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This page sets out the method for harder factorising (ax² + bx + c) 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 Harder factorising (ax² + bx + c) topic page.
When you see a quadratic with a number in front of and no common factor, use the method.
When there is no term and both terms are squares with a minus between them.
When a > 1, the first terms of the brackets multiply to ax².
Look for two brackets whose product gives every term.
4x² − 9 = (2x−3)(2x+3). Both 4x² and 9 are perfect squares.
If the first term is 6x², one bracket can start with 3x and the other with 2x.
Expand your brackets. The middle term has to match.
Once the brackets are right, each factor can be zero.
For 2x^2+7x+3, a=2 and c=3, so ac=6. You need two numbers that multiply to 6 and add to 7. They are 1 and 6. Split the middle: 2x^2+x+6x+3. Group as x(2x+1)+3(2x+1). The shared bracket is 2x+1, and the other bracket is x+3. So 2x^2+7x+3=(2x+1)(x+3). Looking for numbers that multiply to 3 instead of 6 is the mistake that appears as soon as a is not 1.
Check by expanding or by substituting. At x=1, the original is 2+7+3=12 and the factors are 3× 4=12. At x=0 both sides equal 3. Two checks are quick. If the brackets after grouping do not match, the split is wrong or a sign inside a pair is wrong. Go back to the pair of numbers rather than forcing the brackets to look similar.
Method: 1. Multiply a by c. That product is what the pair must multiply to. 2. Find two numbers with that product and with sum b. 3. Split the middle term, or place the numbers in the brackets. 4. Expand to check you return to the original quadratic.
For 6x^2-7x-3, ac=6×(-3)=-18. The pair must multiply to -18 and add to -7. The numbers are -9 and 2, because (-9)× 2=-18 and -9+2=-7. Split: 6x^2-9x+2x-3. Then 3x(2x-3)+1(2x-3)=(3x+1)(2x-3). The +1 is easy to drop. It is the factor in front of the second pair, and it belongs in the bracket.
Check at x=2: factors (6+1)(4-3)=7, and 6(4)-7(2)-3=24-14-3=7. Check at x=0: both sides equal -3. The pair 9 and -2 has the same product -18, but it adds to +7, and the middle term of this quadratic is -7. So the signs of the pair matter. Write the sum next to the product before you split. That one line saves the regrouping.
Method: 1. If c is negative, the two numbers have opposite signs. 2. The larger absolute value takes the sign of b. 3. Check the expansion of the cross terms. 4. A sign error usually shows up in the middle term only.
Solving is factorising plus two small equations. For 4x^2-4x-3=0, ac=-12. The pair -6 and 2 multiplies to -12 and adds to -4. Then 4x^2-6x+2x-3=2x(2x-3)+1(2x-3)=(2x+1)(2x-3). Set each factor to zero: 2x+1=0 gives x=-1/2, and 2x-3=0 gives x=3/2. Substitute both back. 4(3/2)^2-4(3/2)-3=9-6-3=0.
If every term shares a factor, take it out first or the final answer is incomplete. 4x^2+10x+6=2(2x^2+5x+3), and 2x^2+5x+3=(2x+3)(x+1), so the full factorisation is 2(2x+3)(x+1). For an equation equal to zero you may divide by the common factor, because it is not zero. For an expression, the common factor stays in the answer. A difference of squares can sit inside that factor: 8x^2-18=2(4x^2-9)=2(2x-3)(2x+3).
Method: 1. Factorise fully, including a common factor if there is one. 2. Set each factor equal to zero. 3. Solve those linear equations. 4. A common factor of the quadratic is not an extra root unless it gives a factor in x.
9x^2-16 is (3x)^2-4^2, so it is (3x-4)(3x+4). There is no middle term to split. The same idea solves 9x^2-16=0: 3x=4 or 3x=-4, so x=4/3 or x=-4/3. Do not write x=4 and x=-4. The square root of the coefficient of x^2 has to be undone as well.
For 6x^2+7x-3=0, ac=-18. The pair 9 and -2 adds to 7. Split 6x^2+9x-2x-3=3x(2x+3)-1(2x+3)=(3x-1)(2x+3). Then x=1/3 or x=-3/2. Notice the minus in front of the second group: -1(2x+3) is not -2x+3. The bracket is subtracted as a whole. That is the same sign skill as expanding, used in reverse.
Method: 1. Spot a^2-b^2 before you use the full ac search. 2. It factorises as (a-b)(a+b). 3. Then set each factor equal to zero if you are solving. 4. There is no middle term to split.
A quadratic with big numbers is often hiding a common factor. The expression 6x^2+9x-6 looks hard, but every term divides by 3. Take it out first: 3(2x^2+3x-2). The bracket is now a much smaller problem, with ac=2×(-2)=-4 and the pair 4 and -1, giving 3(2x-1)(x+2). If you skip the common factor you can still find a valid factorisation, such as (6x-3)(x+2), but it is not fully factorised because 6x-3 still has a factor of 3.
Exam questions say "factorise fully" or "factorise completely" to remind you of this. Scan for a common factor of all three terms before you do anything else, and check at the end that no bracket has a number that divides into both of its terms. Your answer then has the shape: number out front, then brackets. Expanding your answer is the check, and it should return the expression you started with exactly.
Method: 1. Look for a number or letter that divides every term, and take it out. 2. Factorise the remaining quadratic using the ac method. 3. Write the common factor in front of the two brackets. 4. Expand your answer to confirm it matches the original.
Tip: If one of your brackets has a number that goes into both terms, you have not finished factorising.
For 6x^2-x-12 the product ac is 6×(-12)=-72 and the middle number is -1. List factor pairs of 72 in order: 1 and 72, 2 and 36, 3 and 24, 4 and 18, 6 and 12, 8 and 9. You want two numbers that differ by 1, so 8 and 9. The product is negative, so one is positive and one is negative, and the sum is -1, so the larger one is negative: 8 and -9.
Split the middle term: 6x^2+8x-9x-12. Factorise in two pairs: 2x(3x+4)-3(3x+4). Both brackets match, which is your check that the pair was right. The answer is (3x+4)(2x-3). If the brackets do not match, the pair has the wrong sign or the wrong numbers. Working through factor pairs in order is slower than guessing but it never misses the answer, and you can stop when the difference or sum is correct.
Method: 1. Work out ac and write down the middle coefficient b. 2. List factor pairs of ac in order until the pair has the right sum, using the sign of the product to place signs. 3. Split the middle term and factorise in two pairs. 4. Confirm that the two brackets match before writing the answer.
A difference of two squares does not always look simple. The expression 18x^2-8 has no square coefficients, but it has a common factor of 2. Take it out: 2(9x^2-4). Now 9x^2=(3x)^2 and 4=2^2, so the bracket is (3x-2)(3x+2). The full answer is 2(3x-2)(3x+2). Remember that a plain sum of two squares, such as x^2+4, does not factorise using real numbers.
Powers can hide squares too. The expression x^4-16 is (x^2)^2-4^2, so it factorises to (x^2-4)(x^2+4). The first bracket is itself a difference of two squares, so the complete answer is (x-2)(x+2)(x^2+4). The same idea simplifies algebraic fractions: (4x^2-9)/(2x^2+x-3) becomes ((2x-3)(2x+3))/((2x+3)(x-1))=(2x-3)/(x-1). Factorising is the key step in simplifying, solving and proving.
Method: 1. Take out a common factor first, if there is one. 2. Write each term as a square and use a^2-b^2=(a-b)(a+b). 3. Check whether any bracket can be factorised again. 4. In a fraction, factorise top and bottom, then cancel matching brackets.
You can only use "one bracket must be zero" when the other side of the equation is zero. For 4x^2+4x=15 first subtract 15 from both sides to get 4x^2+4x-15=0. Then ac=-60 and the pair with sum 4 is 10 and -6. Split: 4x^2+10x-6x-15=2x(2x+5)-3(2x+5)=(2x-3)(2x+5). The solutions are x=3/2 and x=-5/2. Writing x(4x+4)=15 and solving x=15 is a very common wrong turn.
Context problems add a final step. If a rectangle has sides (x+2) cm and (3x-1) cm and an area of 40 cm^2, the equation is (x+2)(3x-1)=40. Expand to 3x^2+5x-2=40, so 3x^2+5x-42=0, which factorises as (x-3)(3x+14)=0. A length cannot be negative, so x=3 is the only valid solution. Then use it: the sides are 5 cm and 8 cm. Always say which solution you are rejecting and why.
Method: 1. Collect all terms on one side so that the equation equals zero. 2. Factorise, then set each bracket equal to zero. 3. Solve each linear equation, giving fractions in simplest form. 4. In a context, reject any solution that is negative or does not fit, and give a final answer in context.
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