Question

{x}^{4} - 17x ^{2} + 16factor completely

Answer 1

We can make a u-substitution to make it into a quadratic.

Let

`u=x^2`

Performing this substitution, we have:

`\begin{gathered} x^4-17x^2+16 \\ =(x^2)^2-17(x^2)+16 \\ \text{Making u-substitution,} \\ u^2-17u+16 \\ (u-16)(u-1) \end{gathered}`

Now, taking it back to "x", we have:

`\begin{gathered} (u-16)(u-1) \\ (x^2-16)(x^2-1) \end{gathered}`

We can further break it down by using the rule:

`a^2-b^2=(a+b)(a-b)`

So, let's factor it out completely:

`\begin{gathered} (x^2-16)(x^2-1) \\ ((x)^2-(4)^2)((x)^2-(1)^2) \\ (x-4)(x+4)(x-1)(x+1) \end{gathered}`

Thus, the fully factored form is:

`(x-4)(x+4)(x-1)(x+1)`