Question

find all the zeros of: f(x)= x^3+5x^2-8x-40

Answer 1

given
\[40=2^45\]
the possible zeroes are:
\[\pm 1,\pm 2,\pm 4, \pm 5, \pm 10, \pm 16, \pm 20, \pm 40\]
because it is -40 not +40 either 1 or 3 zeroes must be negative.
because we have +5x^2 not - 5x^2 at least one must be positive.
thise means our answer will look like:
\[(x+a)(x+b)(x-c)\]
5=a+b-c
-8=ab-ac-bc
-40=abc

Answer 2

where did you get 40=2^4(5)

Answer 3

i'm sorry that should be that should be 2^3(5)

Answer 4

oh ok that makes better sense

Answer 5

so would the answer be -5,+or- 2sqrt2

Answer 6

not quite.
that would leave us with a positive 40

Answer 7

only one number can be negative.

Answer 8

that's strange because that's what my teacher put for the answer. I just didn't know how she got it

Answer 9

i'm sorry. my mistake that is correct.

Answer 10

\[-5,\pm2\sqrt{2}\]

Answer 11

yes i had misread the previous one.

Answer 12

in order to reach that you would have to use synthetic division to find the first possible zero.

Answer 13

ok thank you for the help :)

Answer 14

because we have 5x^2 i would suggest 5 as the first possibility

Answer 15

of course :) so synthetic division would leave
\[(x+5)(x^2-8)\]
can you solve it from here?

Answer 16

of course thanks