# Find the remainder on dividing the polynomial \[({{x}^{3}}-3{{x}^{2}}+5x+7)\text{ }\!\!~\!\!\text{ }\] by $(x-3)$. Is \[(x-3)\text{ }\!\!~\!\!\text{ }\] a factor of the polynomial?

Last updated date: 19th Mar 2023

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Answer

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Hint: We have been given polynomial \[({{x}^{3}}-3{{x}^{2}}+5x+7)\text{ }\!\!~\!\!\text{ }\] in which we have to find the remainder. Therefore, the remainder theorem can be used to find the remainder. We also have to find whether$(x-3)$ is a factor of the given polynomial. Therefore, factor theorems can be used.

Complete step-by-step answer:

We have to to find the remainder obtained on dividing the polynomial \[({{x}^{3}}-3{{x}^{2}}+5x+7)\text{ }\!\!~\!\!\text{ }\] by $(x-3)$.

Before proceeding with the question, we must know the remainder theorem. The remainder theorem states that if$p(x)$ is a polynomial of degree one or more and $\alpha $ is any real number, then when \[p(x)\] is divided by \[\text{(}x-\alpha )\] , the remainder is \[p(\alpha )\].

Now, we are given that $p(x)$ is $({{x}^{3}}-3{{x}^{2}}+5x+7)$ and $\text{ }\!\!~\!\!\text{ }(x-\alpha )\text{ }\!\!~\!\!\text{ }$ is $(x-3)$. Therefore, for applying the remainder theorem, we have to find out $p(3)$.

$p(3)$ can found out by just putting the value of $x=3$ in the polynomial $({{x}^{3}}-3{{x}^{2}}+5x+7)$.

Therefore, putting the value of $x=3$ in the polynomial $({{x}^{3}}-3{{x}^{2}}+5x+7)$, we get

$\Rightarrow p(3)=({{3}^{3}})-3({{3}^{2}})+5\times 3+7$

$=27-27+15+7$

$=22.....(i)$

Hence, the remainder of the required polynomial is $22$.

Now, again we have to find whether $x-3$ is a factor of the polynomial $({{x}^{3}}-3{{x}^{2}}+5x+7)$. Therefore, we need to use the factor theorem.

According to the factor theorem, \[(x-\alpha )\] is a factor of\[p(x)\] if and only if $p(\alpha )=0$. Here, we have $(x-\alpha )$ as $(x-3)$ so, $x-3$ would be a factor of$p(x)\text{ }\!\!~\!\!\text{ }$ only if $p(3)$ is equal to zero.

We have already calculated $p(3)$ in equation (i).

Therefore, from there we get $p(3)\ne 0$.

Hence, $(x-3)$ is not a factor of the given polynomial $({{x}^{3}}-3{{x}^{2}}+5x+7)$.

Note: Alternate method of solving this question is just dividing the dividend by divisor by performing simple division. If you get any remainder, then the divisor is not the factor of dividend and if the remainder is 0, then the divisor is a factor of the dividend.

In case you are provided two-degree equation or binomial equation, then in that case factor can be found out by middle term splitting method like for example:

$({{x}^{2}}-5x+6)=({{x}^{2}}-3x-2x+6)$

$=x(x-3)-2(x-3)$

$=(x-2)(x-3)$

Therefore, in this case, factors are 2 and 3.

The easiest way to solve this type of question is by remainder theorem which is already provided above.

Complete step-by-step answer:

We have to to find the remainder obtained on dividing the polynomial \[({{x}^{3}}-3{{x}^{2}}+5x+7)\text{ }\!\!~\!\!\text{ }\] by $(x-3)$.

Before proceeding with the question, we must know the remainder theorem. The remainder theorem states that if$p(x)$ is a polynomial of degree one or more and $\alpha $ is any real number, then when \[p(x)\] is divided by \[\text{(}x-\alpha )\] , the remainder is \[p(\alpha )\].

Now, we are given that $p(x)$ is $({{x}^{3}}-3{{x}^{2}}+5x+7)$ and $\text{ }\!\!~\!\!\text{ }(x-\alpha )\text{ }\!\!~\!\!\text{ }$ is $(x-3)$. Therefore, for applying the remainder theorem, we have to find out $p(3)$.

$p(3)$ can found out by just putting the value of $x=3$ in the polynomial $({{x}^{3}}-3{{x}^{2}}+5x+7)$.

Therefore, putting the value of $x=3$ in the polynomial $({{x}^{3}}-3{{x}^{2}}+5x+7)$, we get

$\Rightarrow p(3)=({{3}^{3}})-3({{3}^{2}})+5\times 3+7$

$=27-27+15+7$

$=22.....(i)$

Hence, the remainder of the required polynomial is $22$.

Now, again we have to find whether $x-3$ is a factor of the polynomial $({{x}^{3}}-3{{x}^{2}}+5x+7)$. Therefore, we need to use the factor theorem.

According to the factor theorem, \[(x-\alpha )\] is a factor of\[p(x)\] if and only if $p(\alpha )=0$. Here, we have $(x-\alpha )$ as $(x-3)$ so, $x-3$ would be a factor of$p(x)\text{ }\!\!~\!\!\text{ }$ only if $p(3)$ is equal to zero.

We have already calculated $p(3)$ in equation (i).

Therefore, from there we get $p(3)\ne 0$.

Hence, $(x-3)$ is not a factor of the given polynomial $({{x}^{3}}-3{{x}^{2}}+5x+7)$.

Note: Alternate method of solving this question is just dividing the dividend by divisor by performing simple division. If you get any remainder, then the divisor is not the factor of dividend and if the remainder is 0, then the divisor is a factor of the dividend.

In case you are provided two-degree equation or binomial equation, then in that case factor can be found out by middle term splitting method like for example:

$({{x}^{2}}-5x+6)=({{x}^{2}}-3x-2x+6)$

$=x(x-3)-2(x-3)$

$=(x-2)(x-3)$

Therefore, in this case, factors are 2 and 3.

The easiest way to solve this type of question is by remainder theorem which is already provided above.

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