Question

What should be added in the polynomial \[{{x}^{3}}+2{{x}^{2}}-9x+1\] so that it is completely divisible by \[x+4\] ?

Answer 1

Hint: To determine what factor to be added to the given polynomial \[{{x}^{3}}+2{{x}^{2}}-9x+1\] so that it is completely divisible by \[x+4\], we will start by using the long division method. We will divide the given polynomial \[{{x}^{3}}+2{{x}^{2}}-9x+1\] by the monomial \[x+4\], which will give us a remainder. The obtained remainder is the factor which is to be added appropriately to the given polynomial to make it completely divisible.

Complete step by step solution:
According to the given question, we have to make the given polynomial \[{{x}^{3}}+2{{x}^{2}}-9x+1\] completely divisible by the monomial \[x+4\]. For this to happen, we will have to add a suitable factor to the polynomial.
We will be using the long division method to determine the same.
\[x+4\overset{{}}{\overline{\left){{{x}^{3}}+2{{x}^{2}}-9x+1}\right.}}\]
Firstly, we will divide \[{{x}^{3}}\] by \[x\] and we get \[{{x}^{2}}\], we will multiply this component by \[x+4\] and compute the value and we get,
\[x+4\overset{{{x}^{2}}}{\overline{\left){\begin{align}
  & {{x}^{3}}+2{{x}^{2}}-9x+1 \\
 & \underline{-({{x}^{3}}+4{{x}^{2}})} \\
 & 0{{x}^{3}}-2{{x}^{2}}-9x+1 \\
\end{align}}\right.}}\]
Next, we will divide \[-2{{x}^{2}}\] by \[x\] and we get \[-2x\], we will multiply this component by \[x+4\]and compute the value and we get,
\[x+4\overset{{{x}^{2}}-2x}{\overline{\left){\begin{align}
  & {{x}^{3}}+2{{x}^{2}}-9x+1 \\
 & \underline{-({{x}^{3}}+4{{x}^{2}})} \\
 & 0{{x}^{3}}-2{{x}^{2}}-9x+1 \\
 & \underline{-(0{{x}^{3}}-2{{x}^{2}}-8x)} \\
 & 0{{x}^{3}}+0{{x}^{2}}-x+1 \\
\end{align}}\right.}}\]
Now, we will divide \[-x\] by \[x\] and we get \[-1\], we will multiply this component by \[x+4\]and compute the value and we get,
\[x+4\overset{{{x}^{2}}-2x-1}{\overline{\left){\begin{align}
  & {{x}^{3}}+2{{x}^{2}}-9x+1 \\
 & \underline{-({{x}^{3}}+4{{x}^{2}})} \\
 & 0{{x}^{3}}-2{{x}^{2}}-9x+1 \\
 & \underline{-(0{{x}^{3}}-2{{x}^{2}}-8x)} \\
 & 0{{x}^{3}}+0{{x}^{2}}-x+1 \\
 & \underline{-(0{{x}^{3}}+0{{x}^{2}}-x-4)} \\
 & \_\_\_\_\_\_\_\_\_\_\_\_\_5\_ \\
\end{align}}\right.}}\]
We can see that the remainder we obtain is 5. This means that the polynomial \[{{x}^{3}}+2{{x}^{2}}-9x+1\] will get completely divisible by the monomial \[x+4\], only if 5 is subtracted from the polynomial \[{{x}^{3}}+2{{x}^{2}}-9x+1\].
Since, the question asked about a factor to be added, so in that case we will add \[-5\] to the given polynomial, we get the new polynomial as:
\[{{x}^{3}}+2{{x}^{2}}-9x+1\] + (\[-5\])
\[\Rightarrow {{x}^{3}}+2{{x}^{2}}-9x-4\]
So, the polynomial \[{{x}^{3}}+2{{x}^{2}}-9x-4\] is completely divisible by \[x+4\].

Note: The long division method should be done carefully, writing down each term distinctly so as to avoid errors. Also, the remainder we obtained was 5 but we added -5 to the polynomial as the question asked for a term to be added to the polynomial to make it completely divisible. If the question had asked for a term to be subtracted, then in that case would directly subtract 5 from the polynomial.