Question

Find the value of ‘t’ for which the polynomial, $p\left(x\right)=x^3+3x^2+2x+t$ gives $3$ as remainder when divided by $(x-1)$.

Answer 1

We have been provided with the cubic polynomial with $3$ as a remainder when divided by $(x-1)$.

Using the remainder theorem, the remainder when $p\left(x\right)$ is divided by $(x-1)$ is $p\left(1\right)$. The remainder is given to be equal to $3$ therefore $p\left(1\right)=3$.

$\Rightarrow p\left(x\right)=x^3+3x^2+2x+t$

$\Rightarrow p\left(1\right)=(1{)}^3+3(1{)}^2+2\left(1\right)+t$

$\Rightarrow p\left(1\right)=1+3+2+t$

$\Rightarrow 3=6+t$

$\Rightarrow t=-3$