Q:

How do you rewrite D(x)=x^4-13x^2+36 as the product of a linear factors?

Accepted Solution

A:
[tex]\boxed{D(x)= (x-2)(x+2)(x-3)(x+3)}[/tex]Explanation:We have the following expression:[tex]D(x)=x^4-13x^2+36[/tex]So we need to rewrite this equation as the product of linear factors. Linear factors are given of the form:[tex]ax+b \\ \\ \\ Where: \\ \\ a \ and \ b \ are \ real \ constants[/tex]Since the degree of the polynomial function is 4, the greatest number of linear factors is also 4. Therefore:Let's change:[tex]Let's \ change: \ u\mathrm{=}x^2 \\ \\ So: \\ \\ x^4-13x^2+36 \ changes \ to \ u^2-13u+36 \\ \\ \\ Factor \ out: \\ \\ u^2-9u-4u+36 \\ \\ u(u-9)-4(u-9) \\ \\ (u-4)(u-9) \\ \\ \\ But \ u=x^2 \\ \\ D(x)=(x^2-4)(x^2-9) \\ \\ \\ Applying \ difference \ of \ squares: \ a^2-b^2=(a-b)(a+b) \\ \\ \boxed{D(x)= (x-2)(x+2)(x-3)(x+3)}[/tex]Learn more:Factor expressions: