Question

What must be subtracted from the polynomial \[f\left( x \right) = {x^4} + 2{x^3} - 13{x^2} - 12x + 21\] the resulting polynomial is exactly divisible by \[{x^2} + 2x - 3\]

Answer 1

Hint: The question is related to the division of a polynomial, First we need to divide the polynomial \[f\left( x \right) = {x^4} + 2{x^3} - 13{x^2} - 12x + 21\] by polynomial \[{x^2} + 2x - 3\] if we get a reminder as a zero then no need to subtract anything in the dividend polynomial suppose the reminder is not a zero if we get a some number or a polynomial then we are subtracting the reminder from the dividend polynomial that resulting polynomial should be exactly divisible by the divisor \[{x^2} + 2x - 3\]

Complete step by step solution:
Consider, the given polynomial,
 \[f\left( x \right) = {x^4} + 2{x^3} - 13{x^2} - 12x + 21\]
\[g\left( x \right) = {x^2} + 2x - 3\]
We have to find the number which subtracted from the \[f\left( x \right)\] which makes the polynomial \[f\left( x \right)\] is exactly divisible by polynomial \[g\left( x \right)\].
First, we need to divide the polynomial \[f\left( x \right)\] by \[g\left( x \right)\] using a sign conversion
\[{x}^{2}+2x-3\overset{{x}^{2}-10}{\overline{\left){\begin{align}
  & {+{x}^{4}}+{{2x}^{3}}-{{13x}^{2}}-12x+21 \\
 & \underline{{{+x}^{4}}+2{{x}^{3}}-3{{x}^{2}}} \\
 & -10{{x}^{2}}-12x +21\\
 & \underline{-10{{x}^{2}}-20x-30} \\
 & 8x-9 \\
\end{align}}\right.}}\]
Here, the reminder is \[r\left( x \right) = 8x - 9\], so we need to subtract \[8x - 9\] from the polynomial \[f\left( x \right) = {x^4} + 2{x^3} - 13{x^2} - 12x + 21\], then we have
\[ \Rightarrow \,\,\,{x^4} + 2{x^3} - 13{x^2} - 12x + 21 - \left( {8x - 9} \right)\]
\[ \Rightarrow \,\,\,{x^4} + 2{x^3} - 13{x^2} - 12x + 21 - 8x + 9\]
On simplification, we get
\[ \Rightarrow \,\,\,{x^4} + 2{x^3} - 13{x^2} - 20x + 30\]
Now, we have to check whether the resultant polynomial \[f\left( x \right) = {x^4} + 2{x^3} - 13{x^2} - 20x + 30\] is exactly divisible by the polynomial \[g\left( x \right) = {x^2} + 2x - 3\] or not i.e.,
\[{x}^{2}+2x-3\overset{{x}^{2}-10}{\overline{\left){\begin{align}
  & {+{x}^{4}}+{{2x}^{3}}-{{13x}^{2}}-20x+30 \\
 & \underline{{{+x}^{4}}+2{{x}^{3}}-3{{x}^{2}}} \\
 & -10{{x}^{2}}-20x +30\\
 & \underline{-10{{x}^{2}}-20x+30} \\
 & 0 \\
\end{align}}\right.}}\]
Here, we get the reminder 0, this means the resultant polynomial \[f\left( x \right) = {x^4} + 2{x^3} - 13{x^2} - 20x + 30\] is exactly divisible by the polynomial \[g\left( x \right) = {x^2} + 2x - 3\]
Hence, we should subtract \[r\left( x \right) = 8x - 9\] to \[f\left( x \right)\] so that the resulting polynomial is exactly divisible by \[g\left( x \right)\].

Note: In this question we saw a division of polynomial which is a long division method, this can also solve and verify easily by using a division algorithm it can be defined as:
Dividend \[ = \] Quotient \[ \times \] divisor \[ + \] remainder.
Here, \[f\left( x \right) = {x^4} + 2{x^3} - 13{x^2} - 12x + 21\] is dividend, \[g\left( x \right) = {x^2} + 2x - 3\] and dividing these two polynomial, we got quotient \[q\left( x \right) = {x^2} - 10\] and reminder \[r\left( x \right) = 8x - 9\] , then the division algorithm can be written as
\[f\left( x \right) = q\left( x \right)g\left( x \right) + r\left( x \right)\]
Or
\[f\left( x \right) - r\left( x \right) = q\left( x \right)g\left( x \right)\]
On substituting, we have
\[{x^4} + 2{x^3} - 13{x^2} - 12x + 21 - \left( {8x - 9} \right) = \left( {{x^2} - 10} \right)\left( {{x^2} + 2x - 3} \right)\]
\[{x^4} + 2{x^3} - 13{x^2} - 12x + 21 - 8x + 9 = {x^4} + 2{x^3} - 3{x^2} - 10{x^2} - 20x + 30\]
\[{x^4} + 2{x^3} - 13{x^2} - 20x + 30 = {x^4} + 2{x^3} - 13{x^2} - 20x + 30\]
Which verifies our answer.