A basic question from factorisation chapter in which we have been asked to find the values of and b If ax3+3x2+bx−3 has a factor (2x+3) and leaves remainder −3 when divided by (x+2), With these values of a and b, factorise the given expression.

ML Aggarwal(avichal publication), Factorisation, chapter 6, question no 26

Let

p(x)=ax3+3x2+bx−3

g(x)=2x+3=0⇒x=−23

f(x)=x+2=0⇒x=−2

Given

g(x) is a factor of f(x)

∴ By factor theorem,

p(−23)=0

⇒a(−23)3+3(−23)2+b(−23)−3=0

⇒a(−827)+3(49)+b(−23)−3=0

⇒−827a+427−23b−3=0

⇒8(−27a+54−12b)=3

⇒−27a+54−12b=24

⇒−3(9a+4b)=24−54=−30

⇒9a+4b=10...(i)

Also, p(x) when divided by f(x) leaves a remainder −3

∴ By remainder theorem,

p(−2)=−3

⇒a(−2)3+3(−2)2+b(−2)−3=−3

⇒−8a+12−2b=0

⇒8a+2b=12

⇒4a+b=6...(ii)

Solving (i) and (ii), we get

a=2 and b=−2

Hence p(x)=2x3+3x2−2x−3

=x2(2x+3)−(2x+3)==(2x+3)(x2−1)

=(2x+3)(x+1)(x−1)