Question

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

Answer 1

By remainder theorem

x+1=0

x=-1

$p(x) = {x}^{3} + {3x}^{2} - 3x - 1$
$p( - 1) = {( - 1)}^{3} + 3 {( - 1)}^{2} - 3( - 1) - 1$
$= - 1 + 3(1) + 3 - 1$
$= - 1 + 3 + 3 - 1$
$= 6 - 2 \\ \\ = 4$
Thus remainder is 4

Answer 2

QUESTION:

The correct question should be:

Find the remainder when $x^3+3x^2-3x-1$ is divided by $x+1$?

ANSWER:

The remainder when $x^3+3x^2-3x-1$ is divided by $x+1$ is 4.

Given,

The polynomial $p(x)=x^3+3x^2-3x-1.$

To find,

The remainder when $p(x)$ is divided by $x+1$.

Solution,

According to the remainder theorem, when a polynomial $p(x)$ is divided by another polynomial $q(x)$, then substituting $x=a$ in $p(x)$ gives the remainder, where $x=a$ is a zero of the polynomial $q(x)$.

Here, the given polynomial is

$p(x)=x^3+3x^2-3x-1$

It has to be divided by

$q(x)=x+1$

Now, it can be seen that putting $q(x)=0$ will give us the zero of the polynomial $q(x)$.

$\implies q(x) = x+1=0$

$\implies x= -1$

Thus, $x=-1$ is a zero of the polynomial $q(x)=x+1.$

So, substituting $x=-1$ in $p(x)$, that is, $p(-1)$ will give us the remainder when $p(x)$ is divided by $q(x)$.

Hence, the remainder will be given as,

$p(-1)=(-1)^3+3(-1)^2-3(-1)-1$

$\implies p(-1)=(-1)+3-3(-1)-1$

$\implies p(-1)=-1+3+3-1$

$\implies p(-1)=6-2$

$\implies p(-1)=4$

⇒ remainder = 4.

Therefore, the remainder when $x^3+3x^2-3x-1$ is divided by $x+1$ is 4.

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