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)