2.11 - The factor theorem for polynomials

Test whether a linear expression is a factor of a polynomial by making one substitution. You will use rational values, work out unknown coefficients and reduce a cubic equation to a quadratic that you already know how to solve.

Testing a factor by substitution

Can you show that $x-1$ is a factor of $x^{100}-1$ without writing out a quotient with a hundred terms? One tiny substitution will settle it. The same idea makes ordinary cubic factorisation much more manageable.

You already know

A factor multiplies another expression to give the original expression. For example, $x-3$ is a factor of $x^2-9=(x-3)(x+3)$ . A root of an equation makes its left-hand side zero; roots are numbers, while factors are expressions.

We need a name for the expressions to which this test applies.

Polynomial

A polynomial in $x$ is a finite sum of constant multiples of non-negative whole-number powers of $x$ . Its degree is its highest power with a non-zero coefficient. For example, $2x^3-5x^2-x+6$ is cubic. The expressions $\frac1x+x$ and $\sqrt{x}+2$ are not polynomials in $x$ .

Write $\mathrm{f}(x)$ for the polynomial. If $\mathrm{f}(x)=(x-a)\mathrm{q}(x)$ , then substituting $x=a$ certainly gives zero. The reverse direction also works: division by $x-a$ leaves a constant remainder $r$ , so $\mathrm{f}(x)=(x-a)\mathrm{q}(x)+r$ . At $x=a$ , this gives $\mathrm{f}(a)=r$ . A zero value therefore means no remainder, and hence an exact factor.

Factor theorem

For a polynomial $\mathrm{f}(x)$ ,

\mathrm{f}(a)=0\quad\Longleftrightarrow\quad x-a\text{ is a factor of }\mathrm{f}(x)

For a linear factor $px-q$ , with $p\ne0$ , test $x=\frac qp$ , the value that makes that factor zero. In this lesson the tested values are rational.
Not on the formulae sheet: learn it.

Substitute the zero of the proposed factor, evaluate fully, then state the conclusion. The final statement connects your arithmetic to the factor you were asked about.

Worked example 1: a monic linear factor

Question. Use the factor theorem to show that $x-2$ is a factor of $2x^3-5x^2-x+6$ .

Plan. Evaluate the polynomial at $x=2$ , where the proposed factor is zero.

\begin{aligned} \mathrm{f}(2)&=2(2)^3-5(2)^2-2+6 && \text{substitute }x=2 \\ &=16-20-2+6 && \text{evaluate powers first} \\ &=0 && \text{collect the terms} \end{aligned}

Why this step? $x-2$ is zero at $2$ , so the factor theorem asks for $\mathrm{f}(2)$ .

Answer. Since $\mathrm{f}(2)=0$ , $x-2$ is a factor, by the factor theorem.

Check. Expanding $(x-2)(2x^2-x-3)$ gives $2x^3-5x^2-x+6$ .

Worked example 2 keeps the polynomial but changes the proposed factor to a non-monic one. Solve the factor's linear equation before substituting.

Worked example 2: a rational value

Question. Use the factor theorem to show that $2x-3$ is a factor of $2x^3-5x^2-x+6$ .

Plan. Solve $2x-3=0$ , then evaluate the polynomial at that rational value.

\begin{aligned} 2x-3=0&\quad\Longrightarrow\quad x=\frac32 && \text{zero of the factor} \\ \mathrm{f}\left(\frac32\right)&=2\left(\frac32\right)^3-5\left(\frac32\right)^2-\frac32+6 && \text{substitute} \\ &=\frac{27}{4}-\frac{45}{4}-\frac64+\frac{24}{4} && \text{common denominator} \\ &=0 \end{aligned}

Why this step? The factor is zero at $\frac32$ , not at $3$ or $-\frac32$ .

Answer. Since $\mathrm{f}(\frac32)=0$ , $2x-3$ is a factor, by the factor theorem.

Check. $(2x-3)(x^2-x-2)$ expands to the stated polynomial.

Degree does not change the test. For $\mathrm{g}(x)=x^5-3x^4-2x+6$ ,

\mathrm{g}(3)=243-243-6+6=0

so $x-3$ is a factor. There is no need to divide the quintic to prove that one claim. Likewise, the opening puzzle has $1^{100}-1=0$ , so $x-1$ is a factor of $x^{100}-1$ . A hundredth-degree polynomial gives way to arithmetic with $1$ .

Watch out

Wrong: to test $x+2$ , substitute $2$ . Right: substitute $-2$ . For the polynomial in the examples, $\mathrm{f}(-2)=-28\ne0$ , so $x+2$ is not a factor.

In a “Use the factor theorem” question, showing a zero substitution and stating the factor conclusion are the required argument. Now test a different rational value.

Using a factor to work out a coefficient

A known factor turns an unknown coefficient into an ordinary equation: the polynomial must equal zero at the factor's root.

Do not try to factorise a polynomial containing an unknown coefficient first. Substitution removes most of the algebra immediately. This use of the factor theorem is not separately supplied on the formulae sheet: learn the connection.

Worked example 3: an unknown middle coefficient

Question. $x-1$ is a factor of $x^3+kx^2-5x+3$ . Work out $k$ .

Plan. Set $\mathrm{f}(1)=0$ , then solve for $k$ .

\begin{aligned} 1+k-5+3&=0 && \text{the known factor gives }\mathrm{f}(1)=0 \\ k-1&=0 && \text{collect constants} \\ k&=1 && \text{add }1 \end{aligned}

Why this step? Knowing that $x-1$ is a factor tells you the value of the entire polynomial at $1$ , not just the value of one term.

Answer. $k=1$ .

Check. With $k=1$ , substitution gives $1+1-5+3=0$ .

Worked example 4 changes only the known factor. A rational root can produce a fractional coefficient; keep the substitution exact.

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2.11 - The factor theorem for polynomials

Testing a factor by substitution

Using a factor to work out a coefficient

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