3. How to factor polynomials

On this page we learn how to factor polynomials with 3 terms (degree 2), 4 terms (degree 3) and 5 terms (degree 4).

We'll make use of the Remainder and Factor Theorems to decompose polynomials into their factors.

What are we looking for?

Example 1

An example of a polynomial (with degree 3) is:

p(x) = 4x³ − 3x² − 25x − 6

The factors of this polynomial are:

(x − 3), (4x + 1), and (x + 2)

Note there are 3 factors for a degree 3 polynomial. When we multiply those 3 terms in brackets, we'll end up with the polynomial p(x).

We'll see how to find those factors below, in How to factor polynomials with 4 terms?

Summary of the process

In general, we'll:

1. Find one factor, by making use of the Remainder Theorem
2. Divide the polynomial by the factor we found, thus giving us a simpler polynomial to work with
3. Find one factor of the simpler polynomial, and divide once again
4. Continue, until we get to a trinomial, which we can usually factor easily.

How to factor polynomials with 3 terms?

Example 2

Here's an example of a polynomial with 3 terms:

q(x) = x² − x + 6

We recognize this is a quadratic polynomial, (also called a trinomial because of the 3 terms) and we saw how to factor those earlier in Factoring Trinomials and Solving Quadratic Equations by Factoring.

We need to find numbers a and b such that

q(x) = x² − 5x + 6 = (x − a)(x − b)

This generally involves some guessing and checking to get the right combination of numbers. The number 6 (the constant of the polynomial) has factors 1, 2, 3, and 6 (and the negative of each one is also possible) so it's very likely our a and b will be chosen from those numbers.

After some trials, we get:

x² − 5x + 6 = (x − 2)(x − 3)

We say the factors of x² − 5x + 6 are (x − 2) and (x − 3).

Notice our 3-term polynomial has degree 2, and the number of factors is also 2.

How to factor polynomials with 4 terms?

Example 3

Above, we discussed the cubic polynomial p(x) = 4x³ − 3x² − 25x − 6 which has degree 3 (since the highest power of x that appears is 3).

Let's find the factors of p(x).

Notice the coefficient of x³ is 4 and we'll need to allow for that in our solution.

We are looking for a solution along the lines of the following (there are 3 expressions in brackets because the highest power of our polynomial is 3):

4x³ − 3x² − 25x − 6 = (ax − b)(cx − d)(fx − g)

The factors of 4 are 1, 2, and 4 (and possibly the negatives of those) and so a, c and f will be chosen from those numbers. This has to be the case so that we get 4x³ in our polynomial.

We observe the −6 as the constant term of our polynomial, so the numbers b, d, and g will most likely be chosen from the factors of −6, which are ±1, ±2, ±3 or ±6.

However, it would take us far too long to try all the combinations so far considered. We'd need to multiply them all out to see which combination actually did produce p(x).

So we go about it a different way.

Finding one factor: We try out some of the possible simpler factors and see if the "work". If we divide the polynomial by the expression and there's no remainder, then we've found a factor.

An easier way is to make use of the Remainder Theorem, which we met in the previous section, Factor and Remainder Theorems. It says:

If a polynomial f(x) is divided by (x − r) and a remainder R is obtained, then f(r) = R.

We go looking for an expression (called a linear term) that will give us a remainder of 0 if we were to divide the polynomial by it.

Trial 1: We try (x − 1) and find the remainder by substituting 1 (notice it's positive 1) into p(x).

p(1) = 4(1)³ − 3(1)² − 25(1) − 6 = 4 − 3 − 25 − 6 = −30 ≠ 0

Trial 2: We try (x + 1) and find the remainder by substituting −1 (notice it's negative 1) into p(x).

p(−1) = 4(−1)³ − 3(−1)² − 25(−1) − 6 = −4 − 3 + 25 − 6 = 12 ≠ 0

Trial 3: We try (x − 2) and find the remainder by substituting 2 (notice it's positive) into p(x).

p(2) = 4(2)³ − 3(2)² − 25(2) − 6 = 32 − 12 − 50 − 6 = −36 ≠ 0

Trial 4: We try (x + 2) and find the remainder by substituting −2 (notice it's negative) into p(x).

p(−2) = 4(−2)³ − 3(−2)² − 25(−2) − 6 = −32 − 12 + 50 − 6 = 0

Since the remainder is 0, we can conclude (x + 2) is a factor.

So we can write p(x) = (x + 2) × ( something )

To find out what goes in the second bracket, we need to divide p(x) by (x + 2). We saw how to divide polynomials in the previous section, Factor and Remainder Theorems.

So we can now write p(x) = (x + 2)(4x² − 11x − 3).

Now, that second bracket is just a trinomial (3-term quadratic polynomial) and we can fairly easily factor it using the process from Factoring Trinomials.

(4x² − 11x − 3) = (4x +1)(x − 3)

So putting it all together, the polynomial p(x) can be written:

p(x) = 4x³ − 3x² − 25x − 6 = (x − 3)(4x + 1)(x + 2)

That is, the factors of p(x) are:

(x − 3), (4x + 1), and (x + 2)

How to factor polynomials with 5 terms?

Example 4

Factor the polynomial r(x) = 3x⁴ + 2x³ − 13x² − 8x + 4.

Solution

Since the degree of this polynomial is 4, we expect our solution to be of the form

3x⁴ + 2x³ − 13x² − 8x + 4 = (3x − a₁)(x − a₂)(x − a₃)(x − a₄)

The first bracket has a 3 (since the factors of 3 are 1 and 3, and it has to appear in one of the brackets.) The remaining unknowns must be chosen from the factors of 4, which are 1, 2, or 4.

Once again, we'll use the Remainder Theorem to find one factor. We'll divide r(x) by that factor and this will give us a cubic (degree 3) polynomial. We'll find a factor of that cubic and then divide the cubic by that factor. Then we are left with a trinomial, which is usually relatively straightforward to factor.

Trial 1: We try substituting x = 1 and find it's not successful (it doesn't give us zero).

r(1) = 3(1)⁴ + 2(1)³ − 13(1)² − 8(1) + 4 = −12

Trial 2: We try substituting x = −1 and this time we have found a factor.

r(1) = 3(−1)⁴ + 2(−1)³ − 13(−1)² − 8(−1) + 4 = 0

We conclude (x + 1) is a factor of r(x). So our factors will look something like this:

3x⁴ + 2x³ − 13x² − 8x + 4 = (3x − a₁)(x + 1)(x − a₃)(x − a₄)

Now we need to divide (x + 1) into r(x).

So

`r_1(x)=(r(x))/(x+1) = 3x^3-x^2-12x+4`

We now need to find the factors of `r_1(x)=3x^3-x^2-12x+4`. We use the Remainder Theorem again:

There's no need to try x = 1 or x = −1 since we already tested them in `r(x)`. (One was successful, one was not).

So we try `x=2`.

`r_1(2) = 3(2)^3-(2)^2-12(2)+4 = 0`

We conclude `(x-2)` is a factor of `r_1(x)`.

We divide `r_1(x)` by `(x-2)` and we get `3x^2+5x-2`. (I will leave the reader to perform the steps to show it's true.)

Finally, we need to factor the trinomial `3x^2+5x-2`. It will clearly involve `3x` and `+-1` and `+-2` in some combination. We arrive at:

`3x^2+5x-2 = (3x-1)(x+2)`

So putting it all together, we have:

r(x) = 3x⁴ + 2x³ − 13x² − 8x + 4 = (3x − 1)(x + 1)(x − 2)(x + 2)

Using a computer algebra system to factor polynomials

What if we needed to factor polynomials like these?

Example 5: x² − 5.2x + 6.751

This trinomial doesn't have "nice" numbers, and it would take some fiddling to factor it by inspection. We could use the Quadratic Formula to find the factors.

Example 6: 2.9x³ − πx² − 4.97x + 43883

The above cubic polynomial also has rather nasty numbers. Finding the first factor and then dividing the polynomial by it would be quite challenging.

Example 7: 3175x⁴ + 256x³ − 139x² − 87x + 480

This quartic polynomial (degree 4) has "nice" numbers, but the combination of numbers that we'd have to try out is immense. The factors of 480 are

{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 32, 40, 48, 60, 80, 96, 120, 160, 240, 480}

We would also have to consider the negatives of each of these.

When a polynomial has quite high degree, even with "nice" numbers, the workload for finding the factors would be quite steep. For example:

Example 8: x⁵ − 4x⁴ − 7x³ + 14x² − 44x + 120

The factors of 120 are as follows, and we would need to keep going until one of them "worked".

{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120}

In fact in this case, the first factor (after trying `+-1` and `-2`) is actually `(x-2)`.

In such cases, it's better to realize the following:

• We are mostly finidng factors so we can solve polynomial equations (where the polynomial is set equal to zero, and we need to find the "roots" of the equation)
• The above techniques are "nice to know" mathematical methods, but are only really useful if the numbers in the polynomial are "nice", and the factors come out easily without too much trial and error

Let's consider a better solution.

Solution using computer algebra systems

Examples 5 and 6 don't really have nice factors, not even when we get a computer to find them for us.

Example 7 has factors (given by Wolfram|Alpha)

`3175,` `(x - 0.637867),` `(x + 0.645296),` ` (x + (0.0366003 - 0.604938 i)),` ` (x + (0.0366003 + 0.604938 i))`

I'm not in a hurry to do that one on paper!

Consider Example 8, which was:

x⁵ − 4x⁴ − 7x³ + 14x² − 44x + 120

Using Wolfram|Alpha, we get

(x − 2)(x − 5)(x + 3)(x² + 4) = 0

Note we don't get 5 items in brackets for this example.

This next one looks scary:

Example 9: x⁴ + 0.4x³ − 6.49x² + 7.244x − 2.112 = 0

Letting Wolfram|Alpha do the work for us, we get:

`0.002 (2 x - 1) (5 x - 6) (5 x + 16) (10 x - 11) `

So while it's interesting to know the process for finding these factors, it's better to make use of available tools.

Coming up

In the next section, we'll learn how to Solve Polynomial Equations.