B6 - Polynomials and the factor theorem | AQA A-Level Mathematics

Learn to expand and factorise polynomials, divide by a linear expression and use the factor theorem to connect factors with roots. You will also use factorisation to explain a pattern that holds for every integer.

Expanding and collecting polynomials

Why does the cube of an integer always differ from that integer by a multiple of $6$ ? For $4$ , the difference is $4^3-4=60=6\times10$ . Checking more integers suggests a pattern, but rewriting the expression will explain every case at once.

Start with a simpler rewriting task. In $(2x-1)(x^2+3x-2)$ , each term in the first bracket multiplies every term in the second. Collecting terms then changes its appearance without changing its value for any $x$ .

Polynomial and degree

A polynomial in $x$ is a finite sum of constant multiples of non-negative integer powers of $x$ . For a nonzero polynomial, its degree is the largest power with a nonzero coefficient after terms are collected. Thus $3x^2-2x+1$ has degree $2$ , while $\dfrac1x$ and $\sqrt{x}$ are not polynomials in $x$ . The zero polynomial has no degree under this convention.

The coefficient multiplies a power, so the coefficient of $x^2$ in $3x^2$ is $3$ . Only matching powers are like terms: $3x^2+2x^2=5x^2$ , but $3x^2+2x$ cannot be collected into one term.

You already know

The distributive law gives $a(b+c)=ab+ac$ . A negative multiplier applies to every term: $-1(x^2+3x-2)=-x^2-3x+2$ . This rule is not supplied in the formula booklet.

Write a separate row of terms for each multiplier before collecting. This makes a missing term or a sign slip easier to spot.

Worked example 1: a linear factor times a quadratic

Question. Expand and simplify $(2x-1)(x^2+3x-2)$ .

Plan. Multiply by $2x$ , multiply by $-1$ , then collect matching powers.

\begin{aligned} (2x-1)(x^2+3x-2) &=2x^3+6x^2-4x-x^2-3x+2 && \text{distribute both terms} \\ &=2x^3+(6-1)x^2+(-4-3)x+2 && \text{collect like powers} \\ &=2x^3+5x^2-7x+2 && \text{simplify coefficients} \end{aligned}

Why this step? The term $-1$ multiplies the whole quadratic, so its final contribution is $+2$ , not $-2$ .

Answer. $2x^3+5x^2-7x+2$ .

Check. At $x=0$ , both forms give $2$ ; at $x=1$ , both give $2$ . The full distribution, rather than these spot checks alone, establishes the identity.

Worked example 2 changes only the constant in the linear bracket from $-1$ to $-3$ . Each contribution from that term changes, including its sign when it meets $-2$ .

Worked example 2: distributing a negative constant

Question. Expand and simplify $(2x-3)(x^2+3x-2)$ .

Plan. Keep the $2x$ row and replace the other multiplier by $-3$ .

\begin{aligned} (2x-3)(x^2+3x-2) &=2x^3+6x^2-4x-3x^2-9x+6 && \text{distribute} \\ &=2x^3+(6-3)x^2+(-4-9)x+6 && \text{collect like powers} \\ &=2x^3+3x^2-13x+6 && \text{descending powers} \end{aligned}

Why this step? Writing $-3(x^2+3x-2)$ as a complete row prevents the negative multiplier being applied to only the first term.

Answer. $2x^3+3x^2-13x+6$ .

Check. At $x=1$ , the product is $(-1)(2)=-2$ ; the collected polynomial gives $2+3-13+6=-2$ .

Addition and subtraction of polynomials use the same collection rule. Put a subtracted polynomial in brackets so its minus sign reaches every term. Degree can fall after collection: $(x^2+x)-(x^2-2x)=3x$ , which has degree $1$ , not $2$ . These expansion and collection rules are not in the formula booklet: learn them.

Watch out

Wrong: $6x^2-x^2=5x$ . Right: $6x^2-x^2=5x^2$ . Collecting changes the coefficient, not the power.

The next question changes the same bracket constant again. Keep all six products until you collect them.

Factors, repeated factors and real roots

Factorisation reverses expansion: a sum becomes a product, and its factors reveal where the polynomial is zero.

For example, $x^3-4x$ has a common factor $x$ , leaving $x^2-4$ . That remaining expression is a difference of two squares. Keep factorising until no factor can be split further over the number system requested.

Factors and zeros

For real numbers $u$ and $v$ ,

uv=0\ \Longleftrightarrow\ u=0\ \text{ or }\ v=0

Also,

a^2-b^2=(a-b)(a+b)

These rules apply to expressions as well as numbers. They do not permit cancelling a potentially zero factor while solving an equation.

Not in the formula booklet: learn it.

A root of $P(x)=0$ is an input at which the polynomial equals zero. A factorisation must retain every factor, even though a repeated root is listed only once when giving distinct solutions.

Worked example 3: take out the common factor

Question. Factorise $x^3-4x$ fully and find all its real roots.

Plan. Extract $x$ , factorise the difference of two squares and apply the zero-product rule.

\begin{aligned} x^3-4x&=x(x^2-4) && \text{common factor }x \\ &=x(x-2)(x+2) && \text{difference of squares} \\ x(x-2)(x+2)=0&\ \Rightarrow\ x=0,\ 2,\ -2 && \text{one factor must vanish} \end{aligned}

Why this step? Dividing the equation by $x$ at the start would silently assume $x\ne0$ and lose a valid root.

Answer. $x(x-2)(x+2)$ ; the real roots are $-2$ , $0$ and $2$ .

Check. Expanding gives $x(x^2-4)=x^3-4x$ . Substitution of each listed root gives zero.

Worked example 4 changes the kind of common factor from a single term to a bracket. Grouping makes that bracket visible, after which the same difference-of-squares step works.

Worked example 4: a common bracket

Question. Factorise $x^3+x^2-4x-4$ fully and find all its real roots.

Plan. Group the first two and last two terms, then extract the shared bracket.

\begin{aligned} x^3+x^2-4x-4 &=x^2(x+1)-4(x+1) && \text{group terms} \\ &=(x+1)(x^2-4) && \text{common bracket }x+1 \\ &=(x+1)(x-2)(x+2) && \text{difference of squares} \end{aligned}

Why this step? Both grouped terms are multiples of the entire bracket $x+1$ , so it can be taken outside just like a numerical common factor.

Answer. $(x+1)(x-2)(x+2)$ ; the real roots are $-1$ , $2$ and $-2$ .

Check. $(x+1)(x^2-4)=x^3+x^2-4x-4$ , recovering every coefficient.

Not every real factorisation ends in three different linear factors. Grouping $x^3+x^2+x+1$ gives $(x+1)(x^2+1)$ ; since $x^2+1>0$ for real $x$ , the only real root is $-1$ . Meanwhile,

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

has one distinct root, $1$ , repeated three times. Its multiplicity is $3$ . These factorisations and the zero-product reasoning are not supplied in the formula booklet.

Watch out

Wrong: replace $(x-1)^3$ by $x-1$ because the root $1$ is listed once. Repeated roots do not erase factors from the polynomial. Right: retain $(x-1)^3$ in the factorisation and list $1$ once in the solution set.

The next grouping problem changes the shared bracket and the square difference. Look for matching brackets before trying possible roots.

Dividing by a linear polynomial

Polynomial division repeatedly removes a multiple of the divisor until the leftover has smaller degree than the divisor.

The ordinary division statement $17=5(3)+2$ distinguishes a quotient from a remainder. Polynomial division does the same job with expressions: the divisor times the quotient, plus the remainder, must recover the dividend exactly.

The division identity

If a polynomial $P(x)$ is divided by $D(x)=mx+n$ , where $m\ne0$ , then

P(x)=D(x)Q(x)+r

Here $Q(x)$ is the polynomial quotient and $r$ is a constant remainder. Stop when the remaining degree is less than $1$ . Remainder $0$ means the divisor is a factor.

Not in the formula booklet: learn it.

At each stage, divide the leading remaining term by the divisor's leading term. Subtract the entire multiple you have just used. A missing power can be written with coefficient zero to keep the columns aligned.

Worked example 5: exact linear division

Question. Divide $x^3-2x^2-5x+6$ by $x-1$ , stating the quotient and remainder.

Plan. Match the leading terms in turn, subtracting each multiple of $x-1$ .

\begin{aligned} x^3\div x&=x^2 && \text{first quotient term} \\ (x^3-2x^2-5x+6)-x^2(x-1)&=-x^2-5x+6 && \text{subtract first multiple} \\ (-x^2-5x+6)-(-x)(x-1)&=-6x+6 && \text{next quotient term }-x \\ (-6x+6)-(-6)(x-1)&=0 && \text{final quotient term }-6 \end{aligned}

Why this step? Each subtraction removes the current highest power, so the process must finish with a constant remainder.

Answer. Quotient $x^2-x-6$ , remainder $0$ .

Check. $(x-1)(x^2-x-6)=x^3-2x^2-5x+6$ , exactly the original polynomial.

Worked example 6 changes only the constant term from $6$ to $7$ . The leading-term decisions stay the same, but there is now a nonzero remainder.

Worked example 6: keeping the remainder

Question. Divide $x^3-2x^2-5x+7$ by $x-1$ , stating the quotient and remainder.

Plan. Use the same leading-term division and retain the final constant left over.

\begin{aligned} (x^3-2x^2-5x+7)-x^2(x-1)&=-x^2-5x+7 && \text{first quotient term }x^2 \\ (-x^2-5x+7)-(-x)(x-1)&=-6x+7 && \text{next term }-x \\ (-6x+7)-(-6)(x-1)&=1 && \text{remainder degree is smaller} \end{aligned}

Why this step? A constant cannot be removed by a further polynomial multiple of the linear divisor, so $1$ is the remainder, not another quotient term.

Answer. Quotient $x^2-x-6$ , remainder $1$ :

x^3-2x^2-5x+7=(x-1)(x^2-x-6)+1

Check. Multiplying out the right side gives the original polynomial, including its constant $7$ .

Inspection, coefficient matching and formal long division are all valid ways to obtain this identity. The useful check is always recombination of divisor, quotient and remainder.

Watch out

Wrong: report quotient $x^2-x-5$ by adding the remainder $1$ to its constant term. That would add the whole divisor when multiplying back, rather than the constant $1$ . Right: keep $x^2-x-6$ as the quotient and $1$ as a separate remainder.

Now change the divisor to $x+1$ . Its sign changes each subtracted multiple, even though the dividend stays fixed.

Using and explaining the factor theorem

The factor theorem tests a proposed linear factor by substituting the input that makes that factor zero.

For example, the zero of $2x-3$ is $x=\dfrac32$ , because $2\left(\dfrac32\right)-3=0$ . Substituting $3$ or $-\dfrac32$ would test a different proposed factor.

Factor theorem

For a polynomial $P$ ,

x-a\text{ is a factor of }P(x)\ \Longleftrightarrow\ P(a)=0

More generally, for $m\ne0$ ,

mx+n\text{ is a factor of }P(x)\ \Longleftrightarrow\ P\left(-\frac nm\right)=0

Not in the formula booklet: learn it.

Here is why it works. Division gives $P(x)=(x-a)Q(x)+r$ . Set $x=a$ : the product term vanishes, so $P(a)=r$ . A zero value is exactly a zero remainder. A nonzero value rules out the factor. For $mx+n$ , the identical argument uses its zero $-\dfrac nm$ ; the remainder is still constant. This remainder relationship is not supplied in the formula booklet.

Worked example 7: a factor with a fractional root

Question. Use the factor theorem to show that $2x-3$ is a factor of $P(x)=2x^3-3x^2-8x+12$ . Hence factorise $P(x)$ fully and find all its real roots.

Plan. Evaluate at $\dfrac32$ , state the factor-theorem conclusion, then divide and factorise the quotient.

\begin{aligned} P\left(\frac32\right) &=2\left(\frac32\right)^3-3\left(\frac32\right)^2-8\left(\frac32\right)+12 && \text{substitute the factor's zero} \\ &=\frac{27}{4}-\frac{27}{4}-12+12=0 && \text{evaluate exactly} \end{aligned}

Therefore $2x-3$ is a factor, by the factor theorem. Division then gives

\begin{aligned} P(x)&=(2x-3)(x^2-4) && \text{linear division} \\ &=(2x-3)(x-2)(x+2) && \text{factorise the quotient} \end{aligned}

Why this step? Obtaining $P\left(\dfrac32\right)=0$ proves the proposed factor, but does not yet identify the other factors or roots.

Answer. $P(x)=(2x-3)(x-2)(x+2)$ , with real roots $\dfrac32$ , $2$ and $-2$ .

Check. Expanding $(2x-3)(x^2-4)$ gives $2x^3-3x^2-8x+12$ , recovering all coefficients.

Worked example 8 makes only the constant coefficient unknown. The same substitution now produces an equation for that coefficient.

Worked example 8: using a factor to find a coefficient

Question. Given that $2x-3$ is a factor of $P(x)=2x^3-3x^2-8x+k$ , find $k$ .

Plan. Set the value at $x=\dfrac32$ equal to zero and solve the resulting linear equation.

\begin{aligned} P\left(\frac32\right)&=0 && \text{factor theorem} \\ \frac{27}{4}-\frac{27}{4}-12+k&=0 && \text{evaluate the known terms} \\ k&=12 && \text{solve} \end{aligned}

Why this step? The given factor fixes a root, so its substitution must give zero rather than an arbitrary remainder.

Answer. $k=12$ .

Check. With $k=12$ , the polynomial becomes the one in Worked example 7, whose full factorisation includes $2x-3$ .

The same idea handles a coefficient of another power. If $x+1$ is a factor of $x^3+kx^2-4x-4$ , substituting $-1$ gives $-1+k+4-4=0$ , hence $k=1$ . Always use the zero of the stated factor.

The opening integer pattern now has a short explanation:

n^3-n=n(n^2-1)=(n-1)n(n+1)

For every integer $n$ , these are three consecutive integers. At least one is even and one is divisible by $3$ , so their product is divisible by $6$ , including when the product is zero. This identity and its divisibility consequence are not supplied in the formula booklet.

Watch out

Wrong: test the factor $2x-3$ by calculating $P(3)$ . Its zero is $\dfrac32$ , so the correct test is $P\left(\dfrac32\right)=0$ . Solve the proposed linear factor equal to zero before substituting.

Choosing the correct input is only part of the argument. The conclusion must connect the zero value to the stated factor.

Exam tip

When AQA specifies “Use the factor theorem”, show the substitution, obtain zero and conclude that the stated expression is a factor. Long division alone does not meet that instruction, even if it finds the correct factor.

In the final question, the unknown moves from the constant to the coefficient of $x$ . Its position changes the linear equation, not the theorem.

Quick Recap

Expand by distributing every term, then collect only matching powers.
Factorise fully and use the zero-product rule without losing a potentially zero factor.
Division gives $P=DQ+r$ ; multiplying back checks both quotient and remainder.
Test a proposed linear factor at its zero, and state the factor-theorem conclusion.

The cube pattern holds for every integer because $n^3-n=(n-1)n(n+1)$ : three consecutive integers always supply factors of both $2$ and $3$ .

Last Accessed

Pure Mathematics

Mechanics

Statistics

B6 - Polynomials and the factor theorem

Expanding and collecting polynomials

Factors, repeated factors and real roots

Dividing by a linear polynomial

Using and explaining the factor theorem

Quick Recap

Learn faster with Nova