Question

Find zeros write as product of linear factors. X^4+10X^2+9 Help please and thanks

Answer 1

Can we factor this? What multiples to make 9?

Answer 2

Of course 3 i just dont know how to go about factoring it (X^2 3 ) (X^2 3 ) ?? not sure what to do or if those threes even go there

Answer 3

3 an 3 multiply to make 9 but add to make 6. Is there another choice for two numbers that multiply to make 9?

Answer 4

9 and 1 to get the middle number 10... How do i go about setting that up though?

Answer 5

(X^2 + 9 ) (X^2 + 1)

do you know the next step?

Answer 6

I thought it was more difficult to write out then that :x.. But I would think i would just set them both equal to zero and solve . which ill get +- 1 and +-3 meaning those are factors but when i try synthetic division None of them work so i thought they where wrong.

Answer 7

Did you notice that

(X^2 + 9 ) and (X^2 + 1)

are not linear? They have a two exponent. We need to factor each some more. When we factor

(X^2 + 9 )

we get

(x - 3i)(x + 3i)

Answer 8

Ok, yes thank you so much are right because the negative radical 9 when i set to zero will get me the i on the outside. Now for the 1 it would be (X-i) ( X+i) Does this mean I cant use synthetic to check if i had to?

Answer 9

Nice job for the 1.

Synthetic division does not work so well for imaginary numbers, probably best to foil. Check;

(x + i)(x - i) = x^2 +i - i +1 = x^2 + 1 :)

Answer 10

Thank you!! for all your help :]]!!!!!

Answer 11

I have one more question :X \[X ^{4}-4X ^{3}+8X ^{2}-16X+16\] i know using some method i get \[\pm 1 \pm2 \pm4 \pm8\div \pm1\] I used X=-2 For synthetic division and got zero perfectly fine and got \[1x ^{3}-2x ^{2}+4x-8 vwhich got me \to x=\pm 2i and x=2\] but the X to the fourth should give me 4 roots right?