Question

What must be subtracted from $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8$ so that the resulting polynomial is exactly divisible ${{x}^{2}}-3x+2$?
[a] x-1
[b] 200x+200
[c] 219x-200
[d] -219x+200

Answer 1

Hint:Use the fact that the remainder obtained on dividing p(x) by x-a is given by $p\left( a \right)$. Hence prove that $\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{x-1}=g\left( x \right)+\dfrac{19}{x-1}$, where g(x) is some polynomial and $\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{x-2}=h\left( x \right)+\dfrac{238}{x-2}$
Use the fact that $\dfrac{1}{{{x}^{2}}-3x+2}=\dfrac{1}{\left( x-2 \right)\left( x-1 \right)}=\dfrac{1}{x-2}-\dfrac{1}{x-1}$
Hence prove that
$\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{\left( {{x}^{2}}-3x+2 \right)}=h\left( x \right)-g\left( x \right)+\left( \dfrac{238}{x-2}-\dfrac{19}{x-1} \right)$
Hence find the remainder obtained on dividing $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8$ by ${{x}^{2}}-3x+2$. Hence find the polynomial that should be subtracted from $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8$ so that the resulting polynomial is divisible by ${{x}^{2}}-3x+2$

Complete step by step answer:

Let r be remainder obtained on dividing $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8$ by $x-1$
Hence, we have
$8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8=g\left( x \right)\left( x-1 \right)+r$ where g(x) is some polynomial.
Put x = 1, we get
$\begin{align}
  & 8+14-2+7-8=g\left( 1 \right)\left( 1-1 \right)+r \\
 & \Rightarrow r=19 \\
\end{align}$
Hence, we have $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8=g\left( x \right)\left( x-1 \right)+19$
Dividing both sides by x-1, we get
$\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{x-1}=g\left( x \right)+\dfrac{19}{x-1}\text{ }\left( i \right)$
Let s be the remainder obtained on dividing $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8$ by x-2.
Hence, we have
$8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8=h\left( x \right)\left( x-2 \right)+s$, where h(x) is some polynomial.
Put x= 2, we get
$\begin{align}
  & 8\times {{2}^{4}}+14\times {{2}^{3}}-2\times {{2}^{2}}+7\times 2-8=h\left( 2 \right)\left( 2-2 \right)+s \\
 & \Rightarrow s=238 \\
\end{align}$
Hence, we have
$8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8=h\left( x \right)\left( x-2 \right)+238$
Divide both sides by $x-2$, we get
\[\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{x-2}=h\left( x \right)+\dfrac{238}{x-2}\text{ }\left( ii \right)\]
Now, we have ${{x}^{2}}-3x+2={{x}^{2}}-2x-x+2$
Taking x common from the first two terms and -1 common from the last two terms, we get
${{x}^{2}}-3x+2=x\left( x-2 \right)-1\left( x-2 \right)$
Take x-2 common, we get
${{x}^{2}}-3x+2=\left( x-2 \right)\left( x-1 \right)$
Hence, we have
$\dfrac{1}{{{x}^{2}}-3x+2}=\dfrac{1}{\left( x-1 \right)\left( x-2 \right)}=\dfrac{x-1-\left( x-2 \right)}{\left( x-1 \right)\left( x-2 \right)}=\dfrac{1}{x-2}-\dfrac{1}{x-1}$
Hence, we have
\[\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{{{x}^{2}}-3x+2}=\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{\left( x-1 \right)\left( x-2 \right)}=\left( 8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8 \right)\left( \dfrac{1}{x-2}-\dfrac{1}{x-1} \right)\]
Hence by distributive property of multiplication over addition, we have
$\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{{{x}^{2}}-3x+2}=\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{x-2}-\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{x-1}$
From equation (i) and equation (ii), we get
$\dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{{{x}^{2}}-3x+2}=h\left( x \right)+\dfrac{238}{x-2}-g\left( x \right)-\dfrac{19}{x-1}$
Hence, we have
$\begin{align}
  & \dfrac{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}{{{x}^{2}}-3x+2}=h\left( x \right)-g\left( x \right)+\dfrac{1}{8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8}\left( 238x-238-19x+38 \right) \\
 & =h\left( x \right)-g\left( x \right)+\dfrac{1}{{{x}^{2}}-3x+2}\left( 219x-200 \right) \\
\end{align}$
Hence the remainder obtained on dividing $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8$ by ${{x}^{2}}-3x+2$ is $219x-200$
Hence the polynomial that must be subtracted from $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8$ to make it divisible by ${{x}^{2}}-3x+2$ us $219x-200$
Hence option [c] is correct

Note:
Verification: We can verify the correctness of our solution by checking that $219x-200$ subtracted from $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8$ is divisible by ${{x}^{2}}-3x+2$
We have $8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}+7x-8-219x+200=8{{x}^{4}}+14{{x}^{3}}-2{{x}^{2}}-212x+192$
Now, we have

Hence our solution is verified to be correct.