Algebraic Methods

What you'll learn

Use the factor theorem to spot factors of cubic polynomials.
Divide a cubic by a known linear factor and factorise completely.
Solve cubic equations and explain how many real roots they have.
Write clear algebraic proofs and disproofs using counterexamples.

1. Polynomials, factors and roots

A lot of Algebraic Methods is about recognising structure. Before you start factorising, make sure you are comfortable substituting values into expressions and collecting like terms carefully.

Definition

Polynomial language

A polynomial in $x$ is an expression made from powers of $x$ , such as $3x^3-2x+5$ .
A coefficient is the number multiplying a power of $x$ , such as 3 in $3x^3$ .
A root, or zero, of $f (x)$ is a value of $x$ that makes $f (x) = 0$ .
A factor is an expression that divides exactly into another expression, with no remainder.

Example

Finding unknown coefficients from two values

Suppose $f(x)=x^3+4x^2+px+q$ , with $f (2) = 0$ and $f (- 3) = 10$ . Find $p$ and $q$ .

Substitute $x = 2$ and simplify:
$\Rightarrow 2p+q=-24$
Substitute $x = - 3$ and simplify:
$\Rightarrow -3p+q=1$
Subtract the second equation from the first:
$\Rightarrow 5p=-25$
So $p = - 5$ . Substitute this into $2 p + q = - 24$ to get $q = - 14$ .

Common Mistake

Substitution signs

A negative input affects every power of $x$ : if $x = - 3$ , then $x^2=9$ but $x^3=-27$ .

2. The factor theorem

The factor theorem is the main shortcut for cubic factorisation.

Key Idea

The factor theorem

For a polynomial $f (x)$ , $(x - a)$ is a factor exactly when $f (a) = 0$ . For a factor such as $(2 x + 3)$ , test $x=−32x=-\frac{3}{2}$ .

A simple flow diagram showing how a linear factor corresponds to the root you substitute into the polynomial.

Once you know one linear factor, you can divide the cubic by it. At AS level, comparing coefficients is often cleaner than long division.

Example

Factorising a cubic completely

Let $f(x)=2x^3-5x^2-4x+12$ . Show that $(x - 2)$ is a factor, then factorise $f (x)$ completely.

Test $x = 2$ :
$f(2)=2(2)^3-5(2)^2-4(2)+12=16-20-8+12=0$
Since $f (2) = 0$ , the factor theorem tells us that $(x - 2)$ is a factor.
Write the remaining quadratic with unknown coefficients:

A coefficient-comparison layout showing how the expanded product matches the original cubic term by term.

$$
f(x)=(x-2)(Ax^2+Bx+C)
$$

4. Expand the right-hand side:

$$
(x-2)(Ax^2+Bx+C)=Ax^3+(B-2A)x^2+(C-2B)x-2C
$$

5. Compare coefficients with $2x^3-5x^2-4x+12$ : $A = 2$ , $B = - 1$ and $C = - 6$ .

Factorise the quadratic factor:
$f(x)=(x-2)(2x^2-x-6)=(x-2)^2(2x+3)$

Tip

Testing possible roots

For an integer cubic, try small values first. Possible rational roots often come from factors of the constant term divided by factors of the leading coefficient.

Common Mistake

Opposite sign

If the factor is $(x + 3)$ , you test $f (- 3)$ , not $f (3)$ . The sign changes because $x + 3 = 0$ gives $x = - 3$ .

Unknown coefficients from factors

If you are told that expressions are factors, turn each one into a root using the factor theorem. This gives simultaneous equations.

Example

Using two factors to find constants

The polynomial $f(x)=px^3-5x^2-9x+q$ has factors $(x + 2)$ and $(x - 3)$ . Find $p$ and $q$ .

Since $(x + 2)$ is a factor, $f (- 2) = 0$ :
$\Rightarrow q=8p+2$
Since $(x - 3)$ is a factor, $f (3) = 0$ :
$\Rightarrow q=72-27p$
Equate the two expressions for $q$ :
$\Rightarrow 35p=70 \Rightarrow p=2$
Substitute $p = 2$ into $q = 8 p + 2$ , giving $q = 18$ .

3. Solving cubics and counting real roots

After factorising, solve each factor equal to zero. A distinct real root is a different real solution; a repeated factor counts as the same root repeated.

Definition

Discriminant

For a quadratic $ax^2+bx+c$ , the discriminant is $b^2-4ac$ .

A visual summary of how the sign of the discriminant changes the number of real roots of a quadratic.

If $b^2-4ac>0$ , there are two distinct real roots.
If $b^2-4ac=0$ , there is one repeated real root.
If $b^2-4ac<0$ , there are no real roots.

A graph helps you connect algebraic factors with roots: a simple root crosses the x-axis, while a repeated root just touches it and turns around.

A cubic graph illustrating the difference between a simple root that crosses the x-axis and a repeated root that touches and turns.

Cubic graph showing a simple root crossing the x-axis and a repeated root touching the x-axis

Example

Solving a cubic after one root is known

Let $r(x)=x^3+6x^2+4x-15$ . Show that $x = - 3$ is a solution, then find the other solutions to two decimal places.

Substitute $x = - 3$ :
$r (- 3) = - 27 + 54 - 12 - 15 = 0$
So $(x + 3)$ is a factor. Dividing gives:
$r(x)=(x+3)(x^2+3x-5)$
Solve the quadratic factor using the quadratic formula:
$x=−3±32−4(1)(−5)2=−3±292x=\frac{-3\pm\sqrt{3^2-4(1)(-5)}}{2}=\frac{-3\pm\sqrt{29}}{2}$
The solutions are $x = - 3$ , $x≈−4.19x\approx -4.19$ and $x≈1.19x\approx 1.19$ .

Example

Showing there is only one real root

Let $s(x)=x^3-3x^2+7x-5$ . Show that $s (x) = 0$ has only one real root.

Test $x = 1$ :
$s (1) = 1 - 3 + 7 - 5 = 0$
Therefore $(x - 1)$ is a factor, and division gives:

A factor tree for the cubic showing that the linear factor gives one real root while the quadratic factor contributes no real roots.

$$
s(x)=(x-1)(x^2-2x+5)
$$

3. Check the discriminant of the quadratic factor:

$$
(-2)^2-4(1)(5)=4-20=-16
$$

4. Since the discriminant is negative, the quadratic has no real roots. The only real root is $x = 1$ .

4. Algebraic proof and counterexamples

A proof is a logical argument showing that a statement is true in every allowed case. A counterexample is one allowed case that makes a statement false.

For parity proofs, use the facts that an even integer can be written as $2 n$ and an odd integer can be written as $2 n + 1$ , where $n$ is an integer.

Example

Writing odd numbers algebraically

Show that adding two neighbouring odd integers always gives a multiple of 4.

A number-line diagram showing two neighbouring odd integers written as 2n+1 and 2n+3.

Let the smaller odd integer be $2 n + 1$ . The next odd integer is $2 n + 3$ .
Add them:
$(2 n + 1) + (2 n + 3) = 4 n + 4 = 4 (n + 1)$
Since $n + 1$ is an integer, $4 (n + 1)$ is a multiple of 4.

Example

Disproving an always statement

Disprove the claim: $n^2-n+3$ is prime for every positive integer $n$ .

Try a small positive integer, such as $n = 3$ .
Substitute it into the expression:
$3^2-3+3=9$
Since $9=3⋅39=3\cdot 3$ , 9 is not prime. So the statement is false.

Example

Proof by odd and even cases

Prove that $n^2+1$ is never a multiple of 4, where $n$ is a positive integer.

If $n$ is even, write $n = 2 k$ :
$n^2+1=(2k)^2+1=4k^2+1$
This is one more than a multiple of 4, so it is not a multiple of 4.
If $n$ is odd, write $n = 2 k + 1$ :
$n^2+1=(2k+1)^2+1=4k^2+4k+2=4(k^2+k)+2$
This is two more than a multiple of 4, so it is not a multiple of 4. Therefore the statement is true in all cases.

5. Proving inequalities

To prove something is positive for all real values, try completing the square. Completing the square means rewriting a quadratic as a squared bracket plus or minus a constant.

A graph of a completed-square quadratic showing why adding a positive constant makes it positive for all real x.

Example

Proving a quadratic inequality

Prove that $x^2+6x>2x-7$ for all real values of $x$ .

Move everything to the left-hand side:
$x^2+6x-(2x-7)=x^2+4x+7$
Complete the square:
$x^2+4x+7=(x+2)^2+3$
Since $(x+2)2≥0(x+2)^2\ge 0$ , we know $x+2)^2+3>0$ for all real $x$ .
Therefore $x^2+6x>2x-7$ is always true.

Common Mistake

Dividing by variables

Only divide an inequality by a variable expression when you know its sign. If $a > 0$ and $b > 0$ , then $ab > 0$ , so dividing by $ab$ does not reverse the inequality sign.

Exam technique

In the exam

For cubics, test likely roots with the factor theorem and show the substitution clearly before dividing.
After factorising, solve every factor and use the discriminant if you need to justify the number of real roots.
For proof, choose the method first: algebraic form for even and odd numbers, a counterexample for false “always” claims, or completing the square for positivity.

Self review

Check yourself

If $(x + 4)$ is a factor of $f (x)$ , what value of $x$ must you substitute?
How can a quadratic factor prove that a cubic has no more real roots?
What single counterexample would disprove “all prime numbers have odd squares”?

Algebraic Methods

What you'll learn

Use the factor theorem to spot factors of cubic polynomials.
Divide a cubic by a known linear factor and factorise completely.
Solve cubic equations and explain how many real roots they have.
Write clear algebraic proofs and disproofs using counterexamples.

1. Polynomials, factors and roots

A lot of Algebraic Methods is about recognising structure. Before you start factorising, make sure you are comfortable substituting values into expressions and collecting like terms carefully.

Definition

Polynomial language

A polynomial in $x$ is an expression made from powers of $x$ , such as $3x^3-2x+5$ .
A coefficient is the number multiplying a power of $x$ , such as 3 in $3x^3$ .
A root, or zero, of $f (x)$ is a value of $x$ that makes $f (x) = 0$ .
A factor is an expression that divides exactly into another expression, with no remainder.

Example

Finding unknown coefficients from two values

Suppose $f(x)=x^3+4x^2+px+q$ , with $f (2) = 0$ and $f (- 3) = 10$ . Find $p$ and $q$ .

Substitute $x = 2$ and simplify:
$\Rightarrow 2p+q=-24$
Substitute $x = - 3$ and simplify:
$\Rightarrow -3p+q=1$
Subtract the second equation from the first:
$\Rightarrow 5p=-25$
So $p = - 5$ . Substitute this into $2 p + q = - 24$ to get $q = - 14$ .

Common Mistake

Substitution signs

A negative input affects every power of $x$ : if $x = - 3$ , then $x^2=9$ but $x^3=-27$ .

2. The factor theorem

The factor theorem is the main shortcut for cubic factorisation.

Key Idea

The factor theorem

For a polynomial $f (x)$ , $(x - a)$ is a factor exactly when $f (a) = 0$ . For a factor such as $(2 x + 3)$ , test $x=−32x=-\frac{3}{2}$ .

A simple flow diagram showing how a linear factor corresponds to the root you substitute into the polynomial.

Once you know one linear factor, you can divide the cubic by it. At AS level, comparing coefficients is often cleaner than long division.

Example

Factorising a cubic completely

Let $f(x)=2x^3-5x^2-4x+12$ . Show that $(x - 2)$ is a factor, then factorise $f (x)$ completely.

Test $x = 2$ :
$f(2)=2(2)^3-5(2)^2-4(2)+12=16-20-8+12=0$
Since $f (2) = 0$ , the factor theorem tells us that $(x - 2)$ is a factor.
Write the remaining quadratic with unknown coefficients:

A coefficient-comparison layout showing how the expanded product matches the original cubic term by term.

$$
f(x)=(x-2)(Ax^2+Bx+C)
$$

4. Expand the right-hand side:

$$
(x-2)(Ax^2+Bx+C)=Ax^3+(B-2A)x^2+(C-2B)x-2C
$$

5. Compare coefficients with $2x^3-5x^2-4x+12$ : $A = 2$ , $B = - 1$ and $C = - 6$ .

Factorise the quadratic factor:
$f(x)=(x-2)(2x^2-x-6)=(x-2)^2(2x+3)$

Tip

Testing possible roots

For an integer cubic, try small values first. Possible rational roots often come from factors of the constant term divided by factors of the leading coefficient.

Common Mistake

Opposite sign

If the factor is $(x + 3)$ , you test $f (- 3)$ , not $f (3)$ . The sign changes because $x + 3 = 0$ gives $x = - 3$ .

Unknown coefficients from factors

If you are told that expressions are factors, turn each one into a root using the factor theorem. This gives simultaneous equations.

Example

Using two factors to find constants

The polynomial $f(x)=px^3-5x^2-9x+q$ has factors $(x + 2)$ and $(x - 3)$ . Find $p$ and $q$ .

Since $(x + 2)$ is a factor, $f (- 2) = 0$ :
$\Rightarrow q=8p+2$
Since $(x - 3)$ is a factor, $f (3) = 0$ :
$\Rightarrow q=72-27p$
Equate the two expressions for $q$ :
$\Rightarrow 35p=70 \Rightarrow p=2$
Substitute $p = 2$ into $q = 8 p + 2$ , giving $q = 18$ .

3. Solving cubics and counting real roots

After factorising, solve each factor equal to zero. A distinct real root is a different real solution; a repeated factor counts as the same root repeated.

Definition

Discriminant

For a quadratic $ax^2+bx+c$ , the discriminant is $b^2-4ac$ .

A visual summary of how the sign of the discriminant changes the number of real roots of a quadratic.

If $b^2-4ac>0$ , there are two distinct real roots.
If $b^2-4ac=0$ , there is one repeated real root.
If $b^2-4ac<0$ , there are no real roots.

A graph helps you connect algebraic factors with roots: a simple root crosses the x-axis, while a repeated root just touches it and turns around.

A cubic graph illustrating the difference between a simple root that crosses the x-axis and a repeated root that touches and turns.

Cubic graph showing a simple root crossing the x-axis and a repeated root touching the x-axis

Example

Solving a cubic after one root is known

Let $r(x)=x^3+6x^2+4x-15$ . Show that $x = - 3$ is a solution, then find the other solutions to two decimal places.

Substitute $x = - 3$ :
$r (- 3) = - 27 + 54 - 12 - 15 = 0$
So $(x + 3)$ is a factor. Dividing gives:
$r(x)=(x+3)(x^2+3x-5)$
Solve the quadratic factor using the quadratic formula:
$x=−3±32−4(1)(−5)2=−3±292x=\frac{-3\pm\sqrt{3^2-4(1)(-5)}}{2}=\frac{-3\pm\sqrt{29}}{2}$
The solutions are $x = - 3$ , $x≈−4.19x\approx -4.19$ and $x≈1.19x\approx 1.19$ .

Example

Showing there is only one real root

Let $s(x)=x^3-3x^2+7x-5$ . Show that $s (x) = 0$ has only one real root.

Test $x = 1$ :
$s (1) = 1 - 3 + 7 - 5 = 0$
Therefore $(x - 1)$ is a factor, and division gives:

A factor tree for the cubic showing that the linear factor gives one real root while the quadratic factor contributes no real roots.

$$
s(x)=(x-1)(x^2-2x+5)
$$

3. Check the discriminant of the quadratic factor:

$$
(-2)^2-4(1)(5)=4-20=-16
$$

4. Since the discriminant is negative, the quadratic has no real roots. The only real root is $x = 1$ .

4. Algebraic proof and counterexamples

A proof is a logical argument showing that a statement is true in every allowed case. A counterexample is one allowed case that makes a statement false.

For parity proofs, use the facts that an even integer can be written as $2 n$ and an odd integer can be written as $2 n + 1$ , where $n$ is an integer.

Example

Writing odd numbers algebraically

Show that adding two neighbouring odd integers always gives a multiple of 4.

A number-line diagram showing two neighbouring odd integers written as 2n+1 and 2n+3.

Let the smaller odd integer be $2 n + 1$ . The next odd integer is $2 n + 3$ .
Add them:
$(2 n + 1) + (2 n + 3) = 4 n + 4 = 4 (n + 1)$
Since $n + 1$ is an integer, $4 (n + 1)$ is a multiple of 4.

Example

Disproving an always statement

Disprove the claim: $n^2-n+3$ is prime for every positive integer $n$ .

Try a small positive integer, such as $n = 3$ .
Substitute it into the expression:
$3^2-3+3=9$
Since $9=3⋅39=3\cdot 3$ , 9 is not prime. So the statement is false.

Example

Proof by odd and even cases

Prove that $n^2+1$ is never a multiple of 4, where $n$ is a positive integer.

If $n$ is even, write $n = 2 k$ :
$n^2+1=(2k)^2+1=4k^2+1$
This is one more than a multiple of 4, so it is not a multiple of 4.
If $n$ is odd, write $n = 2 k + 1$ :
$n^2+1=(2k+1)^2+1=4k^2+4k+2=4(k^2+k)+2$
This is two more than a multiple of 4, so it is not a multiple of 4. Therefore the statement is true in all cases.

5. Proving inequalities

To prove something is positive for all real values, try completing the square. Completing the square means rewriting a quadratic as a squared bracket plus or minus a constant.

A graph of a completed-square quadratic showing why adding a positive constant makes it positive for all real x.

Example

Proving a quadratic inequality

Prove that $x^2+6x>2x-7$ for all real values of $x$ .

Move everything to the left-hand side:
$x^2+6x-(2x-7)=x^2+4x+7$
Complete the square:
$x^2+4x+7=(x+2)^2+3$
Since $(x+2)2≥0(x+2)^2\ge 0$ , we know $x+2)^2+3>0$ for all real $x$ .
Therefore $x^2+6x>2x-7$ is always true.

Common Mistake

Dividing by variables

Only divide an inequality by a variable expression when you know its sign. If $a > 0$ and $b > 0$ , then $ab > 0$ , so dividing by $ab$ does not reverse the inequality sign.

Exam technique

In the exam

For cubics, test likely roots with the factor theorem and show the substitution clearly before dividing.
After factorising, solve every factor and use the discriminant if you need to justify the number of real roots.
For proof, choose the method first: algebraic form for even and odd numbers, a counterexample for false “always” claims, or completing the square for positivity.

Self review

Check yourself

If $(x + 4)$ is a factor of $f (x)$ , what value of $x$ must you substitute?
How can a quadratic factor prove that a cubic has no more real roots?
What single counterexample would disprove “all prime numbers have odd squares”?

Algebraic Methods

What you'll learn

1. Polynomials, factors and roots

2. The factor theorem

Unknown coefficients from factors

3. Solving cubics and counting real roots

4. Algebraic proof and counterexamples

5. Proving inequalities

Recap questions

You've reached the end

Algebraic Methods

What you'll learn

1. Polynomials, factors and roots

2. The factor theorem

Unknown coefficients from factors

3. Solving cubics and counting real roots

4. Algebraic proof and counterexamples

5. Proving inequalities

Recap questions

You've reached the end

Algebraic Methods

What you'll learn

1. Polynomials, factors and roots

2. The factor theorem

Unknown coefficients from factors

3. Solving cubics and counting real roots

4. Algebraic proof and counterexamples

5. Proving inequalities

Recap questions

You've reached the end

Algebraic Methods Revision Guide

Algebraic Methods

What you'll learn

1. Polynomials, factors and roots

2. The factor theorem

Unknown coefficients from factors

3. Solving cubics and counting real roots

4. Algebraic proof and counterexamples

5. Proving inequalities

Recap questions

You've reached the end

Algebraic Methods Revision Guide