Question

If ax3 +3x2 +bx−2 has a factor (2x+3) and leaves remainder 7 when divided by (x+2), find the values of a and b. With these values of a and b, factorise the given expression.

Answer 1

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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,

$p(−23)=0⇒a(−23)3+3(−23)2+b(−23)−3=0⇒a(−827)+3(49)+b(−23)−3=0⇒−827a+427−23b−3=0⇒8(−27a+54−12b)$

$p(−23)=0⇒a(−23)3+3(−23)2+b(−23)−3=0⇒a(−827)+3(49)+b(−23)−3=0⇒−827a+427−23b−3=0⇒8(−27a+54−12b)=3⇒−27a+54−12b=24⇒−3(9a+4b)=24−54=−30⇒9a+4b=10...(i)$

⇒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)