We have been asked to determine the values of a and b from given express f(x)=ax2+bx+2 and g(x)=bx2+ax+1. If x−2 is a factor of f(x) but leaves the remainder −15 when it divides g(x), With these values of a and b, factorise the expression. f(x)+g(x)+4x2+7x

ML Aggarwal, Avichal Publication, Factorisation, chapter 6, question no 27

## Given

f(x)=ax2+bx+2 and

g(x)=bx2+ax+1

x−2 is a factor of f(x)

∴ By factor theorem,

f(2)=0

⇒a(2)2+b(2)+2=0

⇒4a+2b+2=0

By dividing both sides by 2,

⇒2a+b+1=0…(i)

Also given that, g(x) divide dby (x−2), leaves remainder −15

∴ By remainder theorem,

g(2)=−15

⇒b(2)2+2a+1=−15

⇒4b+2a+1+15=0

⇒4b+2a+16=0...(ii)

Now, subtracting equation (i) from equation (ii), we get

(4b+2a+16)–(2a+b+1)=0–0

⇒4b+2a+16−2a−b−1=0

⇒3b+15=0

⇒3b=−15

⇒b=−315=−5

Substituting this value in equation (i), we get

2a+b+1=0

⇒2a−5+1=0

⇒2a−4=0

⇒2a=4

⇒a=2

∴f(x)=ax2+bx+2=2x2–5x+2

g(x)=bx2+ax+1=−5x2+2x+1

Then, f(x)+g(x)+4x2+7x

=2x2–5x+2–5x2+2x+1+4x2+7x

=x2+4x+3

=x2+3x+x+3

=x(x+3)+1(x+3)

=(x+1)(x+3)