i cant figure out how to use the rational root theorem for x^4-x^3-31x^2+25x+150.?

Answer 1

x^4 - x^3 - 31x² + 25x + 150

Starting with Decartes Rule of Signs, because the sign changes 2 times between terms (coefficient on x^4 is positive, then x^3 and x² are negative, then x and 1 are positive), this tells us we will have a maximum of two positive rational roots

And inverting the sign on the positive powers gives us - x^4 - x^3 + 31x² + 25x - 150. Again there are 2 sign changes, so that tells us there are a maximum of two negative rational roots.

By the Rational Root Theorem, if there is a rational root, the root will be factors of the last coefficient (150) divided by factors of the first coefficient (1 on x^4). Since the coefficient on x^4 is 1, all the rational roots will be integers.

Positive factors of 150 (and possible positive roots) are:
1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150

Negative factors (and possible negative roots) are just the opposite of all the ones above.

A little trial and error shows that 3 is a root (the result is zero), so now you can divide by (x - 3).

(x - 3)(x^3 + 2x² - 25x - 50)

More trial and error on the new third order polynomial shows that 5 is another root. So divide by (x - 5).
(x - 3)(x - 5)(x² + 7x + 10)

This last quadratic can easily be factored resulting in:
(x - 3)(x - 5)(x + 2)(x + 5)

The solutions for zeroes would be:
x = 3, x = 5, x = -2 or x = -5

This problem was harder because there are lots of factors to 150 (positive and negative) and 1 and -1 were not roots, nor was 2. (I start with smaller factors first). You could also use a graphing calculator or an online solver to help you out, if you have access to those.

Answer 2

It tells you that any rational root of the polynomial will have to be an integer (since the coefficient of the leading term is 1), and that integer will have to divide (evenly) the constant term, namely 150.

Now you do the rest.