Question

How do you factor ${{x}^{7}}-8{{x}^{4}}-16{{x}^{3}}+128$?

Answer 1

Hint: To factor the given polynomial, we have to use the factor by grouping method. For this, we have to make two pairs from the four terms of the given polynomial as $\left( {{x}^{7}}-8{{x}^{4}} \right)+\left( -16{{x}^{3}}+128 \right)$. Then we have to take the common factors ${{x}^{4}}$ and $-16$ outside of the first and the second pairs. Then on applying the algebraic identities ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ and ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, the given polynomial will be factored completely.

Complete step by step solution:
Let us write the polynomial given in the above question as
$\Rightarrow p\left( x \right)={{x}^{7}}-8{{x}^{4}}-16{{x}^{3}}+128$
Let us make two pairs of the four terms given in the above polynomial by combining the first two and the last two terms to get
$\Rightarrow p\left( x \right)=\left( {{x}^{7}}-8{{x}^{4}} \right)+\left( -16{{x}^{3}}+128 \right)$
Now, we use the factor by grouping method to factor the given polynomial. For this, we take the common factors ${{x}^{4}}$ and $-16$ common from the first and the second pair respectively to get
$\Rightarrow p\left( x \right)={{x}^{4}}\left( {{x}^{3}}-8 \right)-16\left( {{x}^{3}}-8 \right)$
Now, as can be seen, the factor $\left( {{x}^{3}}-8 \right)$ is common. On taking it outside, we get
$\Rightarrow p\left( x \right)=\left( {{x}^{3}}-8 \right)\left( {{x}^{4}}-16 \right)$
Now, putting $8={{2}^{3}}$, $16={{4}^{2}}$, and ${{x}^{2}}={{\left( {{x}^{2}} \right)}^{2}}$ to get
$\Rightarrow p\left( x \right)=\left( {{x}^{3}}-{{2}^{3}} \right)\left( {{\left( {{x}^{2}} \right)}^{2}}-{{4}^{2}} \right)$
Now, applying the algebraic identities ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ and ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ on the first and the second factors respectively, we get
$\begin{align}
  & \Rightarrow p\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}+2x+{{2}^{2}} \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}-4 \right) \\
 & \Rightarrow p\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}-4 \right) \\
\end{align}$
Now we put $4={{2}^{2}}$ in the factor $\left( {{x}^{2}}-4 \right)$ to get
$\Rightarrow p\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}-{{2}^{2}} \right)$
Now, again applying the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the factor $\left( {{x}^{2}}-{{2}^{2}} \right)$, we get
$\begin{align}
  & \Rightarrow p\left( x \right)=\left( x-2 \right)\left( {{x}^{2}}+2x+4 \right)\left( {{x}^{2}}+4 \right)\left( x-2 \right)\left( x+2 \right) \\
 & \Rightarrow p\left( x \right)={{\left( x-2 \right)}^{2}}\left( x+2 \right)\left( {{x}^{2}}+4 \right)\left( {{x}^{2}}+2x+4 \right) \\
\end{align}$
Hence, the given polynomial is factored.

Note: The above polynomial can be factored using the hit and trial method too. But since the degree of the given polynomial is equal to seven, it will be very much hectic to use the hit and trial method. The method of factoring by grouping is very much easy if it can be applied, so it must be preferred for the polynomials having higher degrees.