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# What must be subtracted or added to $p\left( x \right)=8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+8x-12$ so that $4{{x}^{2}}+3x-2$ is a factor of $p\left( x \right)$?

Last updated date: 09th Aug 2024
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Hint: In this question, we have to find what we should subtract from $p\left( x \right)$ so that $q\left( x \right)$ is a factor of $p\left( x \right)$. For this, we just have to divide the polynomials. If the remainder becomes zero then it means that $q\left( x \right)$ is already a factor $p\left( x \right)$, and hence we need not to subtract or add any value. And if the remainder is non-zero, then we have to subtract that remainder from $p\left( x \right)$ so that $q\left( x \right)$ is a factor of $p\left( x \right)$.
We are given $p\left( x \right)=8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+8x-12$ and let $q\left( x \right)=4{{x}^{2}}+3x-2$. We have to find what must be added or subtracted from $p\left( x \right)$ so that $q\left( x \right)$ becomes a factor of $p\left( x \right)$. For this, we will find the remainder when $p\left( x \right)$ is divided by $q\left( x \right)$. Since remainder is an added term in $p\left( x \right)$, we will subtract it from $p\left( x \right)$ to obtain the required answer. Let us use long division method for dividing $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+8x-12$ by $4{{x}^{2}}+3x-2$.
\begin{align} & 4{{x}^{2}}+3x-2\overset{2{{x}^{2}}+2x-1}{\overline{\left){8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+8x-12}\right.}} \\ & ~~~~~~~~~~~~~~~~~~~~-8{{x}^{4}}+6{{x}^{3}}-4{{x}^{2}} \\ & ~~~~~~~~~~~~~~~~~~~~\overline{~~~~0+8{{x}^{3}}+2{{x}^{2}}+8x-12~~~~} \\ & ~~~~~~~~~~~~~~~~~~~~~~~~~-\left( 8{{x}^{3}}+6{{x}^{2}}-4x \right) \\ & ~~~~~~~~~~~~~~~~~~~~~~~~~\overline{~~~~~~~~0-4{{x}^{2}}+12x-12} \\ & ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-\left( -4{{x}^{2}}-3x+2 \right) \\ & ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\overline{~~~~~0+15x-14~~~~~} \\ \end{align}
Explaining the above division step by step, firstly we multiplied $4{{x}^{2}}+3x-2$ by $2{{x}^{2}}$ so that first term become same and can be eliminated while subtracting. Similarly other terms that are subtracted are written. After that $2x$ is multiplied to eliminate the first term. Proceeding similarly we obtained the remainder as $15x-14$. Since the degree of $15x-14$ is less than $4{{x}^{2}}+3x-2$, it cannot be divided further. So $15x-14$ is the required remainder.
Hence, we will have to subtract $15x-14$ from $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+8x-12$ so that $4{{x}^{2}}+3x-2$ become its factor.
Note: Students can make mistakes while performing division. There is a possibility of making mistakes in plus minus signs while dividing. Don’t forget to change signs of all terms while subtracting. The number which is multiplied is a factor of $p\left( x \right)$, that is, $2{{x}^{2}}+2x-1$ is also a factor of $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+8x-12$. Always remember that remainder should always be subtracted from $p\left( x \right)$ to obtain the required answer.