Question

How do you factor $8{{x}^{6}}+7{{x}^{3}}-1$ ?

Answer 1

Hint: Here in this question we have been asked to factorize the given expression $8{{x}^{6}}+7{{x}^{3}}-1$ . For doing that we will assume ${{x}^{3}}=a$ after replacing we will have a quadratic expression to factorize we will use a traditional method. In traditional methods we need to write the coefficient of $a$ as the sum of the factors of the product of the coefficient of ${{a}^{2}}$ and the constant term.

Complete step-by-step solution:
For answering this we have been asked to factorize $8{{x}^{6}}+7{{x}^{3}}-1$ .
Let us assume ${{x}^{3}}=a$
Therefore the equation will be $8{{a}^{2}}+7a-1$
For any quadratic expression it can be factored using the traditional methods.
We need to write the coefficient of $a$ as the sum of factors of the product of the coefficient of ${{a}^{2}}$ and the constant term.
$8\times -1=-8$
$8-1=7$
Therefore the expression can be written as $\Rightarrow 8{{a}^{2}}+7a-1=8{{a}^{2}}+8a-a-1$
By further simplifying we will have
$\begin{align}
  & \Rightarrow 8{{a}^{2}}+8a-a-1=8a\left( a+1 \right)-\left( a+1 \right) \\
 & \Rightarrow \left( 8a-1 \right)\left( a+1 \right) \\
\end{align}$
The expression can be simply expressed as $\left( 8{{x}^{3}}-1 \right)\left( {{x}^{3}}+1 \right)$.
Now we will use the following two factorization formulae to factorize this expression further.
${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$
${{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}}-ab \right)$
By applying this two formulae we can further simplify the expression. After that we will have
$\Rightarrow \left( {{\left( 2x \right)}^{3}}-1 \right)\left( {{x}^{3}}+1 \right)=\left( 2x-1 \right)\left( 4{{x}^{2}}+1-2x \right)\left( x+1 \right)\left( {{x}^{2}}+1+x \right)$ .
Therefore we can conclude that the factors of $8{{x}^{6}}+7{{x}^{3}}-1$ are $\left( 2x-1 \right),\left( 4{{x}^{2}}+1-2x \right),\left( x+1 \right),\left( {{x}^{2}}+1+x \right)$.

Note: While answering questions of this type we should be sure with our concepts we apply and the calculations we perform. We can factorize any quadratic expression in the form $a{{x}^{2}}+bx+c=0$ by using the formulae $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. If the roots are ${{x}_{1}},{{x}_{2}}$ then the factors are $\left( x-{{x}_{1}} \right)\left( x-{{x}_{2}} \right)$ . For this expression $8{{a}^{2}}+7a-1$ we can find the roots by using the formula. After applying we will have
$\begin{align}
  & a=\dfrac{-7\pm \sqrt{49-4\left( 8 \right)\left( -1 \right)}}{2\left( 8 \right)} \\
 & \Rightarrow a=\dfrac{-7\pm \sqrt{49+32}}{16} \\
 & \Rightarrow a=\dfrac{-7\pm \sqrt{81}}{16} \\
 & \Rightarrow a=\dfrac{-7\pm 9}{16} \\
 & \Rightarrow a=\dfrac{-16}{16},\dfrac{2}{16}\Rightarrow a=-1,\dfrac{1}{8} \\
\end{align}$
The roots of the expression are $-1,\dfrac{1}{8}$ then the factors will be $\left( 8a-1 \right)\left( a+1 \right)$.