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How do you factor ${{x}^{3}}+{{x}^{2}}-14x-24$?

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Last updated date: 26th Feb 2024
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IVSAT 2024
Answer
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Hint:We factor the given equation with the help of vanishing method. In this method we find a number $a$ such that for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$. We assume $f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24$ and take the value of $a$ as 4.

Complete step by step solution:
We find the value of $x=a$ for which the function $f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24=0$.
We take $x=a=4$.
We can see $f\left( 4 \right)={{4}^{3}}+{{4}^{2}}-14\times 4-24=64+16-56-24=0$.
So, the root of the $f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24$ will be the function $\left( x-4
\right)$.
This means for $x=a$, if $f\left( a \right)=0$ then $\left( x-a \right)$ is a root of $f\left( x \right)$.
Therefore, the term $\left( x-4 \right)$ is a factor of the polynomial ${{x}^{3}}+{{x}^{2}}-14x-24$.
We can now divide the polynomial ${{x}^{3}}+{{x}^{2}}-14x-24$ by $\left( x-4 \right)$.
\[x-4\overset{{{x}^{2}}+5x+6}{\overline{\left){\begin{align}
& {{x}^{3}}+{{x}^{2}}-14x-24 \\
& \underline{{{x}^{3}}-4{{x}^{2}}} \\
& 5{{x}^{2}}-14x \\
& \underline{5{{x}^{2}}-20x} \\
& 6x-24 \\
& \underline{6x-24} \\
& 0 \\
\end{align}}\right.}}\]
We first tried to equate the highest power of the dividend with the highest power of the divisor and that’s why we multiplied with ${{x}^{2}}$. We get \[{{x}^{3}}-4{{x}^{2}}\]. We subtract it to get
\[5{{x}^{2}}-14x\]. We again equate with the highest power of the remaining terms. We multiply with
$5x$ and subtract to get \[6x-24\]. At the end we had to multiply with 6 to complete the division.
The quotient is \[{{x}^{2}}+5x+6\].
We still can factor \[{{x}^{2}}+5x+6\].
We apply a grouping method where \[{{x}^{2}}+5x+6={{x}^{2}}+3x+2x+6\].
\[\begin{align}
& {{x}^{2}}+3x+2x+6 \\
& =x\left( x+3 \right)+2\left( x+3 \right) \\
& =\left( x+3 \right)\left( x+2 \right) \\
\end{align}\]
Therefore, the factored form of \[{{x}^{2}}+5x+6\] is \[\left( x+3 \right)\left( x+2 \right)\].
The final factorisation is ${{x}^{3}}+{{x}^{2}}-14x-24=\left( x-4 \right)\left( x+3 \right)\left( x+2
\right)$.


Note: We find the value of x for which the function $f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24=0$.
We can see $f\left( -2 \right)={{\left( -2 \right)}^{3}}+{{\left( -2 \right)}^{2}}-14\times \left( -2 \right)-24=-8+4+28-24=0$. So, the root of the $f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24$ will be the function $\left( x+2 \right)$. We can also do this for $\left( x+3 \right)$.