x(x−3)+1(x−3)
6. Factor out the common binomial factor: (6x+ 1)(x−3)
Therefore, the factored form of 6x2−17x−3is (6x+ 1)(x−3).
Question 20
Factor the following algebraic expression:
6x2+ 3x−15
Solution:
12
To factor the given expression 6x2+ 3x−15, we first look for a
common factor among the coefficients of the terms.
Step 1: Factor out the greatest common factor, which is 3.
3(2x2+x−5)
Step 2: To further factor the expression inside the parentheses,
we look for two numbers that multiply to the constant term (-5) and
add up to the coefficient of the linear term (1). The numbers are 5
and -1.
Step 3: Rewrite the expression with the new set of numbers:
3(2x2+ 5x−1x−5)
Step 4: Factor by grouping:
3(2x2+ 5x−1x−5)
= 3(2x(x+ 5) −1(x+ 5))
Step 5: Factor out the common factor in each pair of terms:
3((2x−1)(x+ 5))
Therefore, the factored form of 6x2+3x−15 is 3(2x−1)(x+5).Question
20:
Factor the following algebraic expression:
6x2+ 3x−15
Solution:
To factor the given expression 6x2+ 3x−15, we first look for a
common factor among the coefficients of the terms.
Step 1: Factor out the greatest common factor, which is 3.
3(2x2+x−5)
Step 2: To further factor the expression inside the parentheses,
we look for two numbers that multiply to the constant term (-5) and
add up to the coefficient of the linear term (1). The numbers are 5
and -1.
Step 3: Rewrite the expression with the new set of numbers:
3(2x2+ 5x−1x−5)
Step 4: Factor by grouping:
3(2x2+ 5x−1x−5)
= 3(2x(x+ 5) −1(x+ 5))
13
Step-by-Step Solution: To factor the given expression 4x2−25, we
will utilize the difference of squares formula: a2−b2= (a+b)(a−b).
First, we recognize that 4x2is the square of 2xand 25 is the square
of 5. So, we can rewrite the expression as: 4x2−25 = (2x)2−52.
Now, applying the difference of squares formula, we get: 4x2−25 =
(2x+ 5)(2x−5).
Therefore, the factored form of 4x2−25 is (2x+ 5)(2x−5).
Question 3
Factor the following algebraic expression: 4x2+ 12xy + 9y2
Solution:
To factor the given expression 4x2+ 12xy + 9y2, we will look for two
numbers that multiply to 4×9 = 36 (coefficient of x2and constant
term) and add up to the middle coefficient, which is 12 (coefficient of
xy).
The two numbers are 6and 6as 6×6 = 36 and 6 + 6 = 12.
Now we rewrite the middle term 12xy as 6xy + 6xy:
4x2+ 6xy + 6xy + 9y2
Now, we group the terms and factor by grouping:
(4x2+ 6xy) + (6xy + 9y2)
2x(2x+ 3y)+3y(2x+ 3y)
Finally, we factor out the common binomial factor 2x+ 3y:
(2x+ 3y)(2x+ 3y)
Thus, the factored form of 4x2+ 12xy + 9y2is (2x+ 3y)2.Question 3:
Factor the following algebraic expression: 4x2+ 12xy + 9y2
Solution:
To factor the given expression 4x2+ 12xy + 9y2, we will look for two
numbers that multiply to 4×9 = 36 (coefficient of x2and constant
term) and add up to the middle coefficient, which is 12 (coefficient of
xy).
The two numbers are 6and 6as 6×6 = 36 and 6 + 6 = 12.
Now we rewrite the middle term 12xy as 6xy + 6xy:
4x2+ 6xy + 6xy + 9y2
Now, we group the terms and factor by grouping:
(4x2+ 6xy) + (6xy + 9y2)
2x(2x+ 3y)+3y(2x+ 3y)
2
Finally, we factor out the common binomial factor 2x+ 3y:
(2x+ 3y)(2x+ 3y)
Thus, the factored form of 4x2+ 12xy + 9y2is (2x+ 3y)2.
Question 4
Factor the following algebraic expression: x2−2x−15
Solution:
To factor the expression x2−2x−15, we need to find two numbers
that multiply to −15 and add up to −2. The numbers that satisfy
these conditions are −5and 3.
Therefore, we can rewrite the expression by splitting the middle
term using −5and 3:
x2−2x−15 = x2−5x+ 3x−15
=x(x−5) + 3(x−5)
= (x+ 3)(x−5)
Thus, the factored form of x2−2x−15 is (x+ 3)(x−5).Question 4:
Factor the following algebraic expression: x2−2x−15
Solution:
To factor the expression x2−2x−15, we need to find two numbers
that multiply to −15 and add up to −2. The numbers that satisfy
these conditions are −5and 3.
Therefore, we can rewrite the expression by splitting the middle
term using −5and 3:
x2−2x−15 = x2−5x+ 3x−15
=x(x−5) + 3(x−5)
= (x+ 3)(x−5)
Thus, the factored form of x2−2x−15 is (x+ 3)(x−5).
Question 5
Factor the following algebraic expression:
3x2+ 12x
Solution:
3
To factor the given expression 3x2+12x, we first look for the great-
est common factor (GCF) of the terms 3x2and 12x.
The GCF of 3x2and 12xis 3x.
Next, we factor out the GCF 3xfrom the expression 3x2+ 12x:
3x2+ 12x= 3x(x+ 4)
Therefore, the factored form of the expression 3x2+ 12xis 3x(x+
4).Question 5:
Factor the following algebraic expression:
3x2+ 12x
Solution:
To factor the given expression 3x2+12x, we first look for the great-
est common factor (GCF) of the terms 3x2and 12x.
The GCF of 3x2and 12xis 3x.
Next, we factor out the GCF 3xfrom the expression 3x2+ 12x:
3x2+ 12x= 3x(x+ 4)
Therefore, the factored form of the expression 3x2+ 12xis 3x(x+4).
Question 6
Solution: To factor the given expression 6x2−9xy, we first look for
the greatest common factor (GCF) of the terms.
1. Identify the common factors of the terms: 6x2and 9xy.
The factors of 6x2are 1,2,3,6, x, x2,2x, 3x, 6x, 6x2.
The factors of 9xy are 1,3,9, x, y, xy, 3x, 3y, 9x, 9y, 9xy.
2. Determine the common factor of 6x2and 9xy.
The common factor between 6x2and 9xy is 3x.
3. Factor out the common factor.
6x2−9xy = 3x(2x−3y)
Therefore, the factored form of 6x2−9xy is 3x(2x−3y).Question 6:
Factor the following algebraic expression: 6x2−9xy
Solution: To factor the given expression 6x2−9xy, we first look for
the greatest common factor (GCF) of the terms.
1. Identify the common factors of the terms: 6x2and 9xy.
The factors of 6x2are 1,2,3,6, x, x2,2x, 3x, 6x, 6x2.
The factors of 9xy are 1,3,9, x, y, xy, 3x, 3y, 9x, 9y, 9xy.
2. Determine the common factor of 6x2and 9xy.
The common factor between 6x2and 9xy is 3x.
3. Factor out the common factor.
6x2−9xy = 3x(2x−3y)
Therefore, the factored form of 6x2−9xy is 3x(2x−3y).
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Question 7
Factor the following algebraic expression: 2x3−8x2+ 4x
Solution:
To factor the given algebraic expression 2x3−8x2+ 4x, we first look
for the greatest common factor (GCF) among the terms. In this case,
the GCF is 2x.
2x3−8x2+ 4x= 2x(x2−4x+ 2)
= 2x(x2−2x−2x+ 2)
= 2x[(x2−2x)−2(x−1)]
= 2x[x(x−2) −2(x−1)]
= 2x(x−2)(x−1)
Therefore, the factored form of the given algebraic expression 2x3−
8x2+ 4xis 2x(x−2)(x−1).Question 7:
Factor the following algebraic expression: 2x3−8x2+ 4x
Solution:
To factor the given algebraic expression 2x3−8x2+ 4x, we first look
for the greatest common factor (GCF) among the terms. In this case,
the GCF is 2x.
2x3−8x2+ 4x= 2x(x2−4x+ 2)
= 2x(x2−2x−2x+ 2)
= 2x[(x2−2x)−2(x−1)]
= 2x[x(x−2) −2(x−1)]
= 2x(x−2)(x−1)
Therefore, the factored form of the given algebraic expression 2x3−
8x2+ 4xis 2x(x−2)(x−1).
Question 8
Solution: To factor the given expression 4x2−16xy, we first look
for the greatest common factor of the two terms.
Step 1: Identify the common factor of the terms. The common
factor of 4x2and 16xy is 4x.
Step 2: Factor out the common factor.
4x2−16xy = 4x(x−4y)
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Therefore, the factored form of 4x2−16xy is 4x(x−4y).Question 8:
Factor the following algebraic expression: 4x2−16xy
Solution: To factor the given expression 4x2−16xy, we first look
for the greatest common factor of the two terms.
Step 1: Identify the common factor of the terms. The common
factor of 4x2and 16xy is 4x.
Step 2: Factor out the common factor.
4x2−16xy = 4x(x−4y)
Therefore, the factored form of 4x2−16xy is 4x(x−4y).
Question 9
Solution: To factor the given expression 3x2−12xy + 9y2, we will
look for common factors in each term.
First, we notice that all terms are divisible by 3, so we factor out
3:
3x2−12xy + 9y2= 3(x2−4xy + 3y2)
Next, we observe that the expression x2−4xy + 3y2can be factored
as a perfect square trinomial:
x2−4xy + 3y2= (x−3y)(x−y)
Thus, the fully factored form of the given expression 3x2−12xy +9y2
is:
3x2−12xy + 9y2= 3(x−3y)(x−y)
Question 9: Factor the following expression: 3x2−12xy + 9y2
Solution: To factor the given expression 3x2−12xy + 9y2, we will
look for common factors in each term.
First, we notice that all terms are divisible by 3, so we factor out
3:
3x2−12xy + 9y2= 3(x2−4xy + 3y2)
Next, we observe that the expression x2−4xy + 3y2can be factored
as a perfect square trinomial:
x2−4xy + 3y2= (x−3y)(x−y)
Thus, the fully factored form of the given expression 3x2−12xy +9y2
is:
3x2−12xy + 9y2= 3(x−3y)(x−y)
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Question 10
Factor the following algebraic expression completely:
12x2y−18xy2
Solution:
To factor the given expression 12x2y−18xy2, we first look for the
greatest common factor (GCF) of the two terms.
The GCF of 12x2yand −18xy2is 6xy. So, we factor out 6xy from
the expression:
12x2y−18xy2= 6xy(2x−3y)
Hence, the completely factored form of the expression is 6xy(2x−
3y). Question 10:
Factor the following algebraic expression completely:
12x2y−18xy2
Solution:
To factor the given expression 12x2y−18xy2, we first look for the
greatest common factor (GCF) of the two terms.
The GCF of 12x2yand −18xy2is 6xy. So, we factor out 6xy from
the expression:
12x2y−18xy2= 6xy(2x−3y)
Hence, the completely factored form of the expression is 6xy(2x−
3y).
Question 11
Step-by-step solution: To factor the given algebraic expression
6x2+ 15x, we first identify the greatest common factor (GCF) of the
terms. 1. Identify the GCF of the terms 6x2and 15x: - The GCF of
6 and 15 is 3. - The GCF of x2and xis x. Hence, the GCF of the
terms is 3x.
2. Factor out the GCF from the expression:
6x2+ 15x= 3x(2x+ 5).
Therefore, the factored form of the expression 6x2+ 15xis 3x(2x+
5).Question 11: Factor the following algebraic expression: 6x2+ 15x.
Step-by-step solution: To factor the given algebraic expression
6x2+ 15x, we first identify the greatest common factor (GCF) of the
terms. 1. Identify the GCF of the terms 6x2and 15x: - The GCF of
6 and 15 is 3. - The GCF of x2and xis x. Hence, the GCF of the
terms is 3x.
2. Factor out the GCF from the expression:
6x2+ 15x= 3x(2x+ 5).
Therefore, the factored form of the expression 6x2+15xis 3x(2x+5).
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Question 12
Factor the following algebraic expression: 4x2−9
Solution:
To factor the expression 4x2−9, we will use the difference of squares
formula, which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−9
First, rewrite the expression as a difference of squares: 4x2−32
Now, apply the difference of squares formula:
4x2−32= (2x+ 3)(2x−3)
Therefore, the factored form of the expression 4x2−9is (2x+3)(2x−
3).Question 12:
Factor the following algebraic expression: 4x2−9
Solution:
To factor the expression 4x2−9, we will use the difference of squares
formula, which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−9
First, rewrite the expression as a difference of squares: 4x2−32
Now, apply the difference of squares formula:
4x2−32= (2x+ 3)(2x−3)
Therefore, the factored form of the expression 4x2−9is (2x+3)(2x−
3).
Question 13
Step-by-step solution: To factor the given expression, we can use
the difference of squares formula which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−25
Here, a= 2xand b= 5.
Applying the difference of squares formula: 4x2−25 = (2x+5)(2x−5)
Therefore, the factored form of the expression 4x2−25 is (2x+5)(2x−
5).Question 13: Factor the following algebraic expression: 4x2−25
Step-by-step solution: To factor the given expression, we can use
the difference of squares formula which states that a2−b2= (a+b)(a−b).
Given expression: 4x2−25
Here, a= 2xand b= 5.
Applying the difference of squares formula: 4x2−25 = (2x+5)(2x−5)
Therefore, the factored form of the expression 4x2−25 is (2x+
5)(2x−5).
Question 14
Step-by-step solution: To factor the given expression x2−4y2, we
can recognize that it is a difference of squares because x2is a perfect
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square and 4y2is also a perfect square.
The general form of a difference of squares can be factored as
(a2−b2) = (a+b)(a−b).
In this case, we have a=xand b= 2y. Therefore, we can factor
the expression as:
x2−4y2= (x+ 2y)(x−2y)
Thus, the factored form of the given expression x2−4y2is (x+2y)(x−
2y).Question 14: Factor the following algebraic expression: x2−4y2
Step-by-step solution: To factor the given expression x2−4y2, we
can recognize that it is a difference of squares because x2is a perfect
square and 4y2is also a perfect square.
The general form of a difference of squares can be factored as
(a2−b2)=(a+b)(a−b).
In this case, we have a=xand b= 2y. Therefore, we can factor
the expression as:
x2−4y2= (x+ 2y)(x−2y)
Thus, the factored form of the given expression x2−4y2is (x+
2y)(x−2y).
Question 15
Step-by-step solution: To factor the given expression, we first need
to find the greatest common factor (GCF) of the two terms. The GCF
of 15x2and 10xy is 5x. Therefore, we can rewrite the expression as:
5x(3x+ 2y).
Thus, the factored form of the expression 15x2+ 10xy is 5x(3x+
2y).Question 15: Factor the following algebraic expression: 15x2+10xy.
Step-by-step solution: To factor the given expression, we first need
to find the greatest common factor (GCF) of the two terms. The GCF
of 15x2and 10xy is 5x. Therefore, we can rewrite the expression as:
5x(3x+ 2y).
Thus, the factored form of the expression 15x2+ 10xy is 5x(3x+ 2y).
Question 16
Solution: To factor the given expression 6x2−3x−15, we look for
a common factor for all the terms. In this case, the greatest common
factor is 3. So, we factor out 3:
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6x2−3x−15 = 3(2x2−x−5)
Next, we need to factor the quadratic expression inside the paren-
theses 2x2−x−5. To do this, we find two numbers that multiply to
the leading coefficient (2) times the constant term (-5), which is -10,
and add up to the coefficient of the linear term (-1).
These two numbers are -2 and 5 because −2×5 = −10 and −2+5 = 3.
So, we rewrite the expression as:
2x2−x−5=2x2+ 5x−6x−5
=x(2x+ 5) −1(2x+ 5)
= (x−1)(2x+ 5)
Therefore, the fully factored form of the expression 6x2−3x−15 is
3(x−1)(2x+ 5).Question 16: Factor the following algebraic expression:
6x2−3x−15.
Solution: To factor the given expression 6x2−3x−15, we look for
a common factor for all the terms. In this case, the greatest common
factor is 3. So, we factor out 3:
6x2−3x−15 = 3(2x2−x−5)
Next, we need to factor the quadratic expression inside the paren-
theses 2x2−x−5. To do this, we find two numbers that multiply to
the leading coefficient (2) times the constant term (-5), which is -10,
and add up to the coefficient of the linear term (-1).
These two numbers are -2 and 5 because −2×5 = −10 and −2+5 = 3.
So, we rewrite the expression as:
2x2−x−5=2x2+ 5x−6x−5
=x(2x+ 5) −1(2x+ 5)
= (x−1)(2x+ 5)
Therefore, the fully factored form of the expression 6x2−3x−15 is
3(x−1)(2x+ 5).
Question 17
Solution: To factor the given expression 4x2+ 12xy + 9y2, we look
for two binomials in the form of (ax +by)(cx +dy)where a,b,c, and d
are constants.
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We want to find values of a,b,c, and dsuch that when we multiply
(ax +by)(cx +dy), we get the original expression 4x2+ 12xy + 9y2.
Let’s calculate the product of (ax +by)(cx +dy):
(ax +by)(cx +dy) = acx2+adxy +bcxy +bdy2=acx2+ (ad +bc)xy +bdy2
Now we compare the above expansion with the given expression
4x2+ 12xy + 9y2:
ac = 4
ad +bc = 12
bd = 9
By trying different values of a,b,c, and d, we can determine that
the factorization of 4x2+ 12xy + 9y2is (2x+ 3y)2.
Therefore, the factored form of the expression 4x2+ 12xy + 9y2
is (2x+ 3y)2.Question 17: Factor the following algebraic expression:
4x2+ 12xy + 9y2.
Solution: To factor the given expression 4x2+ 12xy + 9y2, we look
for two binomials in the form of (ax +by)(cx +dy)where a,b,c, and d
are constants.
We want to find values of a,b,c, and dsuch that when we multiply
(ax +by)(cx +dy), we get the original expression 4x2+ 12xy + 9y2.
Let’s calculate the product of (ax +by)(cx +dy):
(ax +by)(cx +dy) = acx2+adxy +bcxy +bdy2=acx2+ (ad +bc)xy +bdy2
Now we compare the above expansion with the given expression
4x2+ 12xy + 9y2:
ac = 4
ad +bc = 12
bd = 9
By trying different values of a,b,c, and d, we can determine that
the factorization of 4x2+ 12xy + 9y2is (2x+ 3y)2.
Therefore, the factored form of the expression 4x2+ 12xy + 9y2is
(2x+ 3y)2.
Question 18
Step-by-step Solution: To factor the expression 6a2b−15ab2+ 9ab,
we will first look for the greatest common factor (GCF) of the terms.
1. Identify the common factor among the terms: The common
factor among the terms 6a2b,−15ab2, and 9ab is 3ab.
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2. Rewrite the expression using the common factor: 3ab(2a−5b+3)
Therefore, the factored form of the expression 6a2b−15ab2+ 9ab is
3ab(2a−5b+ 3).Question 18: Factor the following algebraic expression:
6a2b−15ab2+ 9ab
Step-by-step Solution: To factor the expression 6a2b−15ab2+ 9ab,
we will first look for the greatest common factor (GCF) of the terms.
1. Identify the common factor among the terms: The common
factor among the terms 6a2b,−15ab2, and 9ab is 3ab.
2. Rewrite the expression using the common factor: 3ab(2a−5b+3)
Therefore, the factored form of the expression 6a2b−15ab2+ 9ab is
3ab(2a−5b+ 3).
Question 19
Step-by-step Solution: To factor the expression 6x2−17x−3, we
need to find two numbers that multiply to the coefficient of x2which
is 6 and add up to the coefficient of xwhich is -17. Let’s denote these
two numbers as m and n.
1. Multiply the coefficient of x2by the constant term: 6× −3 = −18
2. Find the factors of -18 that add up to -17 (coefficient of x): -18 =
-18 + 1 3. Rewrite the middle term -17x using the factors found in
step 2: 6x2−18x+ 1x−34. Group the terms: (6x2−18x) + (1x−3) 5.
Factor out the greatest common factor in each group: 6x(x−3)+1(x−3)
6. Factor out the common binomial factor: (6x+ 1)(x−3)
Therefore, the factored form of 6x2−17x−3is (6x+1)(x−3).Question
19: Factor the following algebraic expression: 6x2−17x−3
Step-by-step Solution: To factor the expression 6x2−17x−3, we
need to find two numbers that multiply to the coefficient of x2which
is 6 and add up to the coefficient of xwhich is -17. Let’s denote these
two numbers as m and n.
1. Multiply the coefficient of x2by the constant term: 6× −3 = −18
2. Find the factors of -18 that add up to -17 (coefficient of x): -18 =
-18 + 1 3. Rewrite the middle term -17x using the factors found in
step 2: 6x2−18x+ 1x−34. Group the terms: (6x2−18x) + (1x−3) 5.
Factor out the greatest common factor in each group: 6x(x−3)+1(x−3)
6. Factor out the common binomial factor: (6x+ 1)(x−3)
Therefore, the factored form of 6x2−17x−3is (6x+ 1)(x−3).
Question 20
Factor the following algebraic expression:
6x2+ 3x−15
Solution:
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To factor the given expression 6x2+ 3x−15, we first look for a
common factor among the coefficients of the terms.
Step 1: Factor out the greatest common factor, which is 3.
3(2x2+x−5)
Step 2: To further factor the expression inside the parentheses,
we look for two numbers that multiply to the constant term (-5) and
add up to the coefficient of the linear term (1). The numbers are 5
and -1.
Step 3: Rewrite the expression with the new set of numbers:
3(2x2+ 5x−1x−5)
Step 4: Factor by grouping:
3(2x2+ 5x−1x−5)
= 3(2x(x+ 5) −1(x+ 5))
Step 5: Factor out the common factor in each pair of terms:
3((2x−1)(x+ 5))
Therefore, the factored form of 6x2+3x−15 is 3(2x−1)(x+5).Question
20:
Factor the following algebraic expression:
6x2+ 3x−15
Solution:
To factor the given expression 6x2+ 3x−15, we first look for a
common factor among the coefficients of the terms.
Step 1: Factor out the greatest common factor, which is 3.
3(2x2+x−5)
Step 2: To further factor the expression inside the parentheses,
we look for two numbers that multiply to the constant term (-5) and
add up to the coefficient of the linear term (1). The numbers are 5
and -1.
Step 3: Rewrite the expression with the new set of numbers:
3(2x2+ 5x−1x−5)
Step 4: Factor by grouping:
3(2x2+ 5x−1x−5)
= 3(2x(x+ 5) −1(x+ 5))
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Step 5: Factor out the common factor in each pair of terms:
3((2x−1)(x+ 5))
Therefore, the factored form of 6x2+ 3x−15 is 3(2x−1)(x+ 5).
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