Question

How do you factor by grouping ${{x}^{3}}+3{{x}^{2}}-118x-120$?

Answer 1

Hint: Now to solve the following question we will first write -118x as -120x + 2x. Now we will group the common terms together to form an expression. Now we will again write $3{{x}^{2}}=2{{x}^{2}}+{{x}^{2}}$ . Hence using this we will again simplify the expression. Now on simplifying we will get a quadratic expression. IN the quadratic we will split the middle term such that the product of the terms is the product of the first and last term of quadratic. Now on simplifying the obtained expression we will easily find the factor of the expression.

Complete step by step solution:
Now consider the given expression ${{x}^{3}}+3{{x}^{2}}-118x-120$
Now first let us write 118x as 120x – 2x. Hence we get the expression as,
$\Rightarrow {{x}^{3}}+3{{x}^{2}}-120x+2x-120$
Now taking -120 common from the terms and rearranging the terms we get,
\[\begin{align}
  & \Rightarrow {{x}^{3}}+3{{x}^{2}}-120\left( x+1 \right)+2x \\
 & \Rightarrow {{x}^{3}}+3{{x}^{2}}+2x-120\left( x+1 \right) \\
\end{align}\]
Now let us write $3{{x}^{2}}$ as $2{{x}^{2}}+{{x}^{2}}$ . Hence we get the expression as,
\[\Rightarrow {{x}^{3}}+{{x}^{2}}+2{{x}^{2}}+2x-120\left( x+1 \right)\]
Now taking 2x common from the terms and rearranging we get,
\[\Rightarrow {{x}^{3}}+{{x}^{2}}+2x\left( x+1 \right)-120\left( x+1 \right)\]
Now again taking ${{x}^{2}}$ common from the terms we get,
\[\Rightarrow {{x}^{2}}\left( x+1 \right)+2x\left( x+1 \right)-120\left( x+1 \right)\]
Now let us take x + 1 common from the whole expression. Hence we get,
$\Rightarrow \left( x+1 \right)\left( {{x}^{2}}+2x-120 \right)$
Now consider the quadratic ${{x}^{2}}+2x-120$ .
Now we want to split the middle term such that the product of the terms is equal to the first term and the last term.
Hence we can write 2x as 12x – 10x.
Hence we get the expression as,
$\Rightarrow {{x}^{2}}+12x-10x-120$
Now taking $x$ common from the first two terms and – 10 common from the last two terms we get,
$\Rightarrow x\left( x+12 \right)-10\left( x+12 \right)$
Now again taking x + 12 common we get,
$\Rightarrow \left( x-10 \right)\left( x+12 \right)$
Now substituting the value we get,
$\Rightarrow \left( x+1 \right)\left( x-10 \right)\left( x+12 \right)$
Hence the given expression is factored.

Note: Now note that we can also factorize the expression by finding the roots of the expression. If $\alpha $ is the root of the given expression then $x-\alpha $ is the factor of the expression. Hence if we find the roots then we can easily write the factors of the expression.