y = (x^2+2x-5) (x^3+3x^2-40x)

Find the zeros ( 5 total zeros )

x =
x =
x =
x =
x =

So would it be easier for me to distribute first then factor the whole mess out to find the x-intercepts that are the zeroes?. Or. factor (x^2+2x-5) and (x^3+3x^2-40x) and find the zeroes?

2 answers

Oh no, don't expand, then we would be in a real mess

y = (x^2 + 2x - 5) (x^3 + 3x^2 - 40x)

the first factor does not factor any more, so we use the quadratic equation to find two roots from there.
x^2 + 2x - 5 = 0
x = (-2 ± √24)/2
= (-2 ± 2√6)/2 = -1 ± √6

the cubic factor
x^3 + 3x^2 - 40x)
= x(x^2 + 3x - 40)
= x(x+8)(x-5)
= 0 for x = 0, -8, 5

so there are your 5 roots
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