The factor theorem
In a nutshell
Given a function f(x), a factor is a number a such that f(a)=0. A factor is represented as (x−a). The factor theorem is one of the easiest ways of finding out if a term is a linear factor of a polynomial.
The factor theorem
The factor theorem states that, given a polynomial function f(x), then:
- if (x−a) is a factor of f(x), then f(a)=0; and
- if f(a)=0, then (x−a) is a factor of f(x).
Using the factor theorem to factorise a cubic function
Procedure
| 1.
| Substitute values a into the cubic f(x) until you find a value that gives as a result f(a)=0. |
| 2. | Divide your function f(x) by (x−a). This can be done with long division for example. |
| 3. | Write your cubic in the following form: f(x)=(x−a)(Ax2+Bx+C) where A, B and C are constants. |
| 4. | Factorise the quadratic term (Ax2+Bx+C) if possible, in order to express f(x) a function of linear factors. |
Example 1
Use the factor theorem to prove that (x−4) is a factor of f(x)=−3x3+13x2−6x+8, and fully factorise f(x).
To prove that (x−4) is a factor of f(x)=−3x3+13x2−6x+8, you have to substitute x=4 into your function.
f(4)=−3(4)3+13(4)2−6×4+8=−3×64+13×16−24+8=0
Since f(4)=0, you have proven that (x−4) is a factor of f(x)=−3x3+13x2−6x+8.
Divide your polynomial by the factor, and see if the remainder is 0. Also obtain the quadratic counterpart.
−3x2+x−2x−4)−3x3+13x2−6x+8−3x3+12x2x2−6x+8x2−4x −2x+8−2x+8 0
Now you can write f(x) as
f(x)=(x−4)(−3x2+x+2)
Factorise the term (−3x2+x−2) to present f(x) as a product of linear factors.
f(x)=−(x−4)(x−1)(3x+2)
Example 2
Show that (x−5) is a factor of f(x)=x3−7x2+2x+40, and fully factorise f(x).
Since you are trying to prove that (x−5) is a factor of f(x)=x3−7x2+2x+40, you substitute x=5 into your function.
f(5)=53−7×52+2×5+40=125−175+10+40=0
Because f(5)=0, you have proven that (x−5) is a factor of f(x)=x3−7x2+2x+40.
To find the quadratic factor, you divide your polynomial by the newfound linear factor. This also acts as a double-check, since if the remainder is 0, then your linear factor is indeed a factor of f(x).
x2−2x−8x−5)x3−7x2+2x+40x3−5x2−2x2+2x+40−2x2+10x−8x+40−8x+40 0
So (x−5) is indeed a factor of f(x)=x3−7x2+2x+40.
Hence,
f(x)=(x−5)(x2−2x−8)
Now factorise, if possible, the quadratic term (x2−2x−8) to express f(x) as a product of linear factors. However, you will see that this term does not have any real roots, so you leave it with a quadratic term.
Therefore f(x)=(x−5)(x2−2x+8) is fully factorised.
