Algebraic Manipulation of Polynomials for the TMUA

Q: Does the Factor Theorem work for polynomials with non-integer roots?

Yes, the theorem applies to any real value $a$. If $f(a) = 0$, then $(x - a)$ is a factor, regardless of whether $a$ is an integer, fraction, or irrational number.

Updated August 2025

Master the essential techniques for manipulating polynomials, including expanding brackets and performing long division. This section covers the Factor and Remainder Theorems, which allow you to factorise complex expressions and determine remainders efficiently. These skills are fundamental for solving higher order equations in both TMUA and ESAT assessments.

Core concept

Polynomial manipulation involves using the distributive law to expand brackets and the Factor Theorem ( $f(a) = 0 \iff (x - a)$ is a factor) or Remainder Theorem (remainder is $f(b)$ when dividing by $(x - b)$ ) to simplify and solve algebraic expressions.

In the TMUA and ESAT, you must be proficient in multiplying out brackets and collecting like terms. Collecting like terms refers to the process of grouping all constants together, all $x$ terms together, all $x^2$ terms together, and so on for every power present in the expression.

Algebraic Division

You are expected to perform algebraic long division by both linear expressions, such as $(ax + b)$ , and quadratic expressions, such as $(ax^2 + bx + c)$ . When dividing, it is a good idea to include placeholders for missing powers of $x$ . For instance, if dividing a quartic that has no $x^3$ term, include $0x^3$ in your working to maintain column alignment and avoid errors.

Worked Example: Polynomial Long Division

Consider $x^4 + 2x^2 + 3x - 4$ divided by $x + 3$ . We write this out with a placeholder for the $x^3$ term:

		$x^3$	$-3x^2$	$+11x$	$-30$
$x + 3$	$x^4$	$+0x^3$	$+2x^2$	$+3x$	$-4$
	$x^4$	$+3x^3$
	$0x^4$	$-3x^3$	$+2x^2$	$+3x$	$-4$
		$-3x^3$	$-9x^2$
		$0x^3$	$+11x^2$	$+3x$	$-4$
			$+11x^2$	$+33x$
			$0x^2$	$-30x$	$-4$
				$-30x$	$-90$
				$0x$	$86$

The division shows that $x^4 + 2x^2 + 3x - 4 = (x^3 - 3x^2 + 11x - 30)(x + 3) + 86$ . The quotient is $x^3 - 3x^2 + 11x - 30$ and the remainder is $86$ . This fits the general division form where $11$ divided by $4$ is $2$ remainder $3$ , written as $11 = 2 \times 4 + 3$ .

The Factor Theorem

A factor in algebra is an expression that divides into another exactly, without any remainder. The Factor Theorem states: if $f(x)$ is a polynomial in $x$ , then $f(a) = 0$ if and only if $(x - a)$ is a factor of $f(x)$ . This is a powerful tool for factorising cubics and higher order polynomials by testing values of $a$ .

Worked Example: Factorising a Quadratic

To factorise $f(x) = x^2 + 3x + 2$ , we seek factors $(x - a)$ and $(x - b)$ where $ab = 2$ . We test integer factors of $2$ , such as $1, -1, 2, -2$ .

Testing $a = -1$ : $f(-1) = (-1)^2 + 3(-1) + 2 = 1 - 3 + 2 = 0$ . Since $f(-1) = 0$ , the Factor Theorem tells us that $(x - (-1))$ , which is $(x + 1)$ , is a factor. By inspection or division, the other factor is $(x + 2)$ .

Worked Example: Factorising a Cubic

Factorise $f(x) = x^3 + x^2 - 5x + 3$ . The constant term is $3$ , so we test factors of $3$ : $1, -1, 3, -3$ .

Testing $x = 1$ : $f(1) = 1^3 + 1^2 - 5(1) + 3 = 0$ . Thus, $(x - 1)$ is a factor.

Performing long division or factorising by inspection: $x^3 + x^2 - 5x + 3 = (x - 1)(x^2 + 2x - 3)$ .

The quadratic $x^2 + 2x - 3$ factorises further to $(x - 1)(x + 3)$ . Thus, the full factorisation is $(x - 1)(x - 1)(x + 3)$ , or $(x - 1)^2(x + 3)$ .

The Remainder Theorem

The Remainder Theorem states that when a polynomial $f(x)$ is divided by $(x - b)$ , the remainder is $f(b)$ . This can be seen by writing the division as $f(x) = g(x)(x - b) + R$ . When $x = b$ is substituted, the term $g(b)(b - b)$ becomes zero, leaving $f(b) = R$ .

This principle extends to divisors of the form $(px - q)$ . In this case, the remainder is $f(q/p)$ . For example, if you divide a polynomial by $(2x + 3)$ , you find the remainder by calculating $f(-3/2)$ .

When dividing by a quadratic $ax^2 + bx + c$ , the remainder will be a linear expression $mx + n$ . In general, the degree of the remainder is always at least one less than the degree of the divisor. If the remainder is zero, the divisor is a factor, showing that the Factor Theorem is a special case of the Remainder Theorem.

Key takeaways

The Factor Theorem states $f(a) = 0$ if and only if $(x - a)$ is a factor of the polynomial.
The Remainder Theorem allows you to calculate the remainder of $f(x) \div (x - b)$ by simply evaluating $f(b)$ .
Algebraic division by a quadratic divisor $ax^2 + bx + c$ results in a remainder that is at most linear ( $mx + n$ ).
When performing long division, always use placeholders like $0x^n$ for missing terms to avoid calculation errors.

Tips

When factorising a polynomial $f(x)$ where the constant term is $k$ , always test the factors of $k$ first. For example, if the constant is $6$ , test $\pm 1, \pm 2, \pm 3, \pm 6$ . This trial and error method is often the fastest way to find your first factor in a TMUA question.

Cautions

A common mistake is using the wrong sign in the Remainder Theorem. If the divisor is $(x + 3)$ , you must evaluate $f(-3)$ , not $f(3)$ . Always remember to solve $(divisor) = 0$ to find the value to substitute into $f(x)$ .

Insight

The relationship between the degree of a polynomial and its remainder is fixed. If you divide a polynomial of degree $n$ by one of degree $m$ , the quotient will have degree $n - m$ and the remainder will have a maximum degree of $m - 1$ .

Worked Examples

Example 1

When

(3x^2 + 8x - 3)

is multiplied by

(px - 1)

and the resulting product is divided by

(x + 1)

, the remainder is

24

.
What is the value of

p

A:-4

B:2

C:4

\frac{8}{7}

\frac{11}{4}

Practice Questions

Practice Question 1

Consider the following attempt to prove this true theorem:

Theorem:

a^3 + b^3 = c^3

has no solutions with

a, b

and

c

positive integers.

Attempted proof:
Suppose that there are positive integers

a, b

and

c

such that

a^3 + b^3 = c^3

.
I We have

a^3 = c^3 – b^3

.
II Hence

a^3 = (c - b)(c^2 + cb + b^2)

.
III It follows that

a = c − b

and

a^2 = c^2 + cb + b^2

, since

a < a^2

and

c - b < c^2 + cb + b^2

.
IV Eliminating

a

, we have

(c - b)^2 = c^2 + cb + b^2

.
V Multiplying out, we have

c^2 – 2cb + b^2 = c^2 + cb + b^2

.
VI Hence

3cb = 0

so one of

b

and

c

is zero.
But this is a contradiction to the original assumption that all of

a, b

and

c

are positive. It follows that the equation has no solutions.

Comment on this proof by choosing one of the following options:

A:The proof is correct

B:The proof is incorrect and the first mistake occurs on line I.

C:The proof is incorrect and the first mistake occurs on line II.

D:The proof is incorrect and the first mistake occurs on line III.

E:The proof is incorrect and the first mistake occurs on line IV.

F:The proof is incorrect and the first mistake occurs on line V.

G:The proof is incorrect and the first mistake occurs on line VI.

Frequently asked questions

Does the Factor Theorem work for polynomials with non-integer roots?

Yes, the theorem applies to any real value $a$ . If $f(a) = 0$ , then $(x - a)$ is a factor, regardless of whether $a$ is an integer, fraction, or irrational number.

What happens if I divide by $(2x - 1)$ instead of $(x - a)$ ?

The Remainder Theorem still applies. The remainder is found by setting the divisor to zero: $2x - 1 = 0 \implies x = 1/2$ . Thus, the remainder is $f(1/2)$ .

Can a cubic polynomial have more than three linear factors?

No. The degree of a polynomial determines the maximum number of linear factors. A cubic (degree 3) can have at most three real linear factors, some of which may be repeated.

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