Question

Can someone show me how to find the zeroes of this function? (A.K.A. x intercepts of the function)
f(x) = x^3 + 2x^2 - 4x - 8

Answer 1

Have you tried factoring ?

Answer 2

f(x) = x^3 + 2x^2 - 4x - 8

group first two terms
group last two terms

Answer 3

okay so (x^3+2x^2) -(4x-8)?

Answer 4

oh okay i see what happened there

Answer 5

careful when you put a parenthesis after a minus sign

what you're essentially doing is factoring out a "-1"

Answer 6

f(x) = x^3 + 2x^2 - 4x - 8

= (x^3+2x^2) -1(4x + 8)

Answer 7

Next, factor the GCF from each of the groups

Answer 8

x^2(x+2)

Answer 9

how would i do the other half?

Answer 10

you may factor out 4

Answer 11

f(x) = x^3 + 2x^2 - 4x - 8

= (x^3+2x^2) -1(4x + 8)

= x^2(x+2) - 4(x+2)

Answer 12

Notice that both have "x+2" in common
factor it out

Answer 13

you can do this too just enter that whole equation to your calculator.

Answer 14

The roots (zeros) are the x values where the graph intersects the x-axis. To find the roots (zeros), replace y with 0 and solve for x.
x=−2,2