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

Last updated date: 26th Feb 2024
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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 $\left( x+3 \right)$.