Question

MEDAL!!
The polynomial f(x)=x^3+5x^2-2x-24 has three roots.

Use the calculator to determine the roots.

I don't have a calculator either!

Answer 1

http://xrjunque.nom.es/precis/polycalc.aspx this should help.

Answer 2

X=-4
X=-3
X=2

Answer 3

any rational roots will be one of the positive or negative factors of -24

Answer 4

@EducateMe, how did you find it? What method did you use?

Answer 5

oh, factors of -24? But how do I find the roots? @whpalmer4

Answer 6

@UglyTurtle, it didn't really work.

Answer 7

You can use Descartes' Rule of Signs to determine the sorts of roots to hunt for:

Write the polynomial with the exponents in descending order (you already have that)
Next, starting with the highest term, scan from left to right and count the number of sign changes of the coefficients.

+1 +5 -2 -24 that's one sign change, which means 1 positive, real root

Next, write the polynomial again, except in terms of -x. To do this most easily, just change the signs of the terms with odd exponents:

-1 + 5 +2 -24

Count the sign changes again: 2 of them

This means we have either 2 negative, real roots, or 0 negative, real roots.

Finally, we'll have a total of 3 roots because the highest power exponent is 3. Any roots not "spoken for" by the first count or the second count will be complex conjugate roots, of the form \(a \pm bi\). They always come in pairs in any equation with only real coefficients. Without more knowledge from other sources, we don't know if this equation has:

1 positive real root, and 2 negative real roots
or
1 positive real root, and 2 complex conjugate roots

Answer 8

Now, I said any rational roots would be factors of -24. This comes from the rational root theorem. When we multiply two binomials \((x-a)(x-b)\) we get\[(x-a)(x-b) = x^2 - bx - ax +ab\]Notice that the constant term \(ab\) depends only on the constant terms from the binomials? If we added on any number of additional binomials, the same would still be true. The constant term is just the product of all of the constant terms of the factors (*)

So, we can know that any rational roots will be drawn from the factors of the constant term, with either sign being possible. Unfortunately, if you think about it, there may be many more possible rational roots than actual rational roots if the constant term has many factors! However, it's a starting point. We can test out the possible roots by substituting them into the polynomial and seeing if we get 0 as a result. If we do, that's root. Otherwise, it's on to the next candidate.

When you do find a root, you can divide the polynomial by (x-root) to get a simpler polynomial with the same remaining roots. It's worth immediately checking the latest root again, just to see if you have a repeated root.

(*)I'm only handling the case where the coefficient of the leading term of the binomial is 1.

Answer 9

But how did @EducateMe, you get numbers?

Answer 10

Calculator

Answer 11

If you're lucky, you can guess a root or two and get it down to the point where you can factor by grouping or recognizing the sum of cubes or something like that. If not, you can do a guessing game based on the sign changes of the value of the polynomial. If at x=1 the polynomial is positive, and at x = 2, it is negative, somewhere between x = 1 and x= 2 there must be a root.

Answer 12

@EducateMe what did you put on the calculator?

Answer 13

x^3+5x^2-2x-24 = (x-2)(x+3)(x+4)

Answer 14

I have a special calculator that can find factors

Answer 15

ah okay, so you just factorise the polynomial and get that? @EducateMe

Answer 16

To learn how to find the answer without a calculator you would use synthetic division
You should watch this video https://www.khanacademy.org/math/trigonometry/polynomial_and_rational/synthetic-division/v/synthetic-division

Answer 17

@EducateMe thanks.

Answer 18

When you learn synthetic division, you'll probably also learn synthetic substitution, which will make evaluating the polynomial over and over while checking potential roots easier.

Answer 19

@whpalmer4 Thanks for the advice! :-D