# If the polynomial ${{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10$ is divided by another polynomial ${{x}^{2}}-2x+k$, the remainder comes out to be x+a, then the value of a is

(a) -1

(b) -5

(c) 1

(d) 5

Last updated date: 18th Mar 2023

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Answer

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Hint: To find the remainder when ${{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10$ is divided by ${{x}^{2}}-2x+k$, we perform long division method to divide ${{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10$ and ${{x}^{2}}-2x+k$. We would then equate the answer (that is the remainder) to x+a. This would give us the value of a.

Complete step-by-step answer:

Now, to proceed with long division method to divide ${{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10$ by ${{x}^{2}}-2x+k$, we get -

${{x}^{2}}$-4x+(8-k)

${{x}^{2}}-2x+k$ $\left| \!{\overline {\,

{{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10 \,}} \right. $

-$\left( {{x}^{4}}-2{{x}^{3}}+k{{x}^{2}} \right)$

\[\left| \!{\overline {\,

-4{{x}^{3}}+(16-k){{x}^{2}}-25x+10\text{ } \,}} \right. \]

-$\left( -4{{x}^{3}}+\text{ }8{{x}^{2}}-4kx \right)$

\[\left| \!{\overline {\,

(8-k){{x}^{2}}+(4k-25)x+10 \,}} \right. \]

$-\left( (8-k){{x}^{2}}+(2k-16)x+(8k-{{k}^{2}}) \right)$

\[\]

To understand this, we first write down the divisor (${{x}^{2}}-2x+k$) and dividend (${{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10$) as shown above. Next we start with the highest power of x and accordingly find the first term of quotient. Thus, in this case since ${{x}^{4}}$ was the highest power term in the dividend, we divide this by the highest term in the divisor (${{x}^{2}}$), thus we get, $\dfrac{{{x}^{4}}}{{{x}^{2}}}={{x}^{4-2}}={{x}^{2}}$.

Next, we multiply ${{x}^{2}}-2x+k$ and ${{x}^{2}}$(first quotient term) to get ${{x}^{4}}-2{{x}^{3}}+k{{x}^{2}}$. Then we subtract $\left( {{x}^{4}}-2{{x}^{3}}+k{{x}^{2}} \right)$ from ${{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10$ (which is similar to the long division method). Finally, we obtain \[-4{{x}^{3}}+(16-k){{x}^{2}}-25x+10\text{ }\] (from subtraction). We then apply the same technique again (but now, \[-4{{x}^{3}}+(16-k){{x}^{2}}-25x+10\text{ }\]acts as the dividend).

Thus, we divide highest power of \[-4{{x}^{3}}+(16-k){{x}^{2}}-25x+10\text{ }\] (that is ${{x}^{3}}$) by highest power of divisor. We get, $\dfrac{-4{{x}^{3}}}{{{x}^{2}}}$=-4x (which is the next quotient term). We again follow the same procedure, and then we will get \[(8-k){{x}^{2}}+(4k-25)x+10\] as the dividend. Now, similarly, we multiply by $\dfrac{(8-k){{x}^{2}}}{{{x}^{2}}}$=(8-k). Finally, repeating the steps again, we get the remainder as \[(2k-9)x+(10-8k+{{k}^{2}})\] as the remainder. We stop our process here, since, the highest power of the next dividend (\[(2k-9)x+(10-8k+{{k}^{2}})\]) is less than that of the divisor.

Now, we can equate the remainder to x+a.

Thus,

\[(2k-9)x+(10-8k+{{k}^{2}})\]= x+a

Then, by comparing the coefficient,

2k-9=1

Thus, k=5.

Also, $10-8k+{{k}^{2}}$=a

Since, k=5,

a = 10 – 8(5) +${{5}^{2}}$

a = -5

Hence, the correct answer is (b) -5.

Note: To solve a problem by long division method, one needs to remember that all the terms of divisor, dividend and quotient should be algebraic. Thus, there should not be any logarithmic or exponential term. Further, one should stop the process, when the highest power of the divisor is greater than the highest power of the dividend. Basically there should be no $\dfrac{1}{x}$or $\dfrac{1}{{{x}^{2}}}$ terms in the quotient.

Complete step-by-step answer:

Now, to proceed with long division method to divide ${{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10$ by ${{x}^{2}}-2x+k$, we get -

${{x}^{2}}$-4x+(8-k)

${{x}^{2}}-2x+k$ $\left| \!{\overline {\,

{{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10 \,}} \right. $

-$\left( {{x}^{4}}-2{{x}^{3}}+k{{x}^{2}} \right)$

\[\left| \!{\overline {\,

-4{{x}^{3}}+(16-k){{x}^{2}}-25x+10\text{ } \,}} \right. \]

-$\left( -4{{x}^{3}}+\text{ }8{{x}^{2}}-4kx \right)$

\[\left| \!{\overline {\,

(8-k){{x}^{2}}+(4k-25)x+10 \,}} \right. \]

$-\left( (8-k){{x}^{2}}+(2k-16)x+(8k-{{k}^{2}}) \right)$

\[\]

To understand this, we first write down the divisor (${{x}^{2}}-2x+k$) and dividend (${{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10$) as shown above. Next we start with the highest power of x and accordingly find the first term of quotient. Thus, in this case since ${{x}^{4}}$ was the highest power term in the dividend, we divide this by the highest term in the divisor (${{x}^{2}}$), thus we get, $\dfrac{{{x}^{4}}}{{{x}^{2}}}={{x}^{4-2}}={{x}^{2}}$.

Next, we multiply ${{x}^{2}}-2x+k$ and ${{x}^{2}}$(first quotient term) to get ${{x}^{4}}-2{{x}^{3}}+k{{x}^{2}}$. Then we subtract $\left( {{x}^{4}}-2{{x}^{3}}+k{{x}^{2}} \right)$ from ${{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10$ (which is similar to the long division method). Finally, we obtain \[-4{{x}^{3}}+(16-k){{x}^{2}}-25x+10\text{ }\] (from subtraction). We then apply the same technique again (but now, \[-4{{x}^{3}}+(16-k){{x}^{2}}-25x+10\text{ }\]acts as the dividend).

Thus, we divide highest power of \[-4{{x}^{3}}+(16-k){{x}^{2}}-25x+10\text{ }\] (that is ${{x}^{3}}$) by highest power of divisor. We get, $\dfrac{-4{{x}^{3}}}{{{x}^{2}}}$=-4x (which is the next quotient term). We again follow the same procedure, and then we will get \[(8-k){{x}^{2}}+(4k-25)x+10\] as the dividend. Now, similarly, we multiply by $\dfrac{(8-k){{x}^{2}}}{{{x}^{2}}}$=(8-k). Finally, repeating the steps again, we get the remainder as \[(2k-9)x+(10-8k+{{k}^{2}})\] as the remainder. We stop our process here, since, the highest power of the next dividend (\[(2k-9)x+(10-8k+{{k}^{2}})\]) is less than that of the divisor.

Now, we can equate the remainder to x+a.

Thus,

\[(2k-9)x+(10-8k+{{k}^{2}})\]= x+a

Then, by comparing the coefficient,

2k-9=1

Thus, k=5.

Also, $10-8k+{{k}^{2}}$=a

Since, k=5,

a = 10 – 8(5) +${{5}^{2}}$

a = -5

Hence, the correct answer is (b) -5.

Note: To solve a problem by long division method, one needs to remember that all the terms of divisor, dividend and quotient should be algebraic. Thus, there should not be any logarithmic or exponential term. Further, one should stop the process, when the highest power of the divisor is greater than the highest power of the dividend. Basically there should be no $\dfrac{1}{x}$or $\dfrac{1}{{{x}^{2}}}$ terms in the quotient.

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