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Question: How do you factor \({{x}^{3}}+{{x}^{2}}-14x-24\)?...

How do you factor x3+x2−14x−24{{x}^{3}}+{{x}^{2}}-14x-24?

Explanation

Solution

We factor the given equation with the help of vanishing method. In this method we find a number aa such that for x=ax=a, if f(a)=0f\left( a \right)=0 then (x−a)\left( x-a \right) is a root of f(x)f\left( x \right). We assume f(x)=x3+x2−14x−24f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24 and take the value of aa as 4.

Complete step by step solution:
We find the value of x=ax=a for which the function f(x)=x3+x2−14x−24=0f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24=0.
We take x=a=4x=a=4.
We can see f(4)=43+42−14×4−24=64+16−56−24=0f\left( 4 \right)={{4}^{3}}+{{4}^{2}}-14\times 4-24=64+16-56-24=0.
So, the root of the f(x)=x3+x2−14x−24f\left( x \right)={{x}^{3}}+{{x}^{2}}-14x-24 will be the function (x−4)\left( x-4 \right).
This means for x=ax=a, if f(a)=0f\left( a \right)=0 then (x−a)\left( x-a \right) is a root of f(x)f\left( x \right).
Therefore, the term (x−4)\left( x-4 \right) is a factor of the polynomial x3+x2−14x−24{{x}^{3}}+{{x}^{2}}-14x-24.
We can now divide the polynomial x3+x2−14x−24{{x}^{3}}+{{x}^{2}}-14x-24 by (x−4)\left( x-4 \right).

& {{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{aligned}}\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{aligned} & {{x}^{2}}+3x+2x+6 \\\ & =x\left( x+3 \right)+2\left( x+3 \right) \\\ & =\left( x+3 \right)\left( x+2 \right) \\\ \end{aligned}$$ **Therefore, the factored form of $${{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 $\left( x+3 \right)$.