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To make things simpler, when the sum of coefficients = 0, it means that (x-1) is a factor.

no I dnt know those methods

ok

So is (x-1) a factor to the equation?

yes

2x^3-3x^2+50x-75

Good, now we can rule out the third case (4 complex roots).

ok

what do I divide it by

When you find another root (i.e. another factor) of the cubic.

oh ok, can I use the calculator for that

So that reduces to 12 possibilities only.

Very true, thanks KingGeorge!

You group like so \[2x^3-3x^2+50x-75=x^2(2x-3)+25(2x-3)\]Can evy15 finish the factoring from here?

can you guide me to what to do next

ok hold on

x=i+-5 and 3/2

i mean =-5i

Good job, (+\- 5i).
Now you have all 4 zeroes!

Can you enumerate the four zeroes?

so its 1,+/-5i, and 3/2?

or-1

Excellent! There you are! Not -1. Your list is good.

Big thanks to KingGeorge who cut the work in half!

yes thank you guys so much

yw! :)

But this is not an obvious, nor the easiest, way to do it.

Yep, I will look more carefully for these cases in the future.

can i do the same with x^5-3x^4-24x^3-72x^2+12x=12

sorry wrote the wrong one

x^5-3x^4-24x^3-72x^2-25x+75=0

Your first group will be \[x^4(x-3).\]Knowing this, can you give me the rest of the factorization?

x^4(x-3)-24x^2(x-3)-25(x-3)

Perfect. And can you finish the rest of the factorization?

would it be (x^4-24x^2-25) and (x-3)?

x-25 and x+1