Finding zeros using Decartes Rule of Signs

Question

Question: how many non-real zeros does this equation have?

Equation in question: y=x^4-x^3-28x^2-20x+48

I already found that it has 2 or 0 roots for both positive and negative. How would I find the non-real zeros either using Decartes or on a graph?

Thanks!

Answer 1

Stratton:

We know a polynomial of degree 4 can have at most 4 roots.

To find rational roots, we can use rational root theorem.

+/-1, +/-2, +/-3, +/-4, +/-6, +/-8, +/-12, +/-24, +/- 48 are possible candidates

Plug in x =1 y = 1-1-28-20+48 = 0-48 +48 = 0, so x = 1 is a root.

We can use synthetic division to do a first factorization,

1 | 1 -1 -28 -20 48

1 0 -28 -48

--------------------------------------

1 0 -28 -48 0

So x⁴-x³-28x²-20x+48 = (x-1)*(x³-28x-48)

Use rational root theorem on x³-28x-48 to find possible roots. x=-2 is a root

Use synthetic division again, but on x³-28x-48

-2| 1 0 -28 -48

-2 4 48

-------------------------

1 -2 -24 0

So we can factor x³-28x-48 = (x+2)*(x²-2x-24)

We can use quadratic equation to find roots of x²-2x-24, which are x=6 and x=-4

So our factorization for x⁴-x³-28x²-20x+48 = (x-1)*(x+2)*(x-6)*(x+4)

There are no complex roots; all four roots are real: x= 1, x=-2, x=6, x=-4