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Given f(x)=ax 2 +bx+2 and g(x)=bx 2 +ax+1. If x−2 is a factor of f(x) but leaves the remainder −15 when it divides g(x), find the values of a and b. With these values of a and b, factorise the expression. f(x)+g(x)+4x 2 +7x

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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 x2 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

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1. Given

f(x)=ax2+bx+2 and
g(x)=bx2+ax+1
x2 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 (x2), 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)=00
4b+2a+162ab1=0
3b+15=0
3b=15
b=315=5

Substituting this value in equation (i), we get
2a+b+1=0
2a5+1=0
2a4=0
2a=4
a=2

f(x)=ax2+bx+2=2x25x+2
g(x)=bx2+ax+1=5x2+2x+1

Then, f(x)+g(x)+4x2+7x
=2x25x+25x2+2x+1+4x2+7x
=x2+4x+3
=x2+3x+x+3
=x(x+3)+1(x+3)
=(x+1)(x+3)

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