Question

How do you factor ${\text{y = }}{{\text{x}}^4} - 2{{\text{x}}^3} - 8{\text{x + 16}}$ ?

Answer 1

In this question, we have been given a polynomial equation ${\text{y = }}{{\text{x}}^4} - 2{{\text{x}}^3} - 8{\text{x + 16}}$ and we have been asked to factorise the given polynomial. First, we will find the common factors or terms in the given polynomial and try to separate it in pairs, so that it can be factored again easily. And then. we will try to simplify the terms and factorise it.

Formula used:
$\left( {{{\text{a}}^3} - {{\text{b}}^3}} \right) = ({\text{a - b) (}}{{\text{a}}^2} + {\text{ab + }}{{\text{b}}^2})$

Complete step-by-step answer:
The given polynomial equation is ${\text{y = }}{{\text{x}}^4} - 2{{\text{x}}^3} - 8{\text{x + 16}}$, we will try to find the common digits.
So here, we will try to separate the terms which seems likely to be factored.
${\text{y = }}\left( {{{\text{x}}^4} - 2{{\text{x}}^3}} \right) - \left( {8{\text{x - 16}}} \right)$
So here we can see that, ${{\text{x}}^3}$ is common in first term and $2$ is common in second term, taking the commons out, we get
${{\text{x}}^3}({\text{x - 2) - 8 (x - 2)}}$
As we can see ${\text{(x - 2)}}$ is common here, we will take it out
$({\text{x - 2) ( }}{{\text{x}}^3} - 8)$
So here we get this. Now $8$ can be written as ${2^3}$ because there is ${{\text{x}}^3}$ ,
$({\text{x - 2) ( }}{{\text{x}}^3} - {2^3})$
We can see ${\text{( }}{{\text{x}}^3} - {2^3})$ is now in the form of $\left( {{{\text{a}}^3} - {{\text{b}}^3}} \right) = ({\text{a - b) (}}{{\text{a}}^2} + {\text{ab + }}{{\text{b}}^2})$ , we will use it
$({\text{x - 2) ( x}} - 2)({{\text{x}}^2} + 2{\text{x + }}{{\text{2}}^2})$
$({\text{x - 2) ( x}} - 2)({{\text{x}}^2} + 2{\text{x + 4}})$
And here there are two ${\text{( x}} - 2)$ and can be written as ${{\text{( x}} - 2)^2}$ . So we get,
${{\text{( x}} - 2)^2}({{\text{x}}^2} + 2{\text{x + 4}})$

Therefore $({\text{x - 2) ( x}} - 2)({{\text{x}}^2} + 2{\text{x + 4}})$ are the factors of ${\text{y = }}{{\text{x}}^4} - 2{{\text{x}}^3} - 8{\text{x + 16}}$.

Note:
Alternative method:
The given equation,
${\text{y = }}{{\text{x}}^4} - 2{{\text{x}}^3} - 8{\text{x + 16}}$
${\text{ }}{{\text{x}}^4} - 8{\text{x + 16}} - 2{{\text{x}}^3}$
So here we can see that, ${\text{x}}$ is common in first term and $2$ is common in second term, taking the commons out, we get
${\text{ x}}\left( {{{\text{x}}^3} - 8} \right) - 2({{\text{x}}^3} - 8)$
As we can see $({{\text{x}}^3} - 8)$ is common here, we will take it out
$\left( {{\text{ x}} - 2} \right)({{\text{x}}^3} - 8)$
Now $8$ can be written as ${2^3}$ because there is ${{\text{x}}^3}$ ,
$({\text{x - 2) ( }}{{\text{x}}^3} - {2^3})$
We can see ${\text{( }}{{\text{x}}^3} - {2^3})$ is now in the form of $\left( {{{\text{a}}^3} - {{\text{b}}^3}} \right) = ({\text{a - b) (}}{{\text{a}}^2} + {\text{ab + }}{{\text{b}}^2})$ , we will use it
$({\text{x - 2) ( x}} - 2)({{\text{x}}^2} + 2{\text{x + }}{{\text{2}}^2})$
$({\text{x - 2) ( x}} - 2)({{\text{x}}^2} + 2{\text{x + 4}})$
And here there are two ${\text{( x}} - 2)$ and can be written as ${{\text{( x}} - 2)^2}$ . So we get,
${{\text{( x}} - 2)^2}({{\text{x}}^2} + 2{\text{x + 4}})$
Therefore $({\text{x - 2) ( x}} - 2)({{\text{x}}^2} + 2{\text{x + 4}})$ are the factors of ${\text{y = }}{{\text{x}}^4} - 2{{\text{x}}^3} - 8{\text{x + 16}}$.