Hey I need some help with this problem
Factor the polynomial x^8-98x^4+1 into two non-constant polynomials with
integral coefficients.
I want to see steps on how to do this problem thanks.
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.
Hello, gfbrd!
This takes a little Imagination . . .
Factor $\displaystyle x^8-98x^4+1$ into two non-constant polynomials with integral coefficients.
$\displaystyle \begin{array}{cc}\text{We are given:} & x^8 - 98x^4 + 1 \\ \text{Add/subtract } 100x^4\!: & x^8 - 98x^4 {\color{blue}+ 100x^4} + 1 {\color{blue}- 100x^4} \\ \text{And we have:} & x^8 + 2x^4 + 1 - 100x^4 \\ \text{Factor:} & (x^4 +1)^2 - (10x^2)^2 \\ \text{Factor:} & (x^4+1-10x^2)(x^4+1+10x^2) \\ \text{Therefore:} & (x^4 - 10x^2+1)(x^4+10x^2+1) \end{array}$