Question

f(x)=x^3+3x^2-36x-108
Find x-intercepts of f(x)

Answer 1

to find x-intercepts set f(x)=0 and solve for x

Answer 2

\[x^3+3x^2-36x-108=0\]

Answer 3

You could try to factor by grouping and if that doesn't work we could try to guess a zero using some related theorem.

Answer 4

I'm not sure what to do. I know to add 108 to each side though

Answer 5

Well I suggest factoring by grouping... we don't want to add 108 on both sides

Answer 6

look at x^3 and 3x^2 the first two terms
what factor do they have in common?

Answer 7

oh..i dont know how to factor by grouping

Answer 8

x^2

Answer 9

So x^3+3x^2 can be written as x^2(x+3)

Answer 10

its look at the last two terms
-36x-108

Answer 11

what do and -36 and -108 have in common

Answer 12

-36

Answer 13

coolness so you can write -36x-108 as -36(x+3)

Answer 14

so we now have x^2(x+3)-36(x+3)

Answer 15

hmmm...

Answer 16

like we can now view this as if we have two terms

Answer 17

the first term is x^2(x+3)
and the last term is -36(x+3)

Answer 18

I wonder if they have a common factor

Answer 19

x+3

Answer 20

so you can factor (x+3) from both of those terms

Answer 21

so..(x^2-36)(x+3) ?

Answer 22

bingo

Answer 23

So you can factor x^2-36 more but it really isn't necessary

Answer 24

we are ready to find our zeros

Answer 25

So we have (x^2-36)(x+3)=0
this happens when x^2-36=0 or when x+3=0

Answer 26

you need solve both of those equations

Answer 27

-3, 6

Answer 28

one more solution

Answer 29

-6

Answer 30

yep

Answer 31

x^2-36=0
x^2=36 when x=6 or when x=-6
since 6^2=6(6)=36
and (-6)^2=(-6)(-6)=36

Answer 32

So anyways you found your x-intercepts
(-3,0)
(6,0)
(-6,0)

Answer 33

thank you!!