Question

$P_1\left(x\right)=3x^2+10x+8$ and
$P_2\left(x\right)=x^3+x^2+2x+t$
are two polynomials.
When one of the factors of $P_1\left(x\right)$ divides $P_2\left(x\right)$, 2 is the remainder obtained.

That factor is also a factor of the polynomial $P_3\left(x\right)=2(x+2)$.

Find the value of ‘t’.

• A. 7
• B. 8
• C. 9
• D. 10

Answer 1

The correct option is D 10

Let the factor be $(x-a)$.
Using factor theorem, $P_3\left(a\right)=0$
$P_3\left(x\right)=2(x+2)$
$P_3\left(a\right)=2(a+2)=0$
$⟹a=-2$

So, the factor is $(x+2)$. Verifying that this is also a factor of $P_1\left(x\right)$ using factor theorem,
$P_1(-2)=3\times (-2{)}^2+10\times (-2)+8$
$P_1(-2)=3\times 4-20+8=0$

By remainder theorem, when $P_2\left(x\right)$ is divided by $(x+2)$, the remainder is $P_2(-2)$.
$P_2(-2)=(-2{)}^3+(-2{)}^2+2\times (-2)+t$
$P_2(-2)=-8+4-4+t=t-8$

Given remainder is 2. So,
$P_2(-2)=2$
$⟹t-8=2$
$⟹t=10$