Hi, I m' not sure how to solve this.
I'm asked to find the zeros using synthetic division for function
x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32
I found that one of the zeros is 2 but I can't find the rest. Please help!
First you must understand that this problem requires tail over head.
First the factors of 32 the tail
1,2,4,8,16,32 over the factors of the head which is one
Now you must put all of your factors together you solution can be + or -
here we go:
Which ones are the zeros?
This is your ONLY solution.
I will let someone else show you how to use synthetic division because latex is down and it would be impossible for you understand where I am coming from.
so divide the later by (x-2) to get the quartic that when multiplied by (x-2) will give
you the quintic.
After doing the division gives:
x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32 = (x-2)(x^4 - 6x^3 + 16x^2 - 24x + 16)
Now the rational root theorem suggests we try x=+/-1, +/-2, +/-4, +/-8, +/-16 as
potential roots for the quartic, and we find x=2 is a root (in fact it is the only real root
but I won't use this and will do this the long winded way)so dividing through again
x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32 = (x-2)^2 (x^3 - 4·x^2 + 8·x - 8)
Now the rational root theorem suggests we try x=+/-1, +/-2, +/-4, +/-8 as potential
roots for the cubic, and we find x=2 is a root so dividing through again we find:
x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32 = (x-2)^3 (x^2 - 2x + 4)
Now the discriminant of the quadratc factor will tell us that x^2 - 2x + 4
has no real roots, so the factorisation can proceed no further over the reals.
So we conclude that the only real root is x=2.
Of course we could have concluded earlier that there is a factor of the form:
(x-2)^n, where n=1 or 3, and so saved ourselves some work.