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This page sets out the method for solving quadratics by factorising 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 Solving quadratics by factorising topic page.
When you see in an equation and it factorises, this is the fastest route.
When a shape's area or a product of two numbers is given and the sides involve .
To solve x^2+5x+6=0, write it as (x+2)(x+3)=0. One bracket must be zero, so x is -2 or -3.
Solving x^2-9=0 gives -3 or 3, because x^2 = 9.
Solving x^2+4x=0 gives -4 or 0. Write it as x(x+4)=0, because x is a common factor.
If the question asks for the positive root, give the one above zero.
The product is zero only when one of the factors is zero.
List the roots in increasing order, joined by “or”.
(x-3)(x+5)=0 needs no expansion. Set each bracket equal to zero. x-3=0 gives x=3. x+5=0 gives x=-5. Check both in a bracket if you want: a solution that makes one bracket zero makes the product zero, whatever the other bracket is. Write both solutions. One of them is usually not enough for full marks on a pure algebra question.
The sign change is the part to say out loud. From x+5=0 you subtract 5, so the solution is negative. Reading the bracket (x+5) as "the solution is 5" skips that step and fails the check, because (5-3)(5+5)=20, not 0.
Method: 1. Leave (x-3)(x+5)=0 in factorised form. Do not expand it. 2. Set each bracket equal to zero in turn. 3. The bracket x-3=0 gives the solution x=3.
x^2-8x+15=0 is ready to factorise. The pair 3 and 5 multiplies to 15 and adds to 8, and both signs are negative because the middle term is negative and the constant is positive: (x-3)(x-5)=0. So x=3 or x=5. If the equation arrives as x^2+3x=40, subtract 40 first. x^2+3x-40=0, then (x+8)(x-5)=0.
An equation with no constant term, such as x^2=6x, is dangerous to divide. Dividing by x gives x=6 and loses x=0, which also works: 0^2=6× 0. Move every term to one side and take out the common factor: x^2-6x=0, so x(x-6)=0. The solutions are x=0 and x=6. The same idea solves x^2-16=0 by a difference of two squares, (x-4)(x+4)=0, or by square-rooting once the square is alone: x=4 or x=-4.
Method: 1. Factorise x^2-8x+15=0 directly, because it is already equal to zero. 2. The numbers -3 and -5 multiply to 15 and add to -8, so the factors are (x-3)(x-5). 3. If the equation is given as x^2+3x=40, subtract 40 from both sides before you factorise.
A rectangle with sides x and x+3 and area 40 gives x(x+3)=40. That is not yet equal to zero. Expand and rearrange: x^2+3x-40=0. Factorise: (x+8)(x-5)=0, so x=-8 or x=5. A length cannot be negative, so reject x=-8. The sides are 5 and 8. Check the area: 5× 8=40.
State the rejected root in a sentence. "Length cannot be negative" is the mark, not a private thought. Then answer the question that was asked. If it wants the side lengths, x=5 is unfinished. If both roots are possible, for example two numbers with no physical limit, give both.
A zero root can be a real answer in a number puzzle and an impossible answer in a length puzzle. Read the story before you reject it. x=0 makes a rectangle of area 0, which is not a useful panel or garden, so you would reject it there and say so. In x^2=6x there is no story, and x=0 stays. The algebra and the context are separate decisions. Write both.
Method: 1. Sides of length x and x+3 with area 40 give the equation x(x+3)=40. 2. That equation is not yet a quadratic equal to zero. 3. Expand and rearrange to x^2+3x-40=0, then factorise.
Factorising only helps when one side of the equation is zero. Many exam equations are written in another shape, such as x^2 = 2x + 15 or x^2 + 3x = 28. Collect everything on one side so that the other side is 0. Subtract 2x + 15 from both sides and you get x^2 - 2x - 15 = 0. Keep the x^2 term positive if you can.
Never divide both sides by x. For x^2 = 7x it looks tidy to cancel and write x = 7, but you lose the solution x = 0. Instead move everything to one side, x^2 - 7x = 0, and take out the common factor to get x(x - 7) = 0. Both brackets can be zero, so x = 0 or x = 7. This type of question is often worth three marks, because it is easy to lose one.
Method: 1. Move every term to one side so the equation equals zero. 2. Factorise the quadratic, taking out a common factor first if there is one. 3. Set each bracket equal to zero. 4. Solve each short equation and check both answers in the original.
Tip: If one side is a product and the other side is not zero, expand and rearrange first. Brackets equal to 6 do not give x equals 6.
In a Higher question the quadratic often starts with a number bigger than one, such as 6x^2 + 5x - 6 = 0. Factorise it using the split-the-middle method: multiply 6 × -6 = -36, find the pair that multiplies to -36 and adds to 5, which is 9 and -4, then factorise in pairs. You should reach (3x - 2)(2x + 3) = 0.
Set each bracket to zero. 3x - 2 = 0 gives x = 2/3 and 2x + 3 = 0 gives x = -3/2. The solutions are fractions, and that is quite normal here. Give them as fractions, not rounded decimals, unless the question asks for decimals. Substitute one solution back into the original equation to make sure the sum is zero.
Method: 1. Multiply the first and last numbers and find a pair that gives the middle number. 2. Split the middle term and factorise in pairs. 3. Set each bracket equal to zero and solve each linear equation. 4. Write the solutions as exact fractions and check one in the original.
Not every quadratic factorises neatly. If you cannot find a pair of numbers that works, or the question asks for answers to a number of decimal places or significant figures, use the quadratic formula instead. Write the equation as ax^2 + bx + c = 0 and pick out a, b and c, keeping their signs. The solutions are x = dfrac-b ± √(b^2 - 4ac)2a.
Put each number in brackets when you substitute, especially a negative b or c, so that b^2 and -4ac come out with the right signs. The ± gives two answers: work out the plus one and then the minus one. A negative number under the square root means there are no real solutions. A quadratic with a number in front of x^2, such as 2x^2 + 7x + 3 = 0, can be solved either way: it factorises as (2x + 1)(x + 3), and the formula gives the same two answers.
Method: 1. Write the equation as ax^2 + bx + c = 0 and list a, b and c with their signs. 2. Substitute into the formula using brackets, and work out b^2 - 4ac first. 3. Find the square root, then work out the plus answer and the minus answer separately. 4. Round only at the end, as the question asks, and check one answer in the original equation.
Tip: If the question says "give your answers to 2 decimal places", the numbers will not factorise: go straight to the formula.
A quadratic from a story gives two solutions, but often only one makes sense. A length cannot be negative, and the number of items must be a positive whole number. After solving, look at each answer and decide whether it fits the context. Write a sentence such as "length cannot be negative so x = 5". The mark for rejecting the other root is easy to lose.
Set up the equation carefully. For two consecutive odd numbers, call the first x and the next one x + 2. For a right-angled triangle use Pythagoras with the unknown sides. For a rectangle, area equals length times width. Draw a quick sketch and label the sides. Then form an equation equal to zero, factorise, solve, and reject any solution that is impossible.
Method: 1. Define the unknown clearly and label a sketch. 2. Write an equation from the story and rearrange it to equal zero. 3. Factorise and solve to get two values. 4. Reject any value that does not fit and give the answer with units.
Tip: Always finish by saying which solution you keep and why. "Length cannot be negative" is enough.
An equation like x^2 = 100 has two solutions, x = 10 and x = -10. A common exam question shows a pupil writing only x = 10 and asks you to explain what is missing. Write that squaring a negative number also gives a positive number, so there are two answers, which you can write as x = ± 10. Differences of squares such as x^2 - 36 = 0 give x = ± 6.
Another popular error question shows someone factorising (x + 2)(x - 7) = 0 and giving x = 2 and x = -7. They have copied the numbers from the brackets with the wrong signs. The correct solutions make each bracket zero, so x = -2 and x = 7. Check a claimed solution by substituting it into the brackets and seeing whether one of them becomes zero.
Method: 1. For a squared term on its own, take the square root and give a plus and a minus answer. 2. For brackets, solve each one separately so it equals zero. 3. Substitute your answers back to test them. 4. In an error question, say what was done and what should have been done.
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