Find all the zeros of each equation. x^5 – 3x^4 – 15x^3 + 45x^2 – 16x + 48 = 0
Do you know how to find the possible rational roots?
No, I'm lost.
List out the factors of 48 (the last number) and the factors of 1 (the first coefficient) *** List both the positive and negative factors *** Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48, -1, -2, -3, -4, -6, -8, -12, -16, -24, -48 Factors of 1: 1, -1 Now divide each factor of 48 by each factor of 1 (there are 96 divisions going on here, but they all lead to this list) 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 8, -8, 12, -12, 16, -16, 24, -24, 48, -48
oh made a typo, didn't mean to say 96 divisions, meant to say 20 divisions...but that's irrelevant
anyways, the possible rational roots are 1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 8, -8, 12, -12, 16, -16, 24, -24, 48, -48
Now, to find the actual zeros, you plug in each value into x^5-3x^4-15x^3+45x^2-16x+48 and see which value gives you zero. For example, if x = 1, then x^5-3x^4-15x^3+45x^2-16x+48 (1)^5-3(1)^4-15(1)^3+45(1)^2-16(1)+48 = 60 Since the result is NOT 0, x = 1 is NOT a root for x^5-3x^4-15x^3+45x^2-16x+48 ------------------------------------------------------------ However, if x = 3, then: x^5-3x^4-15x^3+45x^2-16x+48 (3)^5-3(3)^4-15(3)^3+45(3)^2-16(3)+48 = 0 Since the result is 0, x = 3 is a root for x^5-3x^4-15x^3+45x^2-16x+48
So this at least narrows down the possible list of choices
Does that make sense?
Yes, thanks.
yw
tell me what roots you get
4
4 is another root, good there's one more in this list
-4
nice work so the 3 roots are: 3, 4, -4
This is a 5th degree polynomial, so there are 5 roots total. This means that there are 2 more roots we've yet to find.
To find the 2 missing roots, we use either polynomial long division or synthetic division
is the other roots i and -i?
you are 100% correct so the 5 roots are: 3, 4, -4, i, -i
Okay, thanks! (:
yw
Join our real-time social learning platform and learn together with your friends!