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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)$.

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 $\left( x+3 \right)$.

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