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This page sets out the method for expanding & factorising quadratics in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Expanding & factorising quadratics topic page.
When you see two brackets multiplied together, make four products then collect like terms.
When you see a quadratic with a 1 in front of and are asked to factorise, hunt for a number pair.
When you see something squared minus something squared with no middle term, e.g. or .
Expanding (x+3)(x+2) gives x^2+5x+6.
Expanding (x+4)^2 gives x^2+8x+16. The middle term is twice the product.
Factorising x^2+5x+6 gives (x+2)(x+3).
Factorising x^2-9 gives (x-3)(x+3).
Factorising 2a+2b gives 2(a+b).
The numbers in the brackets must multiply to the constant and add to the x coefficient.
(x+3)(x+5) multiplies in four places: x^2, 5x, 3x and 15. The two middle products are like terms, so 5x+3x=8x. The expansion is x^2+8x+15. Write all four products on one line before you collect. A correct set of four can still earn a method mark if the adding goes wrong afterwards. Say the four products out loud if it helps: first, outer, inner, last. Then the collect is a separate job.
Signs belong to the number they sit with. (x-4)(x+2)=x^2+2x-4x-8=x^2-2x-8. The constant is negative because a negative times a positive is negative. (x-4)(x-2)=x^2-6x+8, and the constant is positive because two negatives multiply. Do not decide the constant sign from memory of the bracket signs alone. Multiply them.
Method: 1. From (x+3)(x+5), write the four products x^2, 5x, 3x and 15. 2. Add the like middle terms: 5x+3x=8x. 3. The expanded expression is x^2+8x+15.
(x+5)^2 means (x+5)(x+5), not x^2+25. The middle term is 5x+5x=10x, so the expansion is x^2+10x+25. The same pattern with a minus is (x-5)^2=x^2-10x+25. The constant is still positive. Only the middle sign follows the sign inside the bracket.
x^2-49 is x^2-7^2. There is no x term, which is the clue for a difference of two squares: (x+7)(x-7). Expand that and the middle terms +7x and -7x cancel, leaving x^2-49. It also works with harder squares, such as 4x^2-9=(2x+3)(2x-3). If a middle term is present, it is not a difference of two squares.
Method: 1. Expand (x+5)^2 as (x+5)(x+5). It is not x^2+25. 2. Double the cross term: two lots of 5x make 10x, and the square is x^2+10x+25. 3. Change the middle sign when the bracket is a difference: (x-5)^2 expands to x^2-10x+25.
Pick a small value of x and work out the original brackets and the expanded expression. They must agree. For (x+3)(x+5) at x=2, the brackets give 5× 7=35 and the expansion x^2+8x+15 gives 4+16+15=35. If they disagree, the middle term is the usual suspect: the cross products were missed or their signs were added the wrong way. This check does not prove the algebra for every x, but a mismatch proves the expansion is wrong, which is exactly what you want before you box it.
Use a different value from the one hiding in the question, and avoid x=0 if you want the middle term to show itself. At x=0 a wrong middle term can still give the right constant. x=1 or x=2 is enough. The same check works after factorising: expand the brackets you wrote and compare, either with algebra or with one substituted number. If the constant matches but the middle does not, the factor pair multiplied correctly and added wrongly. Change the pair, not the method. Then expand the new brackets once more before you finish the question.
Method: 1. Pick a small value of x and evaluate both the original brackets and the expansion. 2. At x=2, (x+3)(x+5) gives 5× 7=35 and x^2+8x+15 gives 4+16+15=35. 3. If those two results disagree, check the cross terms and the signs you gave them.
Find two numbers that multiply to c and add to b. For x^2+7x+12, the pairs of 12 are 1 and 12, 2 and 6, 3 and 4. Only 3 and 4 add to 7. So the factorisation is (x+3)(x+4). Jot the pairs in the margin. It is quicker than guessing, and it shows the method.
If c is negative, the two numbers have opposite signs, and the larger absolute value takes the sign of b. For x^2-5x-14, the pairs of 14 are 1 and 14, 2 and 7. Opposite signs: -7 and +2 multiply to -14 and add to -5. The brackets are (x-7)(x+2). Expand to check. A pair that multiplies correctly but adds to the wrong number, such as (x+2)(x+6) for a sum of 7, is a near miss that the check will catch.
Method: 1. Find two numbers that multiply to give c and add to give b. 2. For x^2+7x+12, the factor pairs of 12 are 1 with 12, 2 with 6, and 3 with 4. 3. Keep the pair that adds to 7, which is 3 and 4.
A quadratic such as 2x^2 + 7x - 15 has a number in front of x^2, so the pair-of-numbers trick from x^2 + bx + c needs an extra step. Multiply the first and last numbers: 2 × -15 = -30. Now find two numbers that multiply to -30 and add to the middle number 7. These are 10 and -3. Use them to split the middle term.
Write 2x^2 + 10x - 3x - 15 and factorise the first pair and the second pair separately. You get 2x(x + 5) - 3(x + 5). Both parts now contain the bracket (x + 5), so take it out as a common factor to get (2x - 3)(x + 5). Always expand your answer to check it. If the middle term does not match, change the order of your pair of numbers or recheck the signs.
Method: 1. Multiply the number in front of x^2 by the last number. 2. Find two numbers with that product that add to the middle number. 3. Split the middle term into two terms using those numbers. 4. Factorise in pairs, take out the common bracket and expand to check.
Tip: If the last number is negative, the two numbers have different signs. The larger one takes the sign of the middle term.
An expression with only two terms, one squared term minus another, is a difference of two squares. x^2 - 64 factorises to (x - 8)(x + 8) because 64 = 8^2. The same idea works with coefficients and several letters. For 100a^2 - 81b^2 write each term as a square: (10a)^2 - (9b)^2. The factorised form is (10a - 9b)(10a + 9b).
The harder version asks you to factorise fully, which usually means taking out a common factor first. In 4x^3 - 100x the common factor is 4x, leaving 4x(x^2 - 25). The bracket is a difference of two squares, so the full answer is 4x(x - 5)(x + 5). Also watch for questions like 144 - 4x^2, where a common factor of 4 leaves 4(36 - x^2) and then 4(6 - x)(6 + x).
Method: 1. Check that there are two terms and one is subtracted from the other. 2. Take out any common factor first, including a letter. 3. Write each remaining term as a perfect square. 4. Write the pair of brackets with one plus and one minus, then expand to check.
Three brackets multiplied together produce a cubic expression. Always expand two of the brackets first and simplify, then multiply the result by the third bracket. Choose the pair that looks simplest. For (x - 2)(x - 5)(x + 6) the first pair gives x^2 - 7x + 10. Multiply this by (x + 6), taking each of the three terms in turn.
Writing the multiplication as two separate lines keeps the signs under control. Multiply the quadratic by x first, then multiply it by 6, and add the two results. Collect like terms last. Check by substituting a small number such as x = 3 into the original product and into your answer. If the two values match, the expansion is almost certainly right.
Method: 1. Expand the first two brackets and collect like terms to get a quadratic. 2. Multiply the quadratic by x and then by the number in the last bracket. 3. Add the two lines and collect like terms in order of power. 4. Check with x = 3 in the original and in the final expression.
Some exam questions describe a product of brackets with unknown numbers and ask for the possible values of a coefficient. For instance (x + s)(x - t) expands to x^2 + (s - t)x - st. If you are told the constant term, then s × t is fixed, and you list the factor pairs of that number. For each pair the coefficient of x is s - t.
Read the conditions closely. Positive integers, with a positive coefficient, mean you must have s bigger than t. List every factor pair, work out s - t for each and pick the one the question asks for, such as the smallest. Listing all pairs in a table is safe, because a missing pair is how this type of question loses a mark.
Method: 1. Expand the general brackets to see what the middle and last terms contain. 2. List the factor pairs of the constant term. 3. Work out the middle coefficient for each pair using the conditions given. 4. Pick the value the question asks for and state the pair.
Tip: Put the factor pairs in a small table. A missing pair is the usual way to lose the answer mark.
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