Pure 2.6 - Algebraic manipulation of polynomials | Edexcel A-Level Mathematics

The same polynomial can reveal different information in expanded, factorised or quotient form. You will choose a useful form, divide by linear expressions, use the factor theorem to factorise cubics, and simplify rational expressions while preserving excluded inputs.

See the structure before expanding

How can you factorise $6x^3+11x^2-x-6$ without guessing three brackets at once? One well-chosen substitution will reveal a first factor. Division then turns the remaining problem into a familiar quadratic.

You already know

The distributive law gives $a(b+c)=ab+ac$ . Factorising reverses expansion, and a product is zero if at least one factor is zero.

Before expanding, look for a common bracket. In $(x-2)(x+5)+3x(x-2)$ , both terms contain $x-2$ . Treating that entire bracket as one factor saves work and exposes a root.

Polynomial

A polynomial in $x$ is a finite sum of constant multiples of non-negative integer powers of $x$ . For example, $3x^4-2x+7$ is a polynomial; $x^{-1}+4$ and $\sqrt x+2$ are not polynomials in $x$ . Its degree is its highest power with a non-zero coefficient.

Expanded form is useful for collecting like terms and comparing coefficients. Factorised form is useful for solving equations and simplifying fractions. Neither changes the expression's value. The distributive and factorising rules here are not supplied in the formula booklet.

Worked example 1: a common bracket

Question. Factorise $P(x)=(x-2)(x+5)+3x(x-2)$ , then expand your answer.

Plan. Factor out the shared bracket before expanding the shorter product.

\begin{aligned} P(x)&=(x-2)\big((x+5)+3x\big) &&\text{common factor}\\ &=(x-2)(4x+5) &&\text{collect inside}\\ &=4x^2-3x-10 &&\text{expand} \end{aligned}

Why this step? The shared factor is the whole $x-2$ , so it can come outside one bracket.

Answer. $(x-2)(4x+5)=4x^2-3x-10$ .

Check. At $x=3$ , the original is $8+9=17$ ; both final forms give $17$ .

Change only the joining sign from plus to minus. The sign now belongs to the entire second term.

Worked example 2: a sign before a product

Question. Factorise $(x-2)(x+5)-3x(x-2)$ , then expand.

Plan. Extract $x-2$ , retaining the subtraction inside.

\begin{aligned} (x-2)(x+5)-3x(x-2)&=(x-2)(x+5-3x) &&\text{common factor}\\ &=(x-2)(5-2x) &&\text{collect}\\ &=-2x^2+9x-10 &&\text{expand} \end{aligned}

Why this step? The minus sign changes the contribution of $3x$ to the inner bracket.

Answer. $(x-2)(5-2x)=-2x^2+9x-10$ .

Check. At $x=3$ , the original gives $8-9=-1$ , agreeing with $-18+27-10$ .

In a direct expansion such as $(2x-1)(x+4)-x(x-3)$ , distribute the minus sign after expanding the second product. This is the same sign issue in a less visibly factorised form.

Watch out

Wrong: $-x(x-3)=-x^2-3x$ . Right: $-x(x-3)=-x^2+3x$ . The subtraction applies to both terms.

Use the corrected distributive step to finish this expansion.

Divide by a linear expression

Polynomial division repeats a small cycle: divide the leading terms, multiply the divisor by that quotient term, and subtract. Each subtraction lowers the degree of the remainder. Write zero coefficients for missing powers, such as $x^3+0x^2-7x+6$ , to keep columns aligned.

Division identity

For a polynomial $f$ divided by $ax+b$ , where $a\ne0$ , there are a polynomial quotient $q$ and constant remainder $R$ such that

f(x)=(ax+b)q(x)+R

Not in the formula booklet: learn it.

Only a linear divisor is needed here. A zero remainder means the divisor is a factor. Always check by multiplying back and adding the remainder.

Worked example 3: a monic divisor

Question. Let $f(x)=2x^3-3x^2-11x+6$ . Divide $f(x)$ by $x-3$ .

Plan. Match the leading term, subtract, and repeat.

\begin{aligned} 2x^3/x&=2x^2 &&\text{first quotient term}\\ f(x)-2x^2(x-3)&=3x^2-11x+6 &&\text{subtract product}\\ (3x^2-11x+6)-3x(x-3)&=-2x+6 &&\text{next term }3x\\ (-2x+6)-(-2)(x-3)&=0 &&\text{last term }-2 \end{aligned}

Why this step? Each product cancels the current leading term so the next smaller power can be handled.

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

Check. $(x-3)(2x^2+3x-2)=2x^3-3x^2-11x+6$ .

Change only the divisor to $2x-1$ . Divide by its leading term $2x$ , not by $x$ .

Worked example 4: a non-monic divisor

Question. Let $f(x)=2x^3-3x^2-11x+6$ . Divide $f(x)$ by $2x-1$ .

Plan. Use the same divide-multiply-subtract cycle with leading term $2x$ .

\begin{aligned} 2x^3/(2x)&=x^2 &&\text{first term}\\ f(x)-x^2(2x-1)&=-2x^2-11x+6 &&\text{subtract}\\ (-2x^2-11x+6)-(-x)(2x-1)&=-12x+6 &&\text{next term }-x\\ (-12x+6)-(-6)(2x-1)&=0 &&\text{last term }-6 \end{aligned}

Why this step? Using $2x$ accounts for the divisor's coefficient; ignoring it doubles the first product.

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

Check. $(2x-1)(x^2-x-6)=2x^3-3x^2-11x+6$ .

A remainder need not be zero: $x^2+5x+7=(x+2)(x+3)+1$ . The last $1$ cannot be divided further by a linear expression within polynomial division.

Watch out

Wrong: subtracting $2x^3-6x^2$ as $-2x^3-6x^2$ . Subtract the bracket, changing both signs; otherwise the next remainder is wrong.

Apply the cycle with a new monic divisor.

Use the factor theorem

Division by $x-r$ gives $f(x)=(x-r)q(x)+R$ . Put $x=r$ : the product disappears, leaving $f(r)=R$ . Substitution therefore tells you whether division will leave zero remainder, without doing the division first.

Factor theorem

For a polynomial $f$ ,

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

More generally, $ax-b$ is a factor exactly when $f(b/a)=0$ , for $a\ne0$ .
Not in the formula booklet: learn it.

To test $3x-2$ , substitute $x=2/3$ ; to test $x+1$ , substitute $x=-1$ . First solve the proposed factor equal to zero, then substitute that root in the polynomial.

Worked example 5: a parameter from a factor

Question. Given $g(x)=2x^3-9x^2+(k+5)x+2k$ , and that $x-2$ is a factor, find $k$ .

Plan. Use $g(2)=0$ to obtain a linear equation in $k$ .

\begin{aligned} g(2)&=16-36+2(k+5)+2k &&\text{substitute the root}\\ &=4k-10=0 &&\text{factor theorem}\\ k&=\frac52 &&\text{solve} \end{aligned}

Why this step? “Is a factor” means exactly that the remainder at its root vanishes.

Answer. $k=5/2$ .

Check. $16-36+15+5=0$ when $k=5/2$ .

Change only the proposed factor to $2x-1$ . Its root is a fraction, so substitute $1/2$ , not $1$ .

Worked example 6: a fractional root

Question. For the same $g(x)$ , find $k$ if $2x-1$ is a factor.

Plan. Use $g(1/2)=0$ and collect the $k$ -terms.

\begin{aligned} g\left(\frac12\right)&=\frac14-\frac94+\frac{k+5}{2}+2k &&\text{substitute }\frac12\\ &=\frac12+\frac52k=0 &&\text{collect}\\ k&=-\frac15 &&\text{solve} \end{aligned}

Why this step? The root is the value making the complete linear factor zero, including its coefficient.

Answer. $k=-1/5$ .

Check. $g(1/2)=1/4-9/4+12/5-2/5=0$ .

Testing a non-root gives a non-zero remainder, so it rules out that proposed factor. It does not rule out every other factor.

Watch out

Wrong: “ $2x-1$ is a factor, so $g(1)=0$ .” Right: solve $2x-1=0$ , then use $g(1/2)=0$ .

The explanation below focuses only on why root substitution detects a factor.

Factorise cubics using one root

One root reduces a cubic problem to a quadratic problem. Test sensible small integer values, use the factor theorem for the first factor, divide it out, then factorise the quotient. If rational roots are expected, a fractional candidate may be needed when the leading coefficient is not $1$ .

Worked example 7: a repeated factor

Question. Factorise $x^3+3x^2-4$ fully.

Plan. Test $x=1$ , divide out the resulting factor and factorise the quadratic.

\begin{aligned} f(1)&=1+3-4=0 &&\text{find a root}\\ x^3+3x^2-4&=(x-1)(x^2+4x+4) &&\text{divide by }x-1\\ &=(x-1)(x+2)^2 &&\text{factorise the quotient} \end{aligned}

Why this step? The first factor lowers the degree; you no longer need to guess three brackets.

Answer. $(x-1)(x+2)^2$ .

Check. Expansion gives $x^3+4x^2+4x-x^2-4x-4=x^3+3x^2-4$ .

Change only the constant from $-4$ to $-20$ . A different integer root now leaves a quadratic that does not split over the reals.

Worked example 8: an irreducible quadratic factor

Question. Factorise $x^3+3x^2-20$ over the reals.

Plan. Test $x=2$ , divide out $x-2$ , then inspect the quotient's discriminant.

\begin{aligned} f(2)&=8+12-20=0 &&\text{find a root}\\ x^3+3x^2-20&=(x-2)(x^2+5x+10) &&\text{linear division}\\ D_{\text{quotient}}&=25-40=-15<0 &&\text{no further real roots} \end{aligned}

Why this step? A quadratic factor with negative discriminant has no real linear factors, so factorisation over the reals stops there.

Answer. $(x-2)(x^2+5x+10)$ .

Check. Expanding gives $x^3+5x^2+10x-2x^2-10x-20=x^3+3x^2-20$ .

For a cubic with rational factors, the opening polynomial gives a particularly useful complete calculation. Testing $-1$ gives $-6+11+1-6=0$ . Division gives $6x^2+5x-6$ , and splitting its middle term gives $6x^2+9x-4x-6=(2x+3)(3x-2)$ . Thus

6x^3+11x^2-x-6=(x+1)(2x+3)(3x-2)

The three roots are $-1$ , $-3/2$ and $2/3$ . Multiplying the factors back gives $6x^3+11x^2-x-6$ , so the opening challenge is resolved. If a cubic has an obvious common scalar or variable factor, extract and retain it first: $2x^3-10x^2+12x=2x(x-2)(x-3)$ . Giving roots alone does not answer a factorisation question.

Exam tip

“Factorise fully” requires the quadratic factor to be considered too. If it splits over the requested number system, finish; if it does not, an irreducible quadratic is a valid stopping point.

Try the find-a-root, divide, factorise sequence with a new cubic.

Simplify rational expressions safely

A rational expression is a fraction of polynomials. Factorise before cancelling, and record the original denominator's zeros before anything disappears. Cancellation removes a common factor from the written expression; it cannot restore an input that was originally undefined.

Worked example 9: cancellation with exclusions

Question. Simplify $\dfrac{x^2+5x+6}{x^2-x-6}$ , stating excluded inputs.

Plan. Factorise both polynomials, record restrictions and cancel the shared factor.

\begin{aligned} \frac{x^2+5x+6}{x^2-x-6}&=\frac{(x+2)(x+3)}{(x+2)(x-3)} &&\text{factorise}\\ &=\frac{x+3}{x-3},\qquad x\ne-2,\ 3 &&\text{cancel for allowed inputs} \end{aligned}

Why this step? At $x=-2$ , the original expression is undefined even though the shortened fraction has a value.

Answer. $(x+3)/(x-3)$ , with $x\ne-2$ , $x\ne3$ .

Check. At $x=1$ , both forms give $-2$ .

Change only the sign of the numerator's linear coefficient. Refactorise it; do not assume the old cancellation survives.

Worked example 10: a different common factor

Question. Simplify $\dfrac{x^2-5x+6}{x^2-x-6}$ , stating excluded inputs.

Plan. Factorise the new numerator and retain the original denominator's restrictions.

\begin{aligned} \frac{x^2-5x+6}{x^2-x-6}&=\frac{(x-2)(x-3)}{(x+2)(x-3)} &&\text{factorise}\\ &=\frac{x-2}{x+2},\qquad x\ne-2,\ 3 &&\text{cancel the common factor} \end{aligned}

Why this step? Restrictions come from the original denominator, regardless of which factor cancels.

Answer. $(x-2)/(x+2)$ , with $x\ne-2$ , $x\ne3$ .

Check. At $x=1$ , both forms give $-1/3$ .

When the numerator's degree is at least the denominator's, linear division may give a more useful form. This is just the earlier division identity divided by its non-zero divisor.

Worked example 11: division with a remainder

Question. Rewrite $\dfrac{x^2+5x+7}{x+2}$ as a polynomial plus a proper fraction.

Plan. Divide by $x+2$ , then separate the remainder term.

\begin{aligned} x^2+5x+7&=(x+2)(x+3)+1 &&\text{linear division}\\ \frac{x^2+5x+7}{x+2}&=x+3+\frac1{x+2},\qquad x\ne-2 &&\text{divide the identity} \end{aligned}

Why this step? The remainder is still divided by the original divisor; it is not a separate constant.

Answer. $x+3+1/(x+2)$ , with $x\ne-2$ .

Check. At $x=0$ , both forms give $7/2$ .

Change only the constant in the numerator from $7$ to $8$ . The quotient stays, while the remainder increases by $1$ .

Worked example 12: a changed remainder

Question. Rewrite $\dfrac{x^2+5x+8}{x+2}$ in the same form.

Plan. Keep the quotient and recompute the constant remainder.

\begin{aligned} x^2+5x+8&=(x+2)(x+3)+2 &&\text{multiply back to find remainder}\\ \frac{x^2+5x+8}{x+2}&=x+3+\frac2{x+2},\qquad x\ne-2 &&\text{divide} \end{aligned}

Why this step? Only the part left after the quotient product belongs in the numerator of the remainder fraction.

Answer. $x+3+2/(x+2)$ , with $x\ne-2$ .

Check. At $x=0$ , both forms give $4$ .

There is a useful surprise in Worked example 9: the shortened fraction gives $-1/5$ at $x=-2$ , yet the original has no value there. Two formulas can agree at every allowed input while having different natural domains. State the restriction to make their equivalence precise. These rational-expression methods are not supplied in the formula booklet.

Watch out

Wrong: cancelling the $x$ -terms in $(x+2)/(x+5)$ . Cancellation applies to multiplying factors, not terms joined by addition. Factorise first; if no factor is shared, nothing cancels.

Finish by simplifying a cubic-over-quadratic expression. You can obtain the cubic factor by division by $x-2$ .

Quick Recap

Look for common factors before expanding, and distribute every minus sign.
Division by $ax+b$ gives a polynomial quotient and a constant remainder.
The factor theorem turns a root into a linear factor; divide it out and inspect the remaining quadratic.
Cancel factors in rational expressions and preserve every original denominator exclusion.

The opening cubic is $(x+1)(2x+3)(3x-2)$ : find one root first, then solve the smaller factorisation problem.

Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 2.6 - Algebraic manipulation of polynomials

See the structure before expanding

Divide by a linear expression

Use the factor theorem

Factorise cubics using one root

Simplify rational expressions safely

Quick Recap

Learn faster with Nova