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?

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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To find the zeros of the equation, you can either distribute first and then factor out the expression, or factor the individual terms and then find the zeros. Both methods will give you the same result, so you can choose the approach that you find more comfortable.

Let's go through both methods:

Method 1: Distribute and Factor
1. Distribute the expression: Multiply (x^2+2x-5) and (x^3+3x^2-40x) using the distributive property to get a single expression.
2. Simplify the expression by combining like terms.
3. Factor the simplified expression to find the zeros. Set each factor equal to zero and solve for x.

Method 2: Factor the Factors Individually
1. Factor the expression (x^2+2x-5) and (x^3+3x^2-40x) separately.
2. Set each factor equal to zero and solve for x to find the zeros.

Both methods will eventually give you the same five zeros. Choose the method that you find more manageable or more suited to your preferences.