Q9. - Algebra 1

Question

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

Step-by-Step Solution

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 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 2 x 4 − 32 .

Factorizing the given polynomial.

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

Step 3. State the conclusion.

Notice it is not possible to factorize further.

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