The Factor Theorem

The Factor Theorem is the Remainder Theorem with a punchline. It says: zeros and linear factors are the same information. If $a$ is a zero of $p(x)$, then $(x - a)$ divides $p(x)$, and vice versa. This is how cubics , which look intimidating , get cracked open into a product of a linear factor and a quadratic.

Definition

A factor of a polynomial $p(x)$ is a non-constant polynomial $f(x)$ such that $p(x) = f(x) \cdot g(x)$ for some polynomial $g(x)$. The Factor Theorem is the bridge between zeros and linear factors: $p(a) = 0 \iff (x - a) \text{ is a factor of } p(x).$

Statement, proof, and the procedure

Factor Theorem. Let $p(x)$ be a polynomial of degree $\ge 1$. A real number $a$ is a zero of $p(x)$ if and only if $(x - a)$ is a factor of $p(x)$.

Proof. By the Remainder Theorem, when $p(x)$ is divided by $(x - a)$ the remainder is $p(a)$, so $p(x) = (x - a) q(x) + p(a)$ for some polynomial $q(x)$. If $p(a) = 0$, then $p(x) = (x-a) q(x)$, so $(x - a)$ is a factor. Conversely, if $(x - a)$ is a factor, then $p(x) = (x - a) q(x)$, and substituting $x = a$ gives $p(a) = 0$. Done.

The procedure to factorise a cubic.

1. Hunt for one zero. Try small integers , $\pm 1, \pm 2, \pm 3$ , using the Rational Root Test as a guide. For a polynomial with integer coefficients, the rational roots have the form $\dfrac{\text{factor of constant term}}{\text{factor of leading coefficient}}$. So for $x^3 - 6x^2 + 11x - 6$, possible integer roots are $\pm 1, \pm 2, \pm 3, \pm 6$. Substitute and check.
2. Once a zero $a$ is found, divide $p(x)$ by $(x - a)$. The quotient is a quadratic.
3. Factorise the quadratic. Use the techniques of the next lesson (splitting the middle term) or the quadratic formula.
4. Combine. Write $p(x)$ as a product of linear factors.

Worked walk-through. Factorise $p(x) = x^3 - 6x^2 + 11x - 6$.

• $p(1) = 1 - 6 + 11 - 6 = 0$. So $(x - 1)$ is a factor.
• Divide: $x^3 - 6x^2 + 11x - 6 \div (x-1) = x^2 - 5x + 6$.
• Factorise $x^2 - 5x + 6 = (x - 2)(x - 3)$.
• Final: $p(x) = (x-1)(x-2)(x-3)$. Zeros are $1, 2, 3$.

Long division of polynomials. When you do step 2, you can use long division or the synthetic division short-cut. Here is the long-division way for the cubic above. Write $x^3 - 6x^2 + 11x - 6$ and divide by $(x - 1)$:

• $x^3 \div x = x^2$. Multiply $x^2$ by $(x - 1)$: $x^3 - x^2$. Subtract from the cubic: $-5x^2 + 11x - 6$.
• $-5x^2 \div x = -5x$. Multiply by $(x-1)$: $-5x^2 + 5x$. Subtract: $6x - 6$.
• $6x \div x = 6$. Multiply by $(x-1)$: $6x - 6$. Subtract: $0$.

Quotient: $x^2 - 5x + 6$. Remainder: $0$ (as the Factor Theorem promised).

Divisor of the form $(ax + b)$. The same logic works: $(ax + b)$ is a factor of $p(x)$ iff $p(-\tfrac{b}{a}) = 0$. For example, to test whether $(2x - 1)$ is a factor of $p(x) = 4x^3 - 4x^2 - x + 1$, compute $p(\tfrac{1}{2}) = \tfrac{1}{2} - 1 - \tfrac{1}{2} + 1 = 0$. Yes , $(2x-1)$ is a factor.

Worked examples

Example 1. Is $(x - 2)$ a factor of $p(x) = x^3 - 3x^2 - 4x + 12$?

$p(2) = 8 - 12 - 8 + 12 = 0$. Yes, $(x - 2)$ is a factor.

Example 2. Factorise $p(x) = x^3 + 13x^2 + 32x + 20$.

Try $x = -1$: $p(-1) = -1 + 13 - 32 + 20 = 0$. So $(x + 1)$ is a factor.

Divide $p(x)$ by $(x + 1)$: quotient $x^2 + 12x + 20 = (x + 2)(x + 10)$.

Final: $p(x) = (x + 1)(x + 2)(x + 10)$.

Example 3. Find the value of $k$ for which $(x - 1)$ is a factor of $p(x) = 2x^3 + ax^2 + 2x - 5$.

By Factor Theorem, $p(1) = 0$: $2 + a + 2 - 5 = a - 1 = 0$, so $a = 1$.

Example 4. Show that $(x + 2)$ is a factor of $p(x) = x^3 - 4x + 4$, and factorise.

Test: $p(-2) = -8 + 8 + 4 = 4 \ne 0$. So $(x+2)$ is not a factor here. (A quick reality check that the theorem also discriminates.) Try $p(2) = 8 - 8 + 4 = 4 \ne 0$. Try $p(-1) = -1 + 4 + 4 = 7 \ne 0$. Try $p(1) = 1 - 4 + 4 = 1 \ne 0$. The polynomial has no rational zero of small magnitude; it does not factorise over the rationals in this case. (This is itself useful information.)

Example 5. Use the Factor Theorem to factorise $x^3 + 6x^2 + 11x + 6$.

Try $x = -1$: $p(-1) = -1 + 6 - 11 + 6 = 0$. So $(x + 1)$ is a factor. Divide to get $x^2 + 5x + 6 = (x + 2)(x + 3)$. Therefore $p(x) = (x+1)(x+2)(x+3)$.

Try it yourself

1. Is $(x - 3)$ a factor of $p(x) = x^3 - 7x + 6$?
2. Is $(x + 1)$ a factor of $p(x) = x^4 + x^3 - x - 1$?
3. Factorise $p(x) = x^3 - 23x^2 + 142x - 120$ (one of $1, 2, 3, 5, 8, 10, 12, 15$ is a zero).
4. Find $k$ so that $(x - 2)$ is a factor of $p(x) = x^3 + kx^2 - 4x + 8$.
5. Factorise $x^3 - 3x^2 - x + 3$.
6. Factorise $x^3 + 2x^2 - x - 2$.
7. Test whether $(2x + 1)$ is a factor of $4x^3 + 6x^2 - 1$.
8. Find $a$ and $b$ if $(x - 1)$ and $(x + 2)$ are both factors of $p(x) = ax^3 + bx^2 - 5x + 2$.
9. Factorise $x^3 + x^2 - 4x - 4$.
10. Use the Factor Theorem to show $(x - 1)$ is a factor of $p(x) = x^n - 1$ for every positive integer $n$.

Pitfalls / Insight

• Try small integers first. The Rational Root Test promises: for monic polynomials with integer coefficients, every rational zero is an integer factor of the constant term.
• A remainder of $0$ is the entire requirement. Don't second-guess; the Factor Theorem is iff.
• Negative zeros are common. Always try $-1, -2, -3$ in addition to positive values.

Insight. The Factor Theorem reduces cubic factorisation to "try a few small numbers". The Rational Root Test trims the list. Once a single zero is in hand, polynomial long division finishes the job in three lines. This pattern , guess a zero, divide, factorise the rest , is your standard procedure for every cubic this year and next.