Algebraic expressions (Grade 9) — Mathematics Grades 8–11

Expanding brackets, multiplying and dividing terms, and factorising expressions.

Gr 9 Terms 1 and 3 — Algebraic expressionsTerm 1 and Term 3

Practise Algebraic expressions in the app →

What gets asked

Expand a bracket and simplify the expression.
Multiply out two binomials, such as (x + 2)(x - 3).
Factorise an expression by taking out a common factor.
Factorise a trinomial of the form x squared plus bx plus c.

You must be able to

Multiply every term inside a bracket by the term outside it.
Multiply two binomials using all four products, not just two.
Spot the highest common factor of every term and take it out.
Find two numbers that multiply to c and add to b to factorise a trinomial.

Traps that cost marks

Multiplying only the first term inside a bracket. Every term inside gets multiplied: $3 (x + 4) = 3 x + 12$ , not $3 x + 4$ .
Multiplying only the first and last terms of two binomials. A binomial product needs all four parts: $(x + 2) (x - 3) = x^{2} - 3 x + 2 x - 6$ .
Taking out only part of the common factor. Check every term for the largest shared factor: $6 a^{4} - 4 a^{2} = 2 a^{2} (3 a^{2} - 2)$ .

Worked example

Simplify:

4 (2 x + 5) - 3 x

Expand the bracket: $8 x + 20$ .
Subtract 3x: $8 x + 20 - 3 x$ .
Collect like terms: $5 x + 20$ .

Terms and substitution

What the parts of an expression are called, and how to work out what one is worth.

term: A part of an expression that is added or subtracted. In $3 x + 5$ the terms are $3 x$ and $5$ .
coefficient: The number in front of the letter in a term. In $7 y$ the coefficient is $7$ .
exponent: The small raised number on a letter or number. For a whole-number exponent it says how many times to multiply by itself: $x^{3}$ means $x \times x \times x$ .
like terms: Terms with exactly the same letters, each raised to exactly the same exponent. $3 x y$ and $5 x y$ are like terms.
substitute: To put a number in place of a letter, then work the answer out.

An algebraic expression is built from terms. The plus and minus signs show you where one term ends and the next one starts. In $2 m + 7 n - 4$ there are three terms: $2 m$ , $7 n$ and $- 4$ . A minus sign belongs to the term after it.

Every term has a coefficient and may have an exponent. In $5 x^{2}$ the coefficient is $5$ and the exponent is $2$ . Read them separately. The exponent sits on the letter only, so $5 x^{2}$ means $5 \times x \times x$ , not $5 x \times 5 x$ .

We count terms to name an expression. One term is a monomial, two terms a binomial, three terms a trinomial. Count terms, not letters. So $3 ab$ is one term, a monomial, even though it has two letters in it.

Only like terms can be added or subtracted. $4 a$ and $9 a$ are like terms, so $4 a + 9 a = 13 a$ . So are $3 x y$ and $5 x y$ . But $4 a$ and $9 b$ are not, and neither are $4 a$ and $9 a^{2}$ .

To substitute, write brackets where the letter was, put the number inside, then work it out. The brackets are what keep a minus sign or a power attached to the right thing.

Rules to remember

$x + x = 2 x$
$a + b$ stays $a + b$

Examples

Name the coefficient and the exponent in

- 6 p^{4}

Worked answer

The number in front is $- 6$ , so the coefficient is $- 6$ .
The minus sign belongs to the coefficient, so do not drop it.
The small raised number on $p$ is $4$ , so the exponent is $4$ .

Answer: coefficient $- 6$ , exponent $4$

The sign travels with the coefficient, and the exponent is only ever the raised number.

Simplify:

8 k + 3 - 2 k + 5 k^{2}

Worked answer

Look for terms with exactly the same letter and exponent.
$8 k$ and $- 2 k$ are like terms, so $8 k - 2 k = 6 k$ .
$5 k^{2}$ has a different exponent, so it cannot join them.
$3$ has no letter, so it stays on its own.
The answer is $5 k^{2} + 6 k + 3$ .

Answer: $5 k^{2} + 6 k + 3$

Three terms are left because only two of the four were like each other.

Find the value of

- 3 x^{2}

when

x = 2

Worked answer

Put brackets where $x$ was: $- 3 (2)^{2}$ .
Do the power first: $(2)^{2} = 4$ .
Then multiply: $- 3 \times 4 = - 12$ .

Answer: $- 12$

The exponent sits on $x$ alone, so only the $2$ is squared and the $- 3$ is not.

Traps

Treating $3 x$ and $3 x^{2}$ as like terms because both have an $x$ . Like terms need the same exponent too. $3 x + 3 x^{2}$ cannot be simplified at all.
Giving the coefficient of $- 5 x$ as $5$ . The sign is part of the coefficient. The coefficient of $- 5 x$ is $- 5$ .
Calling $3 x y$ a trinomial because it has three symbols in it. Count terms, not symbols. $3 x y$ is all one term, so it is a monomial.
Squaring the coefficient as well when substituting, so $- 3 x^{2}$ at $x = 2$ becomes $36$ . Only what the exponent sits on gets squared. $- 3 (2)^{2} = - 3 \times 4 = - 12$ .

Learn and practise “Terms and substitution” in the app →

Multiply and divide terms

How to multiply two terms together, and how to divide a whole expression by a single term.

monomial: An expression with only one term in it, such as $5 x^{2}$ or $- 3 ab$ .
base: The number or letter that an exponent sits on. In $x^{5}$ the base is $x$ .
product: The answer you get when you multiply two things together.
quotient: The answer you get when you divide one thing by another.

To multiply two monomials, handle the numbers and the letters separately. Multiply the coefficients first. In $2 x \times 3 x$ the coefficients give $2 \times 3 = 6$ , and the letters give $x \times x = x^{2}$ . The product is $6 x^{2}$ .

Exponents on the same base add when you multiply. $x^{2} \times x^{3} = x^{5}$ , because two $x$ 's and three more $x$ 's make five. The exponents do not multiply, so the answer is not $x^{6}$ .

Dividing works the same way, backwards. Divide the coefficients, and subtract the exponents. So $\frac{8 x ^{3}}{2 x} = 4 x^{2}$ . When the two exponents are equal the letters cancel out, and the quotient is $1$ , not $0$ .

To divide a whole expression by a monomial, divide every single term by it. Miss one term and the answer is wrong. Here $\frac{6 x ^{2} + 1}{x ^{2}} = 6 + \frac{1}{x ^{2}}$ .

You may never divide by $0$ . So an expression like $\frac{6 x + 3}{x}$ only has a meaning while $x$ is not $0$ . Say so when you answer.

Rules to remember

$x^{a} \times x^{b} = x^{a + b}$
$\frac{x ^{a}}{x ^{b}} = x^{a - b}$
$\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}$

Examples

Simplify:

3 x^{2} \times 4 x^{5}

Worked answer

Multiply the coefficients: $3 \times 4 = 12$ .
The base is the same, so add the exponents: $2 + 5 = 7$ .
So the product is $12 x^{7}$ .

Answer: $12 x^{7}$

The coefficients multiply and the exponents add. They are two different jobs.

Simplify:

\frac{8 x ^{3} - 6 x ^{2}}{2 x}

Worked answer

Divide each term on top by $2 x$ , not only the first one.
First term: $\frac{8 x ^{3}}{2 x} = 4 x^{2}$ .
Second term: $\frac{6 x ^{2}}{2 x} = 3 x$ .
The minus sign stays between them.
So the answer is $4 x^{2} - 3 x$ .

Answer: $4 x^{2} - 3 x$

There were two terms on top, so the answer has two terms as well.

Simplify:

\frac{12 a ^{3} b}{4 a ^{3}}

Worked answer

Divide the numbers: $12 \div 4 = 3$ .
For $a$ , subtract the exponents: $3 - 3 = 0$ .
A base divided by itself is $1$ , so $\frac{a ^{3}}{a ^{3}} = 1$ .
The $b$ has nothing under it, so it stays.
The quotient is $3 b$ , and $a$ may not be $0$ .

Answer: $3 b$

The $a$ terms cancelled to $1$ , and multiplying by $1$ changes nothing.

Traps

Adding the coefficients instead of multiplying them, so $2 x \times 3 x$ becomes $5 x^{2}$ . Multiply the coefficients and add the exponents: $2 x \times 3 x = 6 x^{2}$ .
Multiplying the exponents, so $x^{2} \times x^{3}$ becomes $x^{6}$ . Exponents add when the bases are multiplied: $x^{2} \times x^{3} = x^{5}$ .
Writing $x^{3}$ divided by $x^{3}$ as $0$ , because the exponents subtract to zero. Anything divided by itself is $1$ . So $\frac{x ^{3}}{x ^{3}} = 1$ .
Dividing only the first term, so $\frac{6 x ^{2} + 1}{x ^{2}}$ becomes $6 + 1$ . Every term on top is divided: $\frac{6 x ^{2} + 1}{x ^{2}} = 6 + \frac{1}{x ^{2}}$ .

Learn and practise “Multiply and divide terms” in the app →

Expand brackets

Multiplying out brackets, from one bracket to two, and what a squared bracket really means.

expand: To multiply out brackets until none are left.
binomial: An expression with exactly two terms, such as $x + 3$ or $2 y - 5$ .
trinomial: An expression with exactly three terms, such as $x^{2} + 4 x + 4$ .

To expand, multiply what is outside the bracket by every single term inside. Not the first one only. So $3 (x + 4) = 3 x + 12$ . You can always check by looking at each term in turn.

The sign in front travels with the number. In $- x (3 x + 1)$ the $- x$ multiplies both terms inside, and both answers come out negative. The second term does not turn positive on its own.

Two binomials need four products, not two. For $(x + 2) (x - 3)$ , multiply the $x$ by both terms of the second bracket. Then multiply the $2$ by both terms as well. That gives $x^{2} - 3 x + 2 x - 6$ .

Now collect like terms. The two middle terms join: $- 3 x + 2 x = - x$ . So the answer is the trinomial $x^{2} - x - 6$ . Check the sign of each product before you collect.

Squaring a binomial is the same job. $(x + 2)^{2}$ means $(x + 2) (x + 2)$ , so it comes to $x^{2} + 4 x + 4$ . The middle term is the one learners drop.

Rules to remember

$a (b + c) = ab + a c$
$(a + b)^{2} = a^{2} + 2 ab + b^{2}$
$(a - b)^{2} = a^{2} - 2 ab + b^{2}$

Examples

Expand:

- 2 x (3 x - 5)

Worked answer

Multiply by the first term: $- 2 x \times 3 x = - 6 x^{2}$ .
Multiply by the second term: $- 2 x \times (- 5) = 10 x$ .
Two minus signs give a plus, so that term is positive.
The answer is $- 6 x^{2} + 10 x$ .

Answer: $- 6 x^{2} + 10 x$

The minus on $- 2 x$ reached both terms, and it changed the sign of each.

Expand and simplify:

(x + 2) (x - 3)

Worked answer

First terms: $x \times x = x^{2}$ .
Outer terms: $x \times (- 3) = - 3 x$ .
Inner terms: $2 \times x = 2 x$ .
Last terms: $2 \times (- 3) = - 6$ .
Collect the middle pair: $- 3 x + 2 x = - x$ , so the answer is $x^{2} - x - 6$ .

Answer: $x^{2} - x - 6$

All four products were written down before anything was collected.

Expand:

(2 y - 5)^{2}

Worked answer

A square means the bracket times itself: $(2 y - 5) (2 y - 5)$ .
Square the first term: $(2 y)^{2} = 4 y^{2}$ .
Double the product of the two terms: $2 \times 2 y \times (- 5) = - 20 y$ .
Square the last term: $(- 5)^{2} = 25$ .
So the answer is $4 y^{2} - 20 y + 25$ .

Answer: $4 y^{2} - 20 y + 25$

Squaring the two terms on their own would lose the middle term $- 20 y$ .

Traps

Multiplying only the first term inside a bracket. Every term inside gets multiplied: $3 (x + 4) = 3 x + 12$ , not $3 x + 4$ .
Keeping the sign of the first term only, so $- x (3 x + 1)$ becomes $- 3 x^{2} + 1$ . The minus goes onto every term: $- x (3 x + 1) = - 3 x^{2} - x$ .
Multiplying only the first and last terms of two binomials. A binomial product needs all four parts: $(x + 2) (x - 3) = x^{2} - 3 x + 2 x - 6$ .
Writing $(x + 2)^{2}$ as $x^{2} + 4$ , with no middle term. A squared bracket is the bracket times itself: $(x + 2)^{2} = x^{2} + 4 x + 4$ .

Learn and practise “Expand brackets” in the app →

Simplify expressions

Doing several steps in one question, in the right order, and taking a root without splitting it up.

simplify: To write an expression in its shortest correct form, without changing its value.
square root: The number that multiplies by itself to give it. $25 = 5$ , because $5 \times 5 = 25$ .
cube root: The number that multiplies by itself three times to give it. $38 = 2$ .

Work in a fixed order. Powers and brackets first, then multiplying and dividing, then collect like terms at the end. Doing them out of order is where most marks go.

When a whole term is raised to a power, everything inside it is raised. Cube the coefficient and multiply the exponents: $(- 2 x^{2})^{3} = - 8 x^{6}$ . The coefficient is cubed, not tripled.

A power acts before a number in front of it. So $2 (x - 3)^{2}$ means square the bracket first, then multiply by $2$ . It is not the same as $(2 x - 6)^{2}$ .

For a root, simplify under the root sign first. $25 x^{2} - 9 x^{2}$ is not $5 x - 3 x$ . Work out $25 x^{2} - 9 x^{2} = 16 x^{2}$ first, and then take the root, which is $4 x$ . A cube root works the same way: $3 27 x^{3} = 3 x$ .

To divide a long expression by a monomial, divide every term by it. Watch a negative divisor: two minus signs give a plus, so some terms change sign.

Rules to remember

$(ab)^{n} = a^{n} b^{n}$
$(a^{m})^{n} = a^{mn}$
A power is worked out before the number in front of it

Examples

Simplify:

(- 2 x^{2})^{3}

Worked answer

Cube the coefficient: $(- 2)^{3} = - 8$ .
Multiply the exponents: $2 \times 3 = 6$ .
So the answer is $- 8 x^{6}$ .

Answer: $- 8 x^{6}$

The bracket held the whole term, so the $- 2$ was cubed as well.

Simplify:

3 (x + 2) (x - 4)

Worked answer

Expand the two brackets first: $(x + 2) (x - 4) = x^{2} - 2 x - 8$ .
Check the middle pair: $- 4 x + 2 x = - 2 x$ .
Now share the $3$ out over all three terms.
So $3 (x^{2} - 2 x - 8) = 3 x^{2} - 6 x - 24$ .

Answer: $3 x^{2} - 6 x - 24$

The binomial product was done first, and only then was the $3$ used.

Simplify:

\frac{6 x ^{3} - 9 x ^{2}}{- 3 x}

Worked answer

Divide each term on top by $- 3 x$ .
First term: $\frac{6 x ^{3}}{- 3 x} = - 2 x^{2}$ .
Second term: $\frac{- 9 x ^{2}}{- 3 x} = 3 x$ .
Two minus signs give a plus, so that term is positive.
The answer is $- 2 x^{2} + 3 x$ , and $x$ may not be $0$ .

Answer: $- 2 x^{2} + 3 x$

Both terms were divided, and the negative divisor flipped both signs.

Traps

Writing $(- 2 x^{2})^{3}$ as $- 6 x^{5}$ , by tripling and adding. The coefficient is cubed and the exponents multiply: $(- 2 x^{2})^{3} = - 8 x^{6}$ .
Rooting each term, so $25 x^{2} - 9 x^{2}$ becomes $5 x - 3 x$ . Simplify under the root first. $25 x^{2} - 9 x^{2} = 16 x^{2}$ , whose square root is $4 x$ .
Expanding $2 (x - 3)^{2}$ as $(2 x - 6)^{2}$ , by multiplying before squaring. Square the bracket first: $2 (x - 3)^{2} = 2 x^{2} - 12 x + 18$ .
Cancelling one term of the top against the bottom of a fraction. Every term on top is divided: $\frac{6 x ^{3} - 9 x ^{2}}{- 3 x} = - 2 x^{2} + 3 x$ .

Learn and practise “Simplify expressions” in the app →

Factorise

Putting an expression back into brackets, which is expanding done backwards.

factor: Something you multiply by. In $3 \times 4 = 12$ , both $3$ and $4$ are factors of $12$ .
factorise: To write an expression as a product, so it ends up inside brackets.
common factor: A factor that every term shares. In $6 a + 9$ both terms share the factor $3$ .
difference of two squares: An expression of the form $a^{2} - b^{2}$ : one square subtracted from another.

Expanding takes brackets away. Factorising puts them back. If $3 (x + 4) = 3 x + 12$ , then $3 x + 12$ factorises to $3 (x + 4)$ . You can always check by expanding again.

First look for a common factor. Take out the largest one every term shares, numbers and letters together. In $6 a^{4} - 4 a^{2}$ the terms share $2$ and they share $a^{2}$ , so take out $2 a^{2}$ .

Next look for a difference of two squares. Two square terms with a minus between them factorise as $a^{2} - b^{2} = (a + b) (a - b)$ . A plus between them does not factorise at this level.

Grouping is for four terms. Pair them, take a common factor out of each pair, and the same bracket should appear twice. Watch the signs: $(b - a)$ is $- (a - b)$ , so a sign change can make the two brackets match.

Always take out a common factor first. Doing that can turn an expression you could not factorise into one you can.

Rules to remember

$a (b + c) = ab + a c$
$a^{2} - b^{2} = (a + b) (a - b)$

Examples

Factorise:

6 a^{4} - 4 a^{2}

Worked answer

The numbers $6$ and $4$ share $2$ .
Both terms have at least $a^{2}$ , so they share $a^{2}$ .
The common factor is $2 a^{2}$ .
Divide each term by it: $\frac{6 a ^{4}}{2 a ^{2}} = 3 a^{2}$ and $\frac{4 a ^{2}}{2 a ^{2}} = 2$ .
So $6 a^{4} - 4 a^{2} = 2 a^{2} (3 a^{2} - 2)$ .

Answer: $2 a^{2} (3 a^{2} - 2)$

Taking out only $2$ would leave $2 (3 a^{4} - 2 a^{2})$ , which still has a common factor inside.

Factorise:

3 x^{3} - 27 x

Worked answer

Take out the common factor $3 x$ first: $3 x (x^{2} - 9)$ .
Now $x^{2} - 9$ is a difference of two squares, because $9 = 3^{2}$ .
So $x^{2} - 9 = (x + 3) (x - 3)$ .
The full answer is $3 x (x + 3) (x - 3)$ .

Answer: $3 x (x + 3) (x - 3)$

Without the common factor first, the difference of two squares is hidden.

Factorise by grouping:

2 a x - 2 b x + 3 b - 3 a

Worked answer

Four terms, so pair them: $(2 a x - 2 b x)$ and $(3 b - 3 a)$ .
First pair: $2 x (a - b)$ .
Second pair: $3 (b - a)$ , and $(b - a) = - (a - b)$ , so it is $- 3 (a - b)$ .
Both now carry $(a - b)$ : $2 x (a - b) - 3 (a - b)$ .
Take the common bracket out: $(a - b) (2 x - 3)$ .

Answer: $(a - b) (2 x - 3)$

The sign switch on $(b - a)$ is what made the two brackets match.

Traps

Taking out only part of the common factor, so $6 a^{4} - 4 a^{2}$ becomes $2 (3 a^{4} - 2 a^{2})$ . Check every term for the largest shared factor: $6 a^{4} - 4 a^{2} = 2 a^{2} (3 a^{2} - 2)$ .
Trying to factorise $x^{2} + 9$ as though a sum of squares behaved like a difference. Only a difference factorises. $x^{2} - 9$ becomes $(x + 3) (x - 3)$ ; $x^{2} + 9$ stays as it is.
Stopping at $2 x (a - b) - 3 (b - a)$ because the two brackets look different. $(b - a)$ is $- (a - b)$ . Swap it, and the minus in front turns into a plus.

Learn and practise “Factorise” in the app →

Trinomials and fractions

Breaking a three-term expression into two brackets, and using that to shorten a fraction.

algebraic fraction: A fraction with letters in it, such as $\frac{x + 2}{x}$ .
cancel: To divide the top and the bottom of a fraction by the same factor.
restriction: A value the letter is not allowed to take, because it would make the bottom $0$ .

A trinomial like $x^{2} + 7 x + 12$ comes from two brackets multiplied out. To factorise it, look for two numbers. They must multiply to give the last number, and add to give the middle one.

The signs guide you. When the last number is positive, both of your numbers carry the sign of the middle term. When it is negative, one of them is positive and the other is negative.

If every term shares a common factor, take it out before you do anything else. Then factorise the trinomial left inside the bracket. The factor you took out stays in the answer.

To simplify an algebraic fraction, factorise the top and the bottom first. Then cancel any bracket that appears in both. Whole factors cancel this way. Single terms never do.

The bottom of a fraction may never be $0$ . So look at the original bottom line, and write down the restriction it puts on the letter.

Rules to remember

$x^{2} + b x + c$ factorises when two numbers multiply to $c$ and add to $b$
$\frac{k a}{k b} = \frac{a}{b}$ , as long as $k$ is not $0$

Examples

Factorise:

x^{2} + 7 x + 12

Worked answer

Look for two numbers that multiply to $12$ and add to $7$ .
Try $3$ and $4$ : $3 \times 4 = 12$ and $3 + 4 = 7$ .
Both jobs are done, so the brackets are $(x + 3)$ and $(x + 4)$ .
So $x^{2} + 7 x + 12 = (x + 3) (x + 4)$ .

Answer: $(x + 3) (x + 4)$

The same pair had to multiply to $12$ and add to $7$ , not one or the other.

Factorise:

2 x^{2} - 2 x - 24

Worked answer

Every term shares $2$ , so take it out: $2 x^{2} - 2 x - 24 = 2 (x^{2} - x - 12)$ .
Now the pair must multiply to $- 12$ and add to $- 1$ .
Try $3$ and $- 4$ : $3 \times (- 4) = - 12$ and $3 + (- 4) = - 1$ .
So the bracket becomes $(x + 3) (x - 4)$ .
Keep the $2$ in front: $2 (x + 3) (x - 4)$ .

Answer: $2 (x + 3) (x - 4)$

Taking the $2$ out first left a trinomial whose first term was just $x^{2}$ .

Simplify:

\frac{x ^{2} - 9}{x ^{2} + 7 x + 12}

Worked answer

The top is a difference of two squares: $x^{2} - 9 = (x + 3) (x - 3)$ .
The bottom is a trinomial: $x^{2} + 7 x + 12 = (x + 3) (x + 4)$ .
The bracket $(x + 3)$ is in both, so cancel it.
What is left is $\frac{x - 3}{x + 4}$ .
The restriction comes from the original bottom: $x$ may not be $- 3$ or $- 4$ .

Answer: $\frac{x - 3}{x + 4}$

Nothing could cancel until both lines were written as brackets multiplied.

Traps

Adding the pair to get the last number and multiplying to get the middle one. The pair multiplies to give the last number and adds to give the middle one.
Dropping the common factor from the answer, so only $(x + 3) (x - 4)$ is written. The factor you took out stays: $2 x^{2} - 2 x - 24 = 2 (x + 3) (x - 4)$ .
Cancelling single terms, so $\frac{x + 3}{3}$ is written as $x$ . Only a whole factor cancels, never one term. $\frac{x + 3}{3}$ cannot be simplified.
Leaving out the values the letter may not take. Read the restriction off the original bottom line, before anything was cancelled.

Learn and practise “Trinomials and fractions” in the app →

Grade 9 Algebraic expressions

What gets asked

You must be able to

Traps that cost marks

Worked example

Terms and substitution

Rules to remember

Examples

Traps

Multiply and divide terms

Rules to remember

Examples

Traps

Expand brackets

Rules to remember

Examples

Traps

Simplify expressions

Rules to remember

Examples

Traps

Factorise

Rules to remember

Examples

Traps

Trinomials and fractions

Rules to remember

Examples

Traps

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