A or N 80 and 106 68
345-6
Q.80 w2 + 30w + 81
Q.106 ac + xc + aw2 +xw2
Page 353
Q-68 10a2 – 27ab + 5b2
Because 10a^2 has factors 1a 10a and 2a 5a
5b^2 has factors 5b and b
We have four ways to solve
1a 5b 10a 5b 2a 5b 5a 5b
1a 1b 10a 1b 2a 1b 2a 1b
Factorisation means the given polynomial is written as product of . When we multiply the factors
factors – we get back the polynomial back.
Whenever there are two or more terms we try finding the GCF of all the terms, from variables we see
what is common to each term and the numbers are written as product of and then we prime factors
find GCF - then that factor is taken out of all the terms and we get a polynomial in factored form.
Quadratic polynomial can be a also. If its so, then the factorisation becomes very easyperfect square
and we can use the special formula of factorisation.
We use a method splitting the middle term, which gives us four terms and then we use grouping
method to find the factors.
We can do factorisation, using any of the three methods.
Splitting the middle term (ac method)
Completing the Square
Quadratic Formula
Pg. 345-6
80. w2 + 30w + 81 ( the 2’s are to the second power)
Arranging in proper form we get
