Polynomials, factorisation and algebraic division

A polynomial is an expression formed from constants and non-negative integer powers of a variable. Polynomial algebra lets us rewrite such expressions without changing their value. The most useful forms reveal factors, roots, intercepts or long-term behaviour.

Before you start

You should be comfortable with:

• collecting like terms
• the laws of indices
• expanding two brackets
• solving linear equations
• substituting negative and fractional values carefully

Review algebraic fluency if these skills are uncertain. Quadratic-specific methods are developed in quadratics, while decomposing rational expressions is covered in partial fractions.

Polynomial vocabulary

In one variable, a polynomial has the form

$$P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0,$$

where $n$ is a non-negative integer and the coefficients $a_0,a_1,\ldots,a_n$ are constants. If $a_n\ne0$, the degree is $n$ and $a_n$ is the leading coefficient.

For example,

$P(x)=4x^5-3x^2+7$

has degree $5$, leading coefficient $4$ and constant term $7$. Its missing $x^4$, $x^3$ and $x$ terms have coefficient zero:

$P(x)=4x^5+0x^4+0x^3-3x^2+0x+7.$

Writing missing terms explicitly is often helpful during algebraic division.

Expressions such as $x^{-1}+2$, $\sqrt{x}+1$ and $3^x$ are not polynomials in $x$. Their powers are respectively negative, fractional and variable.

Adding and subtracting polynomials

Only like terms can be combined. When subtracting a polynomial, the negative sign changes every term inside its brackets.

Example 1: subtraction

Let

$P(x)=3x^3-2x^2+5x-4,qquad Q(x)=x^3+4x^2-7.$

Then

$$\begin{aligned} P(x)-Q(x) &=(3x^3-2x^2+5x-4)-(x^3+4x^2-7)\\ &=3x^3-2x^2+5x-4-x^3-4x^2+7\\ &=2x^3-6x^2+5x+3. \end{aligned}$$

The absent $x$ term in $Q(x)$ contributes zero, not another power of $x$.

Expanding products

Expansion uses the distributive law. Every term in one factor must multiply every term in the other factor.

Example 2: a linear factor times a quadratic

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

$$\begin{aligned} (2x-3)(x^2+4x-5) &=2x(x^2+4x-5)-3(x^2+4x-5)\\ &=2x^3+8x^2-10x-3x^2-12x+15\\ &=2x^3+5x^2-22x+15. \end{aligned}$$

A quick check is to compare the leading and constant terms. They must be

$2x\cdot x^2=2x^3,qquad (-3)(-5)=15.$

Example 3: three linear factors

Expand $(x-2)(x+1)(x+4)$.

Multiply two factors first:

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

Then multiply by the remaining factor:

$$\begin{aligned} (x^2-x-2)(x+4) &=x^3+4x^2-x^2-4x-2x-8\\ &=x^3+3x^2-6x-8. \end{aligned}$$

The order does not change the answer, so choose a pair that keeps the arithmetic simple.

Useful identities

Several expansions are worth recognising:

$$\begin{aligned} (a+b)^2&=a^2+2ab+b^2,\\ (a-b)^2&=a^2-2ab+b^2,\\ (a+b)(a-b)&=a^2-b^2,\\ (a+b)^3&=a^3+3a^2b+3ab^2+b^3,\\ (a-b)^3&=a^3-3a^2b+3ab^2-b^3. \end{aligned}$$

For instance,

$$(2x-3)^3=8x^3-36x^2+54x-27.$$

The middle terms do not vanish. Cubing each term separately is incorrect.

Factorising polynomials

Factorisation reverses expansion. Always look for a common factor first, then inspect the remaining expression.

Example 4: common factors and difference of two squares

Factorise $12x^3-75x$ fully.

First take out the highest common factor:

$12x^3-75x=3x(4x^2-25).$

The bracket is a difference of squares:

$4x^2-25=(2x)^2-5^2=(2x-5)(2x+5).$

Therefore

$12x^3-75x=3x(2x-5)(2x+5).$

Example 5: factorisation by grouping

Factorise $x^3+2x^2-9x-18$.

Group pairs whose remaining factors match:

$$\begin{aligned} x^3+2x^2-9x-18 &=x^2(x+2)-9(x+2)\\ &=(x+2)(x^2-9)\\ &=(x+2)(x-3)(x+3). \end{aligned}$$

This method is useful when the four terms naturally form two pairs. It is not guaranteed to work for every cubic.

Factors and roots

If

$P(x)=(x-a)Q(x),$

then $x-a$ is a factor. Setting $P(x)=0$ gives

$x-a=0quad\text{or}quad Q(x)=0,$

so $x=a$ is a root. Notice the sign:

• the factor $x-4$ gives the root $4$
• the factor $x+4$ gives the root $-4$
• the factor $2x-3$ gives the root $\frac32$

Repeated factors produce repeated roots. For example,

$P(x)=(x-2)^2(x+5)$

has a repeated root $x=2$ and a root $x=-5$.

Algebraic division by a linear expression

Polynomial division has the same structure as integer division:

$$\text{dividend}=(\text{divisor})(\text{quotient})+\text{remainder}.$$

When a polynomial $P(x)$ is divided by a non-zero linear expression, the quotient has degree one less than $P(x)$ and the remainder is a constant.

Example 6: division with no remainder

Divide $2x^3+3x^2-11x-6$ by $x+3$.

The leading term $2x^3$ divided by $x$ gives $2x^2$. Multiply back and subtract:

$$\begin{array}{r|rrrr} &2x^2&-3x&-2\\ x+3&2x^3&+3x^2&-11x&-6\\ &2x^3&+6x^2&&\\ \cline{2-5} &&-3x^2&-11x&\\ &&-3x^2&-9x&\\ \cline{3-5} &&&-2x&-6\\ &&&-2x&-6\\ \cline{4-5} &&&&0 \end{array}$$

Hence

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

The quadratic factor can be factorised further:

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

Therefore

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

Example 7: division with a remainder

Divide $x^3-4x+7$ by $x-2$.

Include the missing square term:

$x^3+0x^2-4x+7.$

Long division gives quotient $x^2+2x$ and remainder $7$, so

$$x^3-4x+7=(x-2)(x^2+2x)+7.$$

Check by expanding:

$$(x-2)(x^2+2x)+7=x^3+2x^2-2x^2-4x+7=x^3-4x+7.$$

Because the remainder is not zero, $x-2$ is not a factor.

Dividing by $ax-b$

Do not treat $2x-1$ as if it were $x-1$. The leading term of the divisor matters.

For example, divide $6x^3+x^2-4x+1$ by $2x-1$.

$$\begin{aligned} 6x^3\div2x&=3x^2,\\ (6x^3+x^2)-(6x^3-3x^2)&=4x^2,\\ 4x^2\div2x&=2x,\\ (4x^2-4x)-(4x^2-2x)&=-2x,\\ -2x\div2x&=-1. \end{aligned}$$

The subtraction leaves no remainder, so

$$6x^3+x^2-4x+1=(2x-1)(3x^2+2x-1).$$

The factor theorem

The factor theorem states:

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

More generally,

$$ax-b\text{ is a factor of }P(x)\quad\Longleftrightarrow\quad P\left(\frac ba\right)=0.$$

This works because division by $x-a$ gives

$P(x)=(x-a)Q(x)+r.$

Substituting $x=a$ produces $P(a)=r$. This is also the remainder theorem: the remainder on division by $x-a$ is $P(a)$.

Example 8: test a proposed factor

Show that $x-2$ is a factor of

$P(x)=x^3+x^2-8x-4.$

Evaluate $P(2)$:

$$P(2)=2^3+2^2-8(2)-4=8+4-16-4=-8.$$

Since $P(2)\ne0$, the claim is false: $x-2$ is not a factor. This example matters because a request to “show” something does not make it true if the polynomial has been copied incorrectly. Trust the mathematics and check the signs.

However,

$$P(-2)=(-2)^3+(-2)^2-8(-2)-4=-8+4+16-4=8,$$

so $x+2$ is not a factor either.

Example 9: factorise a cubic completely

Factorise

$P(x)=2x^3-5x^2-4x+3,$

given that $x-3$ is a factor.

First verify the information:

$P(3)=54-45-12+3=0.$

Divide by $x-3$:

$$P(x)=(x-3)(2x^2+x-1).$$

Then factorise the quadratic:

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

Thus

$$P(x)=(x-3)(2x-1)(x+1),$$

and the roots are

$x=3,qquad x=\frac12,qquad x=-1.$

Example 10: find an unknown coefficient

The polynomial

$P(x)=2x^3+kx^2-13x+6$

has factor $x-2$. Find $k$.

The factor theorem gives $P(2)=0$:

$$2(2)^3+k(2)^2-13(2)+6=0.$$

Therefore

$$16+4k-26+6=0,$$

so

$4k-4=0quad\Longrightarrowquad k=1.$

Example 11: two factor conditions

The polynomial

$P(x)=x^3+ax^2+bx+6$

has factors $x-1$ and $x+2$. Find $a$ and $b$.

From $x-1$:

$P(1)=1+a+b+6=0,$

so

a+b=-7.\tag{1}

From $x+2$:

$P(-2)=-8+4a-2b+6=0,$

so

2a-b=1.\tag{2}

Adding equations $(1)$ and $(2)$ gives

$3a=-6,$

hence $a=-2$ and then $b=-5$. Therefore

$P(x)=x^3-2x^2-5x+6.$

Since two linear factors are known, its final factor can be found by division:

$P(x)=(x-1)(x+2)(x-3).$

Finding possible rational roots

If an integer-coefficient polynomial has a rational root $\frac pq$ in lowest terms, then $p$ divides the constant term and $q$ divides the leading coefficient. This result narrows the candidates but does not prove that any candidate is a root.

For

$P(x)=2x^3-3x^2-8x+12,$

possible rational roots include

$$\pm1,\ \pm2,\ \pm3,\ \pm4,\ \pm6,\ \pm12, \quad \pm\frac12,\ \pm\frac32.$$

Testing simple values first,

$P(2)=16-12-16+12=0.$

Thus $x-2$ is a factor. Division gives

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

At A-level, a question may give one factor or expect you to spot a small root. Systematic testing is safer than guessing randomly.

Polynomial identities

An equation is true only for particular values of its variable. An identity is true for every permitted value and is written with $\equiv$.

For example,

$$(x+2)(ax^2+bx+c)\equiv 3x^3+4x^2-5x+2.$$

Expanding the left side gives

$$ax^3+(b+2a)x^2+(c+2b)x+2c.$$

Equal polynomials have equal coefficients of every power, so

$$a=3,qquad b+2a=4,qquad c+2b=-5,qquad 2c=2.$$

These give $a=3$, $b=-2$ and $c=1$, which satisfy all four equations. Comparing coefficients is especially useful when an identity contains unknown constants.

Common mistakes

Losing signs during subtraction

In

$P(x)-(x^2-3x+4),$

the second polynomial becomes

$-x^2+3x-4.$

Every sign changes.

Omitting zero-coefficient terms

When dividing $x^3-5x+2$, write

$x^3+0x^2-5x+2.$

The placeholder keeps powers aligned.

Testing the wrong value

For factor $x+3$, test $P(-3)$, not $P(3)$. Solve the factor equation $x+3=0$ first if uncertain.

Stopping before full factorisation

The expression

$2x(x^2-9)$

is factorised, but not fully factorised. Continue to

$2x(x-3)(x+3).$

Confusing a factor with a root

$x-5$ is an expression and may be a factor. $5$ is a number and may be a root. They encode the same information but are not interchangeable objects.

Assuming all cubics have three real roots

A cubic has three roots over the complex numbers when multiplicity is counted, but it may have only one real root. Polynomial factorisation over the real numbers can therefore leave an irreducible quadratic factor.

Self-check

Try these without looking back. Expand or factorise fully unless stated otherwise.

1. Expand $(3x+2)(x^2-2x+5)$.
2. Factorise $18x^3-8x$.
3. Find the remainder when $P(x)=2x^3-x^2+4x-7$ is divided by $x+1$.
4. Given that $x-2$ is a factor of $x^3+kx^2-9x+2$, find $k$.
5. Factorise $x^3-2x^2-5x+6$ fully.
6. The polynomial $P(x)=2x^3+ax^2+bx-3$ has factors $x-1$ and $2x+1$. Find $a$ and $b$.

Answers

1. $3x^3-4x^2+11x+10$.
2. $2x(3x-2)(3x+2)$.
3. $P(-1)=-2-1-4-7=-14$.
4. $P(2)=8+4k-18+2=0$, so $k=2$.
5. $(x-1)(x+2)(x-3)$.
6. $P(1)=0$ gives $a+b=1$. Since $2x+1=0$ at $x=-\frac12$, $P(-\frac12)=0$ gives $a-2b=13$. Hence $a=5$ and $b=-4$.

What to learn next

• Study quadratics for quadratic roots, discriminants and graph structure.
• Study functions to connect polynomial formulae with domains, ranges and mappings.
• Study partial fractions to reverse the process of combining rational expressions.
• Revisit this lesson before polynomial differentiation, integration and numerical methods, where correct algebra is essential.