## DeadShot Group Title Determine the zeros of f(x) = x4 - x3 + 7x2 - 9x - 18. one year ago one year ago

1. shubhamsrg Group Title

-1 fits in.

2. shubhamsrg Group Title

Got it, break 7 as 9-2 /

so $\pm 1$ is the answer?

4. shubhamsrg Group Title

nopes.

5. shubhamsrg Group Title

What do get after writing 7 as 9-2 ?

$x^4 - x^3 + 9 - 2x^2 - 9x -18$ ?

7. shubhamsrg Group Title

x^4 - x^3 + 9x^2 -2x^2 - 9x - 18.=0 =>(x^4 + 9x^2) - (x^3 -9x) -(2x^2 -18) =0 You see what should be the next step ?

yeah, factor each binomial, right?

9. shubhamsrg Group Title

yep.

(x^2 + 1) (x^2 + 9) for the first one?

11. shubhamsrg Group Title

try again.

I'm confused

13. amoodarya Group Title

x1 = 2 x2 = -1 x3 = 0 + 3i x4 = 0 - 3i

So, substitute (0 - 3i) and -1 for x^2, making it ((0 - 3i) + 9(-1))?

so, then it would become (-3i - 9)

16. amoodarya Group Title

So, divide x^4 - x^3 + 7x^2 - 9x - 18 by x^2 + 9 ?

so it would be x^2 - x + 7 - x - 2 ?

combine like terms, and it would be x^2 - 2x + 5 , right?

20. zordoloom Group Title

Hmm.. Looks like you have 2 real zeros and 2 imaginary.

21. zordoloom Group Title

Do you still need help?

yes, I don't know what to do next

I got it! Thanks!