Question

When $2{{x}^{3}}-9{{x}^{2}}+10x-p$ is divided by $\left( x+1 \right)$, the remainder is $-24$. Find the value of $p$.

Answer 1

Hint: We first explain the remainder theorem for polynomial where we take divisor, dividend, quotient and remainder as $A\left( x \right),a\left( x \right),q\left( x \right),r\left( x \right)$ respectively and get the formula as $A\left( x \right)=a\left( x \right)\times q\left( x \right)+r\left( x \right)$. Then we use the root of the equation of the dividend or the quotient to get the remainder. We use that to solve the equation and find the value of $p$.

Complete step by step answer:
We follow the remainder theorem for polynomial where we assume the divisor, dividend, quotient and remainder as $A\left( x \right),a\left( x \right),q\left( x \right),r\left( x \right)$ respectively. Then we say,
$A\left( x \right)=a\left( x \right)\times q\left( x \right)+r\left( x \right)$.
For the given problem we assume,
$A\left( x \right)=2{{x}^{3}}-9{{x}^{2}}+10x-p,a\left( x \right)=\left( x+1 \right),r\left( x \right)=-24$.

We now have to find the value of the $p$.Putting the values, we get
$2{{x}^{3}}-9{{x}^{2}}+10x-p=q\left( x \right)\times \left( x+1 \right)+\left( -24 \right)$.
We now put the value of $x=-1$. This value is the root value of the equation $A\left( x \right)-r\left( x \right)$.
So, the value of $\left( x+1 \right)=-1+1=0$.
\[2{{x}^{3}}-9{{x}^{2}}+10x-p=q\left( x \right)\times \left( x+1 \right)+\left( -24 \right) \\
\Rightarrow 2{{\left( -1 \right)}^{3}}-9{{\left( -1 \right)}^{2}}+10\times \left( -1 \right)-p=q\left( -1 \right)\times 0+\left( -24 \right) \\ \]
Simplifying we get
\[2{{\left( -1 \right)}^{3}}-9{{\left( -1 \right)}^{2}}+10\times \left( -1 \right)-p=-24
\Rightarrow -2-9-10-p=-24 \\
\therefore p=-21+24=3 \\ \]
Therefore, the value of $p$ is 3.

Note: If we find the root of the dividend or the quotient as $t$, then we have to always use the value to find the remainder as $A\left( t \right)$. The root value of $t$ makes the function $A\left( x \right)-r\left( x \right)$ equal to 0 as it also becomes the root of $a\left( x \right)\times q\left( x \right)$.