Algebra quiz
salehxp
P.3Polynomials.pdf
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SECTION P.3 POLYNOMIALS
Definitions:
a. A monomial is a constant, a variable, or the product of a constant and one or more variables,
with the variables having only nonnegative integer exponents.
b. Numerical coefficient or coefficient – constant multiplying a variable.
c. Degree of a monomial – sum of exponents of the variables. The constant zero has no degree.
d. Polynomial – sum of finite number of monomials. Degree of polynomial is the greatest of the degrees of the terms.
e. Like or similar terms are terms that have exactly same variables raised to the same power. 7x2y and 5yx2 are like terms. The terms -3xy and 6xy2 are not like terms.
f. Binomial – polynomial with two terms. Trinomial – polynomial with three terms.
A nonzero constant is called constant polynomial.
Operations of Polynomials
a. To add or subtract polynomials, add/subtract the coefficients of the like terms.
b. To multiply polynomials, use distributive property (horizontally or vertically),
or FOIL method
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FOIL method.
Special Product Formulas
Evaluate a Polynomial
To evaluate a polynomial, substitute the given value or values for the variable or variables and then
perform the indicated operations using the Order of Operations Agreement.
In Exercises 1 to 10, match the descriptions, labeled A to J, with the appropriate examples.
A. 𝑥3𝑦 + 𝑥𝑦 B. 7𝑥2 + 5𝑥 − 11
C. 1
2 𝑥2 + 𝑥𝑦 + 𝑦2 D. 4𝑥𝑦
E. 8𝑥3 − 1 F. 3 − 4𝑥2 G. 8 H. 3𝑥5 − 4𝑥2 + 7𝑥 − 11
I. 8𝑥4 − √5𝑥3 + 7 J. 0
2. A binomial of degree 3 → E
4. A binomial with a leading coefficient of −4 → F
6. A fourth-degree polynomial that has a third-degree term → I
8. A trinomial in x and y → C
10. A fourth-degree binomial → A
In Exercises 15 to 20, for each polynomial, determine its
a. standard form, b. degree, c. coefficients, d. leading coefficient, and e. terms.
17. 𝑥3 − 1
a. Standard form: 𝑥3 − 1 b. Degree: 3
c. Coefficients: 𝟏, −𝟏 d. Leading coefficient: : 𝟏 e. Terms:: 𝑥3, −1
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20. 3𝑥2 − 5𝑥3 + 7𝑥 − 1
a. Standard form: −5𝑥3 + 3𝑥2 + 7𝑥 − 1 Degree: 3 b. Coefficients: −𝟓, 𝟑, 𝟕, −𝟏 c. Leading coefficient: −𝟓 d. Terms: −5𝑥3, 3𝑥2, 7𝑥, −1
In Exercises 21 to 26, determine the degree of the given polynomial.
24. −9𝑥5𝑦 + 10𝑥𝑦4 − 11𝑥2𝑦2 Degree: 6
26. 5𝑥2𝑦 − 𝑦4 + 6𝑥𝑦 Degree: 4
In Exercises 27 to 44, perform the indicated operation and simplify if possible by combining like
terms. Write the result in standard form.
27. (3𝑥2 + 4𝑥 + 5) + (2𝑥2 + 7𝑥 − 2)
= (3𝑥2 + 2𝑥2) + (4𝑥 + 7𝑥) + (5 − 2)
= 𝟓𝒙𝟐 + 𝟏𝟏𝒙 + 𝟑
32. (7𝑠2 − 4𝑠 + 11) − (−2𝑠2 + 11𝑠 − 9)
= 7𝑠2 − 4𝑠 + 11 + 2𝑠2 − 11𝑠 + 9
= 7𝑠2 + 2𝑠2 − 4𝑠 − 11𝑠 + 11 + 9
= 𝟗𝒔𝟐 − 𝟏𝟓𝒔 + 𝟐𝟎
38. (3𝑥 − 4)(𝑥2 − 6𝑥 − 9)
= 3𝑥(𝑥2 − 6𝑥 − 9) − 4(𝑥2 − 6𝑥 − 9)
= 3𝑥3 − 18𝑥2 − 27𝑥 − 4𝑥2 + 24𝑥 + 36
= 𝟑𝒙𝟑 − 𝟐𝟐𝒙𝟐 − 𝟑𝒙 + 𝟑𝟔
In Exercises 45 to 58, use the FOIL method to find the indicated product.
54. (3𝑎 − 5𝑏 )(4𝑎 − 7𝑏)
= 3𝑎(4𝑎) + 3𝑎(−7𝑏) − 5𝑏(4𝑎) − 5𝑏(−7𝑏)
= 12𝑎2 − 21𝑎𝑏 − 20𝑎𝑏 + 35𝑏2
= 𝟏𝟐𝒂𝟐 − 𝟒𝟏𝒂𝒃 + 𝟑𝟓𝒃𝟐
In Exercises 59 to 66, use the special product formulas to perform the indicated operation.
60. (4𝑥2 − 3𝑦 )(4𝑥2 + 3𝑦 )
= (4𝑥2)2 − (3𝑦)2 = 𝟏𝟔𝒙𝟒 − 𝟗𝒚𝟐
46.
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62. (6𝑥 + 7𝑦 )2
= (6𝑥)2 + 2(6𝑥)(7𝑦) + (7𝑦)2 = 𝟑𝟔𝒙𝟐 + 𝟖𝟒 𝒙𝒚 + 𝟒𝟗𝒚𝟐
64. (3𝑥 − 5𝑦2 )2
= (3𝑥)2 − 2(3𝑥)(5𝑦2) + (5𝑦2)2 = 𝟗𝒙𝟐 − 𝟑𝟎𝒙𝒚𝟐 + 𝟐𝟓𝒚𝟒
66. [(𝑥 − 2𝑦) + 7][(𝑥 − 2𝑦) – 7]
= (𝑥 − 2𝑦)2 − 72
= 𝒙𝟐 − 𝟒𝒙𝒚 + 𝟒𝒚𝟐 − 𝟒𝟗
In Exercises 67 to 74, evaluate the given polynomial for the indicated value of the variable.
72. 5𝑥3 − 𝑥2 + 5𝑥 − 3, 𝑓𝑜𝑟 𝑥 = −1
= 5(−1)3 − (−1)2 + 5(−1) − 3
= 5(−1) − 1 − 5 − 3 = −14
86. The temperature, in degree Fahrenheit, of a patient after receiving a certain medication is given by
3 2Temperature 0.0002 0.0114 0.0158 104t t t ,
where t is the number of minutes after receiving the medication.
a. What was the patient’s temperature just before the medication was given?
b. What was the patient’s temperature 10 minutes after the medication was given?
Solution:
a. Evaluate, temperature at t = 0
3 2
3 2
Temperature 0.0002 0.0114 0.0158 104
0.0002 0 0.0114 0 0.0158 0 104
104o
t t t
F
The patient’s temperature was 104 o F before taking the medication.
b. Evaluate, temperature at t = 10
Temperature = 0.0002𝑡3 − 0.0114𝑡2 + 0.0158𝑡 + 104
= 0.0002(10)3 − 0.0114(10)2 + 0.0158(10) + 104
= 0.2 − 1.14 + 0.158 + 104
= 103.2180𝐹
The patient’s temperature was about103.2180𝐹, 10 minutes after taking the medication.
Practice Exercise: Solve problems 1,3,5,7,9, 19, 23, 29, 31, 37, 45, 57, 63, 65, 71, 73, 83
