Expanding and Factorising Brackets
Expanding Brackets
Expanding One Bracket
To expand a bracket, multiply the term outside the bracket by each term inside the bracket. This removes the brackets. For example, 3x(x + 2) expands to 3x \times x + 3x \times 2, which simplifies to 3x^2 + 6x.
When dealing with negative signs, remember the basic rules of multiplication:
- \times - = +
- \times + = -
It's helpful to put brackets around negative terms to avoid errors.
Example:
Expand 4x(2x - 3).
Multiply the term outside the brackets by both terms inside: 4x \times 2x + 4x \times (-3)
Simplify: 8x^2 - 12x
Expand -7x(4 - 5x).
Multiply the term outside the brackets by both terms inside: -7x \times 4 + (-7x) \times (-5x)
Simplify: -28x + 35x^2
Expand & Simplify
When simplifying an expression with more than one set of brackets, follow these steps:
Expand each set of brackets separately.
Collect like terms.
For example, to expand and simplify 4(x + 7) + 5x(3 - x), first expand each set of brackets:
4(x + 7) = 4x + 28
5x(3 - x) = 15x - 5x^2
So, 4(x + 7) + 5x(3 - x) = 4x + 28 + 15x - 5x^2. Then, collect like terms: 19x + 28 - 5x^2.
Examples:
Expand and simplify 2(x + 5) + 3x(x - 8).
Expand: 2x + 10 + 3x^2 - 24x
Collect like terms: 3x^2 - 22x + 10
Expand and simplify 3x(x + 2) - 7(x - 6).
Expand: 3x^2 + 6x - 7x + 42
Collect like terms: 3x^2 - x + 42
Expanding Double Brackets
To expand two brackets, each term in the first bracket must be multiplied by each term in the second bracket.
FOIL Method
FOIL is a mnemonic for remembering the order of multiplications:
First: Multiply the first terms in each bracket.
Outside: Multiply the first term in the first bracket by the second term in the second bracket.
Inside: Multiply the second term in the first bracket by the first term in the second bracket.
Last: Multiply the last terms in each bracket.
For example, to expand (x + 1)(x + 3), use FOIL:
First: x \times x = x^2
Outside: x \times 3 = 3x
Inside: 1 \times x = x
Last: 1 \times 3 = 3
So, (x + 1)(x + 3) = x^2 + 3x + x + 3 = x^2 + 4x + 3.
Grid Method
Alternatively, you can use a grid to keep track of the multiplications. For example, to expand (x + 1)(x + 3), write the brackets as headings in a grid:
x | +1 | |
|---|---|---|
x | x^2 | x |
+3 | 3x | 3 |
Add together all the terms inside the grid: x^2 + x + 3x + 3 = x^2 + 4x + 3.
Expanding a Bracket Squared
To expand a bracket squared, rewrite it as two separate brackets multiplied together. For example, (x + 3)^2 = (x + 3)(x + 3). Then, use one of the methods above.
Caution: Avoid the common mistake of saying (x + 3)^2 = x^2 + 3^2. This is incorrect. For instance, if x = 1, (1 + 3)^2 = 4^2 = 16, but 1^2 + 3^2 = 1 + 9 = 10.
Examples:
Expand (2x - 3)(x + 4).
Using FOIL: (2x \times x) + (2x \times 4) + (-3 \times x) + (-3 \times 4) = 2x^2 + 8x - 3x - 12
Simplify: 2x^2 + 5x - 12
Expand (x - 3)(3x - 5).
Using FOIL: (x \times 3x) + (x \times -5) + (-3 \times 3x) + (-3 \times -5) = 3x^2 - 5x - 9x + 15
Simplify: 3x^2 - 14x + 15
Expand (2x + 3)^2.
Rewrite: (2x + 3)(2x + 3)
Using FOIL: (2x \times 2x) + (2x \times 3) + (3 \times 2x) + (3 \times 3) = 4x^2 + 6x + 6x + 9
Simplify: 4x^2 + 12x + 9
Expanding Triple Brackets
To expand three brackets:
Multiply out any two brackets using a standard method (FOIL or grid) and simplify.
Replace the two brackets with the expanded result.
Expand this long bracket with the third (unused) bracket. This often looks like (x + a)(x^2 + bx + c).
Every term in the first bracket must be multiplied with every term in the second bracket. This leads to six terms. A grid can help keep track of all six terms.
For example, to expand (x + 2)(x^2 + 3x + 1), use a grid:
x^2 | +3x | +1 | |
|---|---|---|---|
x | x^3 | 3x^2 | x |
+2 | 2x^2 | 6x | 2 |
Add all the terms inside the grid: x^3 + 3x^2 + x + 2x^2 + 6x + 2 = x^3 + 5x^2 + 7x + 2.
Examples:
Expand (2x - 3)(x + 4)(3x - 1).
Expand the first two brackets: (2x - 3)(x + 4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12
Rewrite: (2x^2 + 5x - 12)(3x - 1)
Multiply all the terms: 6x^3 - 2x^2 + 15x^2 - 5x - 36x + 12
Simplify: 6x^3 + 13x^2 - 41x + 12
Expand (x - 3)(x + 2)(2x - 1).
Expand the first two brackets: (x - 3)(x + 2) = x^2 + 2x - 3x - 6 = x^2 - x - 6
Rewrite: (x^2 - x - 6)(2x - 1)
Multiply all the terms: 2x^3 - x^2 - 2x^2 + x - 12x + 6
Simplify: 2x^3 - 3x^2 - 11x + 6
Factorising
Basic Factorising
Factorisation is the opposite of expanding brackets. It involves writing an expression as the product (multiplication) of two or more terms (factors). For example, 3(x + 2) is factorised as it is 3 \times (x + 2), while 3x + 6 is not factorised as it is "something" + "something".
To factorise two terms:
Find the highest common factor (HCF) of the coefficients.
Find the highest common factor of the variables.
Multiply both to get the common factor.
Rewrite each term as "common factor × something".
Take out the common factor by writing it outside brackets.
Put the remaining terms inside the brackets.
For example, to factorise 12x^2 + 18x:
The HCF of 12 and 18 is 6.
The HCF of x^2 and x is x.
The common factor is 6x.
Rewrite: 6x \times 2x + 6x \times 3
Take out the common factor: 6x(2x + 3)
Exam Tip: You can always check your factorisation by expanding the brackets in your answer.
Examples:
Factorise 5x + 15.
The HCF of 5 and 15 is 5.
Rewrite: 5 \times x + 5 \times 3
Take out the 5: 5(x + 3)
Factorise fully 30x^2 - 24x.
The HCF of 30 and 24 is 6.
The HCF of x^2 and x is x.
The common factor is 6x.
Rewrite: 6x \times 5x - 6x \times 4
Take out the 6x: 6x(5x - 4)
Factorising by Grouping
To factorise expressions with common brackets, treat the whole bracket as a common factor. For example, to factorise 3x(t + 4) + 2(t + 4), the common bracket is (t + 4). So, the factorised expression is (t + 4)(3x + 2).
Some questions may require you to form the common bracket yourself. For example, to factorise xy + px + qy + pq, group the first pair of terms (xy + px) and factorise to get x(y + p). Then group the second pair of terms (qy + pq) and factorise to get q(y + p). Now factorise x(y + p) + q(y + p) to get (y + p)(x + q).
This is called factorising by grouping. The groupings are not always the first pair of terms and the second pair of terms, but two terms with common factors.
Exam Tip: As always, once you have factorised something, expand it by hand to check your answer is correct.
Example:
Factorise ab + 3b + 2a + 6.
Method 1: Factorise b from (ab + 3b) and 2 from (2a + 6) to get b(a + 3) + 2(a + 3).
Factorise the common bracket (a + 3) to get (a + 3)(b + 2).
Method 2: Factorise a from (ab + 2a) and 3 from (3b + 6) to get a(b + 2) + 3(b + 2).
Factorise the common bracket (b + 2) to get (b + 2)(a + 3).
Factorising Quadratics
Factorising Simple Quadratics
A quadratic expression is in the form ax^2 + bx + c (as long as a \neq 0). If a = 1, it can be called a “monic” quadratic expression. If a \neq 1, it can be called a “non-monic” quadratic expression.
Method 1: Factorising "by inspection"
To factorise x^2 - 2x - 8, find a pair of numbers that multiply to -8 and add to -2. The numbers -4 and +2 satisfy these conditions. Write these numbers in a pair of brackets: (x + 2)(x - 4).
Method 2: Factorising "by grouping"
To factorise x^2 - 2x - 8, find a pair of numbers that multiply to -8 and add to -2. The numbers 2 and -4 satisfy these conditions. Rewrite the middle term by using 2x and -4x: x^2 + 2x - 4x - 8. Group and factorise the first two terms, using x as the highest common factor, and group and factorise the second two terms, using -4 as the factor: x(x + 2) - 4(x + 2). Note that these now have a common factor of (x + 2) so this whole bracket can be factorised out: (x + 2)(x - 4).
Method 3: Factorising "by using a grid"
To factorise x^2 - 2x - 8, find a pair of numbers that multiply to -8 and add to -2. The numbers -4 and +2 satisfy these conditions. Write the quadratic equation in a grid, splitting the middle term as -4x and 2x.
x | -4 | |
|---|---|---|
x | x^2 | -4x |
+2 | +2x | -8 |
Write a heading for the first row, using x as the highest common factor of x^2 and -4x. Then, fill in the remaining row heading using the same idea. Read off the factors from the column and row headings: (x + 2)(x - 4).
Which method should I use for factorising simple quadratics?
The first method, by inspection, is by far the quickest, so it is recommended in an exam for simple quadratics (where a = 1).
However, the other two methods (grouping or using a grid) can be used for harder quadratic equations where a \neq 1, so you should learn at least one of them too.
Exam Tip: As a check, expand your answer and make sure you get the same expression as the one you were trying to factorise.
Examples:
Factorise x^2 - 4x - 21.
By inspection: (x + 3)(x - 7)
Factorise x^2 - 5x + 6.
By splitting the middle term and grouping: (x - 2)(x - 3)
Factorise x^2 - 2x - 24.
By using a grid: (x + 4)(x - 6)
Factorising Harder Quadratics
Factorising a \neq 1 "by grouping"
To factorise 4x^2 - 25x - 21, find a pair of numbers that multiply to 4 \times -21 = -84 and add to -25. The numbers -28 and +3 satisfy these conditions. Rewrite the middle term using -28x and +3x: 4x^2 - 28x + 3x - 21. Group and factorise the first two terms, using 4x as the highest common factor, and group and factorise the second two terms, using 3 as the factor: 4x(x - 7) + 3(x - 7). Note that these terms now have a common factor of (x - 7) so this whole bracket can be factorised out, leaving 4x + 3 in its own bracket: (x - 7)(4x + 3).
Factorising a \neq 1 "by using a grid"
To factorise 4x^2 - 25x - 21, find a pair of numbers that multiply to 4 \times -21 = -84 and add to -25. The numbers -28 and +3 satisfy these conditions. Write the quadratic equation in a grid, splitting the middle term as -28x and +3x.
4x | +3 | |
|---|---|---|
x | 4x^2 | +3x |
-7 | -28x | -21 |
Write a heading for the first row, using 4x as the highest common factor of 4x^2 and -28x. Then, fill in the remaining row heading using the same idea. Read off the factors from the column and row headings: (x - 7)(4x + 3).
Exam Tip: As a check, expand your answer and make sure you get the same expression as the one you were trying to factorise.
Factorise 6x^2 - 7x - 3.
By splitting the middle term and grouping: (3x + 1)(2x - 3)
Factorise 10x^2 + 9x - 7.
By using a grid: (2x - 1)(5x + 7)
Difference Of Two Squares
When a "squared" quantity is subtracted from another "squared" quantity, you get the difference of two squares. For example,
a^2 - b^2
9^2 - 5^2
(x + 1)^2 - (x - 4)^2
4m^2 - 25n^2, which is (2m)^2 - (5n)^2
To factorise the difference of two squares, use the formula:
a^2 - b^2 = (a + b)(a - b)
It is fine to write the second bracket first, (a - b)(a + b), but the a and the b cannot swap positions. a^2 - b^2 must have the a's first in the brackets and the b's second in the brackets.
Exam Tip: The difference of two squares is a very important rule to learn, as it often appears in harder questions involving factorisation, e.g. in algebraic fractions. The word difference in maths means a subtraction; it should remind you that you are subtracting one squared term from another. You should be able to recognise factorised difference of two squares expressions.
Examples:
Factorise 9x^2 - 16.
Recognise that (3x)^2 and 4^2 are both squared terms and the second term is subtracted from the first term - you can factorise using the difference of two squares.
(3x + 4)(3x - 4)
Factorise 4x^2 - 25.
Recognise that (2x)^2 and 5^2 are both squared terms and the second term is subtracted from the first term - you can factorise using the difference of two squares.
(2x + 5)(2x - 5)
Quadratics Factorising Methods
How do I know if it factorises?
Method 1: Use a calculator to solve the quadratic expression equal to 0. If the solutions are integers or fractions (without square roots), then the quadratic expression factorises.
Method 2: Find the value under the square root in the quadratic formula, b^2 – 4ac (called the discriminant). If this number is a perfect square number, then the quadratic expression factorises.
Which factorisation method should I use for a quadratic expression?
Does it have 2 terms only?
Yes, like x^2 - 7x: Use "basic factorisation" to take out the highest common factor x(x - 7).
Yes, like x^2 - 9: Use the "difference of two squares" to factorise (x + 3)(x - 3).
Does it have 3 terms?
Yes, starting with x^2, like x^2 - 3x - 10: Use "factorising simple quadratics" by finding two numbers that add to -3 and multiply to -10. (x + 2)(x - 5).
Yes, starting with ax^2, like 3x^2 + 15x + 18: Check to see if the 3 in front of x^2 is a common factor for all three terms (which it is in this case), then use "basic factorisation" to factorise it out first 3(x^2 + 5x + 6). The quadratic expression inside the brackets is now x^2 + …, which factorises more easily 3(x + 2)(x + 3).
Yes, starting with ax^2, like 3x^2 - 5x - 2: The 3 in front of x^2 is not a common factor for all three terms. Use "factorising harder quadratics", for example, factorising by grouping or factorising using a grid (3x + 1)(x - 2).
Example:
Factorise -8x^2 + 100x - 48.
Spot the common factor of -4 and put it outside a set of brackets: -4(2x^2 - 25x + 12).
Check the discriminant for the expression inside the brackets, b^2 - 4ac = (-25)^2 - 4(2)(12) = 625 - 96 = 529, to see if it will factorise. 529 = 23^2, it is a perfect square, so the expression will factorise.
Proceed with factorising 2x^2 - 25x + 12 as you would for a harder quadratic, where a \neq 1.
"+12" means the signs will be the same. "-25" means that both signs will be negative.
The only numbers which multiply to give 24 and follow the rules for the signs above are:
-24 and -1
-12 and -2
-8 and -3 but only the first pair add to give -25.
Split the -25x term into -24x - x.
Group and factorise the first two terms, using 2x as the highest common factor, and group and factorise the last two terms, using -1 as the highest common factor.
These factorised terms now have a common term of (x - 12), so this can now be factorised out.
-4(2x^2 - 25x + 12) = -4(2x^2 - 24x - x + 12) = -4[2x(x - 12) - 1(x - 12)] = -4(2x - 1)(x - 12)