# Math Help - Find All Real Zeros

1. ## Find All Real Zeros

Use Descartes' Rule of Signs and the Rational Zeros Theorem to find all the real zeros of the polynomial function given below. Use the zeros to factor f over the real numbers.

f(x) = 4x^5 + 12x^4 - x - 3

I had a lot of trouble with this question. Help!

2. Originally Posted by magentarita
Use Descartes' Rule of Signs and the Rational Zeros Theorem to find all the real zeros of the polynomial function given below. Use the zeros to factor f over the real numbers.

f(x) = 4x^5 + 12x^4 - x - 3

I had a lot of trouble with this question. Help!
Let us use the Rational Zeros Theorem to find all possible solutions of this function.

Factors of leading coefficient (4): 1, 2, 4
Factors of lone constant (3): 1, 3

Possible solutions: $1, 2, 4, \frac{1}{3}, \frac{2}{3}, \frac{4}{3}$

Using Descartes' Rule of Signs, note there is one changes in sign when going through the equation. (+ to -, - to -... + to - is only change of sign). This means there is one positive x-intercept for this function, and it is one of the six given through the Rational Zeros Theorem.

3. Originally Posted by magentarita
Use Descartes' Rule of Signs and the Rational Zeros Theorem to find all the real zeros of the polynomial function given below. Use the zeros to factor f over the real numbers.

f(x) = 4x^5 + 12x^4 - x - 3

I had a lot of trouble with this question. Help!
The signs of the coefficients change sign once, so there is exactly 1 positive real root.

f(-x)=-4x^5+12x^4+x-3

This changes sign twice so there are either two or zero negative roots.

By the rational roots theorem all the rational roots are amoung:

+/-1, +/-3, +/- 1/4, +/-3/4, +/-1/2, +/-3/2

Checking these shows the only rational root is x=-3, so (x+3) is a factor of f(x), so either by inspection or long division we find:

f(x)=(x+3)(4x^4-1)=(x+3)(2x^2-1)(2x^2+1)=(x+3)(sqrt(2)x+1)(sqrt(2)x-1)(2x^2+1)

So the remaining real roots are:

x=+/- 1/sqrt(2)

From what we saw above that must be all the real roots so the others roots are complex.

RonL

4. ## Great job!

I thank you both. I will read up more on such topics.