AL3, AL4 - Dividing polynomials and the factor theorem

Polynomial division reverses the multiplication you already know. AL3 shows you how to find a quotient and remainder; AL4 turns a single substitution into a test for a linear factor. You will use the two ideas together to factorise cubics and find missing coefficients.

Quotients and remainders

Can one small calculation tell you whether a whole cubic has a factor? For example, substituting $x=1$ into $x^3-2x^2-5x+6$ gives $0$ . That will eventually reveal a bracket without any trial expansion. First, we need to see what division leaves behind.

You already know

Multiplying $(x-2)(x^2+4x+3)$ gives $x^3+2x^2-5x-6$ . Subtract a whole polynomial by changing every sign in its bracket. A missing power has coefficient $0$ .

Ordinary division gives $23=5\times4+3$ : dividend equals divisor times quotient plus remainder. The polynomial version has the same structure. In long division, remove the highest remaining power at each step, then subtract the complete product.

Quotient and remainder

When a polynomial $\mathrm{f}(x)$ is divided by a non-zero polynomial $\mathrm{d}(x)$ , write

\mathrm{f}(x)\equiv\mathrm{d}(x)\mathrm{q}(x)+\mathrm{r}(x)

The polynomial $\mathrm{q}(x)$ is the quotient. The remainder $\mathrm{r}(x)$ is zero or has degree smaller than the divisor. A linear divisor therefore leaves a constant remainder; a quadratic divisor may leave a linear remainder.

The division identity is not on the formulae sheet: learn it. Dividing it by $\mathrm{d}(x)$ gives a quotient plus a fraction, but that fraction is not itself the remainder.

Worked example 1: exact division

Question. Divide $x^3+2x^2-5x-6$ by $x-2$ , stating the quotient and remainder.

Plan. Divide the leading terms, multiply the divisor by that result, subtract, then repeat.

\begin{aligned} x^3\div x&=x^2 && \text{first quotient term}\\ (x^3+2x^2-5x-6)-x^2(x-2)&=4x^2-5x-6 && \text{subtract the product}\\ 4x^2\div x&=4x && \text{next quotient term}\\ (4x^2-5x-6)-4x(x-2)&=3x-6 && \text{subtract again}\\ 3x\div x&=3 && \text{last quotient term}\\ (3x-6)-3(x-2)&=0 && \text{nothing remains} \end{aligned}

Why divide the leading terms? The chosen product cancels the highest power, so each subtraction reduces the degree of what remains.

Answer. Quotient $x^2+4x+3$ ; remainder $0$ .

Check. $(x-2)(x^2+4x+3)=x^3+2x^2-5x-6$ , recovering the dividend.

Now change only the constant term, from $-6$ to $-2$ . The earlier cancellations stay the same, but the final subtraction no longer gives zero.

Worked example 2: a non-zero remainder

Question. Divide $x^3+2x^2-5x-2$ by $x-2$ , stating the quotient and remainder.

Plan. Use the same leading-term cancellations and stop when the remainder is constant.

\begin{aligned} (x^3+2x^2-5x-2)-x^2(x-2)&=4x^2-5x-2 && \text{remove the cubic term}\\ (4x^2-5x-2)-4x(x-2)&=3x-2 && \text{remove the quadratic term}\\ (3x-2)-3(x-2)&=4 && \text{remove the linear term} \end{aligned}

Why stop at $4$ ? A constant has smaller degree than a linear divisor; another polynomial term cannot remove it.

Answer. Quotient $x^2+4x+3$ ; remainder $4$ . Equivalently, for $x\ne2$ ,

\frac{x^3+2x^2-5x-2}{x-2}=x^2+4x+3+\frac4{x-2}

Check. $(x-2)(x^2+4x+3)+4=x^3+2x^2-5x-2$ .

The same leading-term method works with a quadratic divisor. For example, write a missing term explicitly before dividing:

\begin{aligned} x^3+0x^2-3x+5-x(x^2+1)&=-4x+5 && \text{subtract the leading product}\\ x^3-3x+5&\equiv x(x^2+1)+(-4x+5) && \text{degree }1\text{ is below }2 \end{aligned}

Here the quotient is $x$ and the remainder is $-4x+5$ . A remainder need not be a number: its permitted degree depends on the divisor.

Watch out

Wrong: “The remainder in Worked example 2 is $\frac4{x-2}$ .” That is the fractional term after division. The remainder is $4$ , as the identity $\mathrm{f}(x)\equiv(x-2)\mathrm{q}(x)+4$ shows. State the quotient and remainder separately when asked.

Try another change to the constant term; keep the final remainder separate from its fractional form.

Why a zero output reveals a factor

A factor means exact division, so a zero remainder is the link between substitution and factorisation.

Divide a polynomial by $x-a$ . Its remainder is a constant $r$ , giving

\begin{aligned} \mathrm{f}(x)&\equiv(x-a)\mathrm{q}(x)+r && \text{division identity}\\ \mathrm{f}(a)&=(a-a)\mathrm{q}(a)+r=r && \text{substitute }x=a \end{aligned}

The whole quotient disappears when $x=a$ . That is why a single substitution finds the remainder without doing the division. In particular, zero output means zero remainder.

Factor theorem

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

This holds for a polynomial $\mathrm{f}(x)$ and a real number $a$ . Both directions matter: a factor gives a zero output, and a zero output gives a factor.
Not on the formulae sheet: learn it.

For example, if $\mathrm{f}(-3)=0$ , the factor is $x-(-3)=x+3$ . If $\mathrm{f}(2)=7$ , then $x-2$ is not a factor: division by it leaves remainder $7$ . The number substituted is a root, while the bracket is a factor; they are connected but are different objects.

A bracket such as $2x-3$ becomes zero at $x=\frac32$ , so test $\mathrm{f}(\frac32)$ . Since $2x-3=2(x-\frac32)$ , the same zero-output test applies; the factor's coefficient need not be $1$ .

Key idea

Set the proposed linear factor equal to zero to find the input to test. A zero output proves that factor; a non-zero output disproves it.

The sign inside the bracket is the common trap. Locate it in this attempted use of the theorem.

Finding all the factors of a cubic

Once a cubic has one linear factor, division leaves a quadratic, which you can factorise using GCSE methods.

For a monic polynomial with integer coefficients, any integer root divides the constant term. Indeed, if $a^3+ba^2+ca+d=0$ , then $d=-a(a^2+ba+c)$ . Try the positive and negative divisors of $d$ first. This gives useful candidates, not a promise that every cubic has an integer root. If the constant term is zero, factor out $x$ immediately. This candidate rule is not on the formulae sheet: learn how to use it.

Worked example 3: a monic cubic

Question. Using the factor theorem, factorise $\mathrm{f}(x)=x^3-2x^2-5x+6$ .

Plan. Test a divisor of $6$ , divide out the factor obtained, then factorise the quotient.

\begin{aligned} \mathrm{f}(1)&=1-2-5+6=0 && \text{test a candidate}\\ \mathrm{f}(x)&\equiv(x-1)(x^2-x-6) && \text{divide out }x-1\\ x^2-x-6&=(x-3)(x+2) && \text{product }-6\text{ and sum }-1 \end{aligned}

The division can be checked directly: $x(x-1)$ is not the quotient; the quotient must be quadratic. Long division gives quotient terms $x^2$ , $-x$ , $-6$ and remainder $0$ .

Why test $1$ first? It divides the constant and makes substitution quick. The written equality $\mathrm{f}(1)=0$ shows that the factor theorem, rather than an unsupported guess, produced the factor.

Answer. $\mathrm{f}(x)\equiv(x-1)(x-3)(x+2)$ .

Check. Multiplying the last two brackets gives $x^2-x-6$ ; multiplying by $x-1$ recovers the cubic.

The next example changes the leading coefficient from $1$ to $2$ . The zero-input test is unchanged, but the remaining quadratic needs a non-monic factorisation.

Worked example 4: a non-monic quotient

Question. Using the factor theorem, factorise $\mathrm{g}(x)=2x^3-x^2-7x+6$ .

Plan. Test $1$ , divide by $x-1$ , then split the middle term of the quadratic.

\begin{aligned} \mathrm{g}(1)&=2-1-7+6=0 && \text{so }x-1\text{ is a factor}\\ \mathrm{g}(x)&\equiv(x-1)(2x^2+x-6) && \text{polynomial division}\\ 2x^2+x-6&=2x^2+4x-3x-6 && \text{split the middle term}\\ &=2x(x+2)-3(x+2) && \text{factor each pair}\\ &=(2x-3)(x+2) && \text{take out }x+2 \end{aligned}

Why split $x$ as $4x-3x$ ? The two coefficients add to $1$ and multiply to $2\times(-6)=-12$ , allowing the same bracket to appear in each pair.

Answer. $\mathrm{g}(x)\equiv(x-1)(2x-3)(x+2)$ .

Check. $\mathrm{g}(\frac32)=\frac{27}{4}-\frac94-\frac{21}{2}+6=0$ , consistent with the factor $2x-3$ .

This answers the opening question: $\mathrm{f}(1)=0$ exposes the factor $x-1$ , and division exposes the rest. A further surprise is that $x^3-3x+2\equiv(x-1)^2(x+2)$ . A cubic can have only two distinct real roots because the same linear factor can occur twice. Testing $1$ once establishes one factor; division is what reveals its repetition.

Watch out

Wrong: after finding $\mathrm{f}(1)=0$ , writing $x+1$ as the factor. Right: the factor is $x-1$ , because that bracket becomes zero at the tested input $1$ . Divide by the bracket established by the substitution, keeping its sign unchanged.

That first bracket starts the factorisation, but the quotient is still needed to justify the remaining factors.

Exam tip

When OCR says “Using the factor theorem” with detailed reasoning, show a substitution giving $0$ , identify its factor, and obtain the quadratic quotient before writing the full factorisation. A list of calculator roots does not show that argument.

Use the same chain with a new cubic. You may choose any successful candidate.

Using a factor to find a coefficient

A stated factor is an equation in disguise: substitute the input that makes the factor zero, then solve for the unknown coefficient.

Here the unknown is a coefficient such as $k$ , rather than a root. Keep those roles separate: the proposed factor already tells you the input to use.

Worked example 5: an unknown constant term

Question. Given that $x-2$ is a factor of $\mathrm{f}(x)=x^3-4x+k$ , determine $k$ .

Plan. Use $\mathrm{f}(2)=0$ to form a linear equation in $k$ .

\begin{aligned} \mathrm{f}(2)&=8-8+k && \text{substitute the known root}\\ 0&=k && \text{use the stated factor} \end{aligned}

Why set the output to zero? Exact divisibility by $x-2$ means precisely $\mathrm{f}(2)=0$ ; it does not mean $\mathrm{f}(x)=2$ .

Answer. $k=0$ .

Check. $x^3-4x=x(x-2)(x+2)$ , which contains the stated factor.

Now move the unknown coefficient from the constant term to the $x$ term. It is multiplied by the substituted input, so it cannot be copied unchanged.

Worked example 6: an unknown linear coefficient

Question. Given that $x-2$ is a factor of $\mathrm{g}(x)=x^3+kx+6$ , determine $k$ .

Plan. Substitute $2$ into every occurrence of $x$ and set the output equal to zero.

\begin{aligned} \mathrm{g}(2)&=8+2k+6=0 && \text{factor theorem}\\ 2k&=-14 && \text{collect the constants}\\ k&=-7 && \text{divide by }2 \end{aligned}

Why does $kx$ become $2k$ ? Only $x$ is replaced by $2$ ; $k$ is the fixed coefficient we are trying to find.

Answer. $k=-7$ .

Check. At $x=2$ , $x^3-7x+6=8-14+6=0$ . In fact, $x^3-7x+6\equiv(x-2)(x-1)(x+3)$ .

You can combine coefficient-finding with division if a question then asks for all the factors. First find the coefficient, replace it in the original polynomial, and only then divide. This prevents carrying an unnecessary unknown through the division.

Watch out

Wrong: for a factor $x+3$ , using $\mathrm{f}(3)=0$ . Right: use $\mathrm{f}(-3)=0$ . Also keep powers and coefficients distinct: at $x=-3$ , $kx^2=9k$ , whereas $kx=-3k$ .

The next coefficient lies on the square term, so decide its multiplier before solving.

Quick Recap

Division gives dividend equal to divisor times quotient plus remainder. Stop when the remainder's degree is smaller than the divisor's.
The remainder on dividing by $x-a$ is $\mathrm{f}(a)$ ; it is a factor exactly when that output is $0$ .
To factorise a cubic, establish a linear factor, divide it out, then factorise the quadratic quotient.
A stated factor supplies an input and a zero output, giving an equation for a missing coefficient.

The opening calculation $\mathrm{f}(1)=0$ proves that $x-1$ divides the entire cubic, leading to $x^3-2x^2-5x+6\equiv(x-1)(x-3)(x+2)$ .

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AL3, AL4 - Dividing polynomials and the factor theorem

Quotients and remainders

Why a zero output reveals a factor

Finding all the factors of a cubic

Using a factor to find a coefficient

Quick Recap

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