Question

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, find k and a.

Answer 1

Hint: The polynomial which has to be divided must have a higher degree of polynomial as compared to another polynomial which divides the first polynomial.
The polynomial which gets divided is known as dividend, the polynomial which divides dividend is divisor, the multiple obtained is quotient and the remaining polynomial is known as remainder.

Complete step by step answer:
The polynomial p(x) is equal to\[{{x}^{4}}-6{{x}^{3}}+16{{x}^{^{2}}}-25x+10\].
The divisor polynomial is taken to be g(x) =${{x}^{2}}-2x+k$.
When p(x) is divided by g(x), the quotient obtained is to be termed as q(x) and remainder to be termed as r(x).
The p(x) is divided by g(x) as follows:
\[\begin{matrix}
  {{x}^{2}}-2x+k\overline{\left){\begin{align}
  & {{x}^{4}}-6{{x}^{3}}+16{{x}^{2}}-25x+10\text{ } \\
 & \underline{-{{x}^{4}}+2{{x}^{3}}-k{{x}^{2}}} \\
\end{align}}\right.}({{x}^{2}}-4x+8-k) \\
  0-4{{x}^{3}}+16{{x}^{2}}-k{{x}^{2}}-25x+10 \\
  \underline{+\text{ }4{{x}^{3}}-8{{x}^{2}}+4xk} \\
  8{{x}^{2}}-k{{x}^{2}}-25x+4kx+10 \\
  -\underline{8{{x}^{2}}+k{{x}^{2}}+16x-2kx-8k+{{k}^{2}}} \\
  -9x+2kx+10-8k+{{k}^{2}} \\
\end{matrix}\]
The remainder r(x) is equal to x + a.
Remainder r(x) is equal to \[-9x+2kx+10-8k+{{k}^{2}}\] and it is compared with given remainder to find values of k and a.
The values are compared as:
\[-9x+2kx+10-8k+{{k}^{2}}\]= x + a
→ \[(-9+2k)x+(10-8k+{{k}^{2}})\]= x + a
The coefficients of x are compared and equalized to each other and similarly, coefficients of constant terms are equalized to find values of k and a.
Coefficients of linear variable x are compared as follows:
$\begin{align}
  & -9+2k=1 \\
 & 2k=1+9 \\
 & k=\dfrac{10}{2} \\
 & k=5
\end{align}$
The value of k is determined to be equal to 5.
Now, the coefficients of constant terms are compared to get value of a. The coefficients of constants are compared as:
\[\begin{align}
  & 10-8k+{{k}^{2}}=a \\
 & 10-(8\times 5)+{{(5)}^{2}}=a \\
 & a=-5
\end{align}\]
Therefore, k =$5$, a =$-5$.

Note: For finding two unknown values as here, k and a, it is essential that there are two relations available. Two equations are required for finding two variables.