Question

Factor$\mathbf{2}{\mathit{x}}^{\mathbf{4}}\mathbf{-}\mathbf{32}$completely.

Short answer

The factored form of$2x^4-32$ is $\mathbf{2}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{)}$ .

Step-by-step solution

Step 1. State the concept of ‘factorization of polynomial’.

The process of simplifying the polynomial and breaking it into products of its factors is called as the factorization of polynomials.

Step 2. Factor the polynomial  2x4−32.

Factorizing the given polynomial.

$\begin{array}{l}2x^4-32=2(x^4-16)\\ =2\left(x^4-2^4\right)\left[2^4=16\right]\\ =2\left(x^{2\times 2}-2^{2\times 2}\right)\\ =2\left[{\left(x^2\right)}^2-{\left(2^2\right)}^2\right]\left[{\text{Using\hspace{0.17em}\hspace{0.17em}x}}^{m\times n}={\left(x^m\right)}^n\right]\\ =2\left[(x^2-2^2)(x^2+2^2)\right]\left[\text{Using\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{a}^2-b^2=(a-b)(a+b)\right]\\ =2\left[(x-2)(x+2)(x^2+2^2)\right]\left[\text{Againby\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{a}^2-b^2=(a-b)(a+b)\right]\\ =2\left[(x-2)(x+2)(x^2+4)\right]\\ =2(x-2)(x+2)(x^2+4)\end{array}$

Step 3. State the conclusion.

Notice it is not possible to factorize further.

Therefore, the factorization of $2x^4-32$is $2(x-2)(x+2)(x^2+4)$.