Need help factoring polynomials?

y = x^3 - 3x^2 - 13x + 15

Try x=1 --> 1^3 - 3*1^2 - 13 *1 + 15 = 1 - 3 - 13 + 15 = 0

That gives the (x-1).

. . . . . x^2. - 2x . . - 15

. . . . ________________________

x - 1 | x^3 - 3x^2 - 13x + 15

. . . . | x^3 - x^2

. . . . | -------------

. . . . | . . . .-2x^2 .- 13x

. . . . | . . . . -2x^2 + 2x

. . . . | . . . .----------------

. . . . | . . . . . . . . . .-15x + 15

. . . . | . . . . . . . . . .-15x + 15

. . . . | . . . . . . . . .---------------

. . . . | . . . . . . . . . . . . . . . . 0

y = (x-1)(x^2 - 2x - 15)

. .= (x-1)(x-5)(x+3)

I usually do it by guessing one of the roots. And there is a method to this guessing.

Notice how if you look at (x - 1)(x + 3) (x - 5) the 3 constants multiply together to make +15

For any polynomial whose highest degree term has a coeffient of 1, the constants in its factors must multiply together to give you then constant. term of the polynomial. This is actually a "special case"

of a more general rule described in the Rational Root Theorem.

Anyhow, when approach a polynomial with this form. You know that the roots have to be factors of 15

and that limits the candidates. Since 1 is always one of the choices and is really easy to try, you

plug in 1 and get 1^3 - 3(1)^2 - 13(1) + 15 = 1 - 3 - 13 +15 = 0

And when 1 is a root (x - 1) is a factor.

At this point you've got two ways to go. You can keep trying factors of 15 until you find another root, which will get you the 2nd factor. If you do that the 3rd factor will include the number that makes are 3 constants

multiply to get 15. So if you found (x - 1) and (x + 3), the two constants so far multiply to get

-1 x 3 = -3, you'd need to multiply by -5 to make + 15, so the missing factor has to be (x - 5)

The other method is to divide the original polynomial by (x-1) to get down to a 2nd degree polynomial.

You can either factor it, or use the quadratric formula to find the roots. This actually works better when

a polynomial has multiple roots, which isn't as easy to spot when guessing the roots.

You use synthetic division to find your first factor, and then you will be left with a quadratic which you can factorise to establish the other two.

Create a table using the coefficients of the cubic, squared, normal and constant components of the expression ...

1, -3, -13, 15

And try out various numbers which you expect to be factors, e.g. try "1".

So you use "1" and apply it. If you are unfamiliar with synthetic division, here's an explanation:

Work out the sum of your initial remainder and your cubic coefficient

1, -3, -13, 15

0

1 (because 1 + 0 = 1),

multiply this by the number you are applying (1) and set it as the second remainder like so ...

1, -3, -13, 15

0, 1,

1

And then work out the sum of the coefficient of the square (-3) and the new remainder

1, -3, -13, 15

0, 1,

1, -2,

Repeat as before

1, -3, -13, 15

0, 1, -2, -15

1, -2, -15, 0

Because you've shown that the final remainder is zero, you have proven that (x-1) is a factor of the expression that you're trying to factorise.

The quotient (i.e. the quadratic [in this example] that you are left with) is "x^2 -2x -15"

This factorises to give:

(x+3)(x-5)

And so therefore x^3 -3x^2 -13 + 15 = (x-1)(x+3)(x-5)

I assume you have met the Factor Theorem i.e. if (x - h) is a factor of f(x) then x = h is a solution of the equation f(x) = 0, so all we have to do is to find just one value of x that makes y = 0, and just by looking at it we can see that if x = 1, y = o, so x - 1 is a factor. Now, either by long division, or better still, Synthetic Division we can find the quotient when y is divided by x - 1, and this is x^2 - 2x - 15. The easiest way to deal with that is to factorise the quadratic as normal to (x + 3)(x - 5) so in full it is y = (x - 1)(x + 3)(x - 5)