1. ## Factoring help

Hey I need some help with this problem

Factor the polynomial x^8-98x^4+1 into two non-constant polynomials with
integral coefficients.

I want to see steps on how to do this problem thanks.

2. ## Re: Factoring help

Start by letting $\displaystyle \displaystyle X = x^4$, then the equation becomes \displaystyle \displaystyle \begin{align*} X^2 - 98X + 1 \end{align*}. This doesn't factorise over the integers, but does factorise over the real numbers by completing the square.

3. ## Re: Factoring help

I completed the square and got
(x-49)^2 +2400
what do I do next?

4. ## Re: Factoring help

That can't possibly be right - expand it out and check...

5. ## Re: Factoring help

sorry it should be -2400
(x-49)^2 - 2400 = (x-49)(x-49) - 2400 = x^2 -49x - 49x + 2401 -2400 = x^2 - 98x + 1
ok so when completing the square i get

(x-49)^2 - 2400
what do I do next?

6. ## Re: Factoring help

Write it as \displaystyle \displaystyle \begin{align*} (X - 49)^2 - \left( 20\sqrt{6} \right)^2 \end{align*} and then use the Difference of Two Squares rule to factorise it.

7. ## Re: Factoring help

so it is ((x-49)^2 + (20sqrt(6))^2)((x-49)^2 - (20sqrt(6))^2))
what do I do from here?

8. ## Re: Factoring help

Hello, gfbrd!

This takes a little Imagination . . .

Factor $\displaystyle x^8-98x^4+1$ into two non-constant polynomials with integral coefficients.

$\displaystyle \begin{array}{cc}\text{We are given:} & x^8 - 98x^4 + 1 \\ \text{Add/subtract } 100x^4\!: & x^8 - 98x^4 {\color{blue}+ 100x^4} + 1 {\color{blue}- 100x^4} \\ \text{And we have:} & x^8 + 2x^4 + 1 - 100x^4 \\ \text{Factor:} & (x^4 +1)^2 - (10x^2)^2 \\ \text{Factor:} & (x^4+1-10x^2)(x^4+1+10x^2) \\ \text{Therefore:} & (x^4 - 10x^2+1)(x^4+10x^2+1) \end{array}$

9. ## Re: Factoring help

lol sorry my imagination isn't really as good as yours
thanks a lot soroban