Question

Find the remainder when \[p\left( x \right)={{x}^{3}}-6{{x}^{2}}+14x-3\] is divisible by \[g\left( x \right)=1-2x\] and verify the result by long division.

Answer 1

Hint: In this problem, we have to divide the given equation and the factor. We can use the polynomial long division method to divide the given problem by dividing the highest order term in the dividend to the highest order term in the divisor step by step until we get the remainder.

Complete step by step solution:
We know that the given division is,
\[\dfrac{{{x}^{3}}-6{{x}^{2}}+14x-3}{-2x+1}\]
Now we can set up the polynomials to be divided in long division, we get
\[-2x+1\overset{{}}{\overline{\left){{{x}^{3}}-6{{x}^{2}}+14x-3}\right.}}\]
Now we can divide the highest order term in the dividend \[{{x}^{3}}\] by the highest order term in the divisor x, we get
\[-2x+1\overset{-\dfrac{1}{2}{{x}^{2}}}{\overline{\left){{{x}^{3}}-6{{x}^{2}}+14x-3}\right.}}\]
We can now multiply the quotient term to the divisor, we get
\[-2x+1\overset{-\dfrac{1}{2}{{x}^{2}}}{\overline{\left){\begin{align}
  & {{x}^{3}}-6{{x}^{2}}+14x-3 \\
 & {{x}^{3}}-\dfrac{1}{2}{{x}^{2}} \\
\end{align}}\right.}}\]
We know that the expression is to be subtracted in the dividend, so we can change the sign in \[{{x}^{3}}-\dfrac{1}{2}{{x}^{2}}\], we get
\[\begin{align}
  & -2x+1\underline{\overset{-\dfrac{1}{2}{{x}^{2}}}{\overline{\left){\begin{align}
  & {{x}^{3}}-6{{x}^{2}}+14x-3 \\
 & -{{x}^{3}}+\dfrac{1}{2}{{x}^{2}} \\
\end{align}}\right.}}} \\
 & \text{ }-\dfrac{11}{2}{{x}^{2}}+14x \\
\end{align}\]
Now we can bring down the next term from the dividend to the current dividend and multiply the new quotient to the divisor and write it, we get
\[\begin{align}
  & -2x+1\underline{\overset{-\dfrac{1}{2}{{x}^{2}}+\dfrac{11}{4}x}{\overline{\left){\begin{align}
  & {{x}^{3}}-6{{x}^{2}}+14x-3 \\
 & -{{x}^{3}}+\dfrac{1}{2}{{x}^{2}} \\
\end{align}}\right.}}} \\
 & \text{ }-\dfrac{11}{2}{{x}^{2}}+14x \\
 & \text{ }-\dfrac{11}{2}{{x}^{2}}+\dfrac{11}{4} \\
\end{align}\]
We can now multiply the new quotient to the divisor, then subtract those expression to get the next dividend value, we get
\[\begin{align}
  & -2x+1\underline{\overset{-\dfrac{1}{2}{{x}^{2}}+\dfrac{11}{4}x}{\overline{\left){\begin{align}
  & {{x}^{3}}-6{{x}^{2}}+14x-3 \\
 & -{{x}^{3}}+\dfrac{1}{2}{{x}^{2}} \\
\end{align}}\right.}}} \\
 & \text{ }-\dfrac{11}{2}{{x}^{2}}+14x \\
 & \text{ }\underline{+\dfrac{11}{2}{{x}^{2}}-\dfrac{11}{4}x} \\
 & \text{ }\dfrac{45}{4}x-3 \\
\end{align}\]
Now we can bring down the next term to the current dividend, we can divide the highest order term in the dividend 31x by the divisor x and we can multiply the new quotient term by the divisor.
\[\begin{align}
  & -2x+1\underline{\overset{-\dfrac{1}{2}{{x}^{2}}+\dfrac{11}{4}x-\dfrac{45}{8}}{\overline{\left){\begin{align}
  & {{x}^{3}}-6{{x}^{2}}+14x-3 \\
 & -{{x}^{3}}+\dfrac{1}{2}{{x}^{2}} \\
\end{align}}\right.}}} \\
 & \text{ }-\dfrac{11}{2}{{x}^{2}}+14x \\
 & \text{ }\underline{+\dfrac{11}{2}{{x}^{2}}-\dfrac{11}{4}x} \\
 & \text{ }\dfrac{45}{4}x-3 \\
 & \text{ }\underline{\dfrac{45}{4}x-\dfrac{45}{8}} \\
\end{align}\]
We can now Subtract the expression to get the remainder
\[\begin{align}
  & -2x+1\underline{\overset{-\dfrac{1}{2}{{x}^{2}}+\dfrac{11}{4}x-\dfrac{45}{8}}{\overline{\left){\begin{align}
  & {{x}^{3}}-6{{x}^{2}}+14x-3 \\
 & -{{x}^{3}}+\dfrac{1}{2}{{x}^{2}} \\
\end{align}}\right.}}} \\
 & \text{ }-\dfrac{11}{2}{{x}^{2}}+14x \\
 & \text{ }\underline{+\dfrac{11}{2}{{x}^{2}}-\dfrac{11}{4}x} \\
 & \text{ }\dfrac{45}{4}x-3 \\
 & \text{ }\underline{-\dfrac{45}{4}x+\dfrac{45}{8}} \\
 & \text{ }\dfrac{21}{8} \\
\end{align}\]

Therefore, the remainder is \[\dfrac{21}{8}\].

Note: Students make mistakes while finding the answer using polynomial long division. We should know how to find the answer using polynomial long division step by step. We will also make mistakes while hanging the signs in order to cancel the step consequently.