Question

find the remainder when x³ + 3x² +3x+1 is divided by x-¹2​

Answer 1

Question :-

Find the Remainder when the Polynomial $\mathtt{p(x) = x^3 + 3x^2 + 3x + 1}$ is divided by $\mathtt{g(x) = x - 2}$

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Solution :-

Remainder Theorem :-

Let $\mathtt{p(x)}$ be a Polynomial of Degree 1 or more and $\alpha$ be any Real Number.

If $\mathtt{p(x)}$ is divided by $\mathtt{(x - \alpha)}$, then the Quotient is $\mathtt{p(\alpha)}$ .

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So, we have -

$\mathtt{g(x)=0 \Rightarrow (x - 2) = 0 \Rightarrow x=2}$

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By the Remainder Theorem,

When $\mathtt{p(x)}$ is divided by $\mathtt{(x - 2)}$, then the Remainder is $\bf p(2)$.

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Now,

$\begin {aligned}\bold {p(2)} &= \mathtt{(2^3 + 3\times2^2 + 3\times2 + 1)}\\\\&\Rightarrow \mathtt{(8+12+ 6 + 1)}\\\\\therefore\ \bold {p (2)}&=\underline {\boxed {\bf 27}}\ \diamond \end{aligned}$

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Hence, the Required Remainder is 27.

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Additional Information :-

PROOF for Remainder Theorem :-

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Let $\mathtt{p(x)}$ be a Polynomial of Degree 1 or more.

Suppose, When $\mathtt{p(x)}$ is divided by $\mathtt{(x - \alpha)}$, then the Quotient is $\mathtt{q(x)}$ and Remainder is $\mathtt{r(x)}$ .

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By Division Algorithm, we have -

$\begin {aligned} &\mathtt{p(x) = (x-\alpha) \times q(x) + r(x)},\\\\&\text{Where Degree r(x) < Degree (x - \alpha ) = 1}\ \boxed {\because\ (x - \alpha) = 1} \end{aligned}$

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But,

$\begin{aligned} \mathtt {Degree\: r(x) < 1} &\Rightarrow \mathtt{Degree\: r(x) = 0}\\\\&\Rightarrow \mathtt{r(x)\: is\: a\: constant, equal\: to\: r}\ (say)\\\\\therefore\ \mathtt{p(x)} &= \mathtt{(x - \alpha) \times q(x) + r}\ ...(1)\end{aligned}$

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Putting $\mathtt{x-\alpha}$ on both sides of $...(1)$, we get -

$\mathtt{p(\alpha) = 0 \times q(x) + r \Rightarrow r = p(\alpha)}$ .

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Thus, when $\mathtt{p(x)}$ is divided by $\mathtt{(x-\alpha)}$, then the Remainder is $\mathtt{p(\alpha)}$ .