Question

f(5)=0 and f(-4)=0
factor completely
x^4-27x^2-14x+120

Answer 1

it must factor as
\[(x-5)(x+4)(something)\] i guess you can find the something either by division, or by thinking

Answer 2

i can't write division here, but we can use the "think" method if you like

Answer 3

\[x^4-27x^2-14x+120=(x-5)(x+4)(whatever)\] or multiplying out you get
\[(x^2-x-20)(whatever)=x^4-27x^2-14x+120\]

Answer 4

we know it has to be
\[(x^2-x-20)(ax^2+bx+c)=x^4-27x^2-14x+120\] what we need are \(a,b,c\)
\(a\) and \(c\) should be pretty obvious, especially \(a\) since you need \(x^4\) so \(a=1\)

Answer 5

also the constant is \(120\) and you have a constant of \(-20\) so \(c=120\div(-20)=-6\)

Answer 6

\[(x^2-x-20)(x^2+bx-6)=x^4-27x^2-14x+120\]

Answer 7

the only real work is finding \(b\)