Thread: Finding Zeros

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!
Thanks

Originally Posted by cuteisa89

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!
Thanks

I love these kinds of problems:

First you must understand that this problem requires tail over head.

32=tail
1=head

First the factors of 32 the tail

1,2,4,8,16,32 over the factors of the head which is one

1

Now you must put all of your factors together you solution can be + or -

here we go:

1
2
4
8
16
32

-1
-2
-4
-8
-16
-32

Which ones are the zeros?

2

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.

Originally Posted by cuteisa89

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!
Thanks

As one of the zeros is 2 (x-2) is a factor of x^5 - 8x^4 + 28x^3 - 56x^2 + 64x - 32,
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
we find:

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.

RonL