Solving quadratic equations by factorising
Rearrange a quadratic equation so one side is zero, factorise it, then use the fact that a product is zero only when one factor is zero to read off both solutions.
What a learner can do afterwards
- Rearrange x² + 5x = 14 into x² + 5x - 14 = 0 before factorising
- Apply the zero product rule only when a product equals zero, and find all distinct solutions
- Reject a solution that cannot fit the context, such as a negative length
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Factorise quadratics with the pair method, special patterns or splitting the middle term to solve equations. A product is zero only when a factor is zero, so solve a zero case from each bracket. For 2x²+7x+3=0, the pair's product is the leading coefficient times the constant: 2 times 3=6. Its sum is 7, so use 6 and 1. Split: 2x²+6x+x+3=0. Factor each pair: 2x(x+3)+1(x+3)=0. Take out x+3: (2x+1)(x+3)=0. Solve 2x+1=0 or x+3=0: x=-0.5 or x=-3. Add: -0.5+(-3)=-3.5.
If AB = 0, then A = 0 or B = 0. Solve these alternative cases separately. From (x - 3)(x + 4) = 0, solve x - 3 = 0 or x + 4 = 0, giving x = 3 or x = -4. Do not set the factors equal to each other. If a product equals a nonzero number, neither factor can be zero, and the zero product rule does not apply.
Solve x² + 3x - 10 = 0. The pair 5 and -2 has product -10 and sum 3, so (x + 5)(x - 2) = 0. Set each factor to zero separately: x + 5 = 0 or x - 2 = 0. Thus x = -5 or x = 2. Check both in the original equation: 25 - 15 - 10 = 0 and 4 + 6 - 10 = 0.
The equation 3x squared = 12x is not yet set to zero. Move every term to one side: 3x squared - 12x = 0. Take out the common factor 3x: 3x(x - 4) = 0. The rule gives x = 0 or x = 4. Do not divide both sides by x here: that assumes x is not zero and drops the solution x = 0.
A garden has width x metres, length x + 3 metres and area 40 square metres. Area = width times length, so x(x + 3) = 40. Rearrange: x² + 3x - 40 = 0. The pair 8 and -5 gives (x + 8)(x - 5) = 0, so x = -8 or x = 5. A length cannot be negative. Keep x = 5: the garden is 5 by 8 metres, and 5 times 8 = 40.
Two factors may give the same solution. For (x - 9)² = 0, both factors are x - 9, so both zero cases give x = 9. Count distinct values, not brackets: this equation has one distinct solution. A nonzero constant factor does not add a solution, since that constant cannot equal zero.
You can build an equation from its roots. A root r gives a factor x - r. Roots 4 and -8 give (x - 4)(x + 8) = 0, which expands to x² + 4x - 32 = 0. For roots r and s, expanding (x - r)(x - s) gives x² - (r + s)x + rs. Thus the middle coefficient is the negative of the roots' sum.
Translate a number story before solving. If a number is 6 less than twice its square, write x = 2x² - 6. Move all terms to one side: 2x² - x - 6 = 0. Factorise: (2x + 3)(x - 2) = 0, so x = -3/2 or x = 2. If the story asks for the positive number, keep 2. Read 'less than' carefully: subtract 6 from twice the square.
A difference of squares uses A² - B² = (A - B)(A + B). Expanding gives A² + AB - AB - B², so the cross terms cancel. For x² - 121 = 0, 121 = 11², giving (x - 11)(x + 11) = 0. Solve x - 11 = 0 or x + 11 = 0: x = 11 or x = -11. Both check because 11² = (-11)² = 121.
To solve x² = 11x - 24, subtract 11x and add 24 on both sides: x² - 11x + 24 = 0. For a quadratic with x² coefficient 1, find two numbers whose product is the constant and whose sum is the x coefficient. Here -3 and -8 have product 24 and sum -11, so (x - 3)(x - 8) = 0. Solve each zero case: x = 3 or x = 8. Check in the original: 9 = 33 - 24 and 64 = 88 - 24.
A squared bracket means two identical brackets. For (x + 3)² = 25, expand (x + 3)(x + 3): x² + 3x + 3x + 9 = 25. Combine like terms and subtract 25 from both sides: x² + 6x - 16 = 0. The pair 8 and -2 has product -16 and sum 6, so (x + 8)(x - 2) = 0. Solve x + 8 = 0 or x - 2 = 0: x = -8 or x = 2. Their sum is -8 + 2 = -6. Check: (-8 + 3)² = 25 and (2 + 3)² = 25.
Move every term to one side so the other is zero, factorise, set each factor to zero, and keep the solutions that fit the context.
The practice questions and readings were written for this dataset. The referenced video is by Khan Academy.
More to read
Why zero matters
A factorisation is another way to write an expression. It can be useful whether or not the expression appears in an equation.
The zero product rule has a separate condition: the product must equal zero. For x² + 2x = 15, subtract 15 to get x² + 2x - 15 = 0. Factorising gives (x + 5)(x - 3) = 0. Solve x + 5 = 0 or x - 3 = 0, so x = -5 or x = 3.
Check each value in the original equation. For -5, 25 - 10 = 15. For 3, 9 + 6 = 15.
From a split to solutions
Consider 2x² + 7x - 4 = 0. The product target is -8. Splitting 7x into 8x - x gives 2x(x + 4) - 1(x + 4) = 0, so (2x - 1)(x + 4) = 0.
There are two alternative cases. From 2x - 1 = 0, x = 1/2. From x + 4 = 0, x = -4. A solution need not be an integer.
Do not equate the two factors. Their product being zero tells you that at least one factor is zero; it does not tell you that the factors are equal.
Before dividing by a variable
Division by x is allowed only when x is nonzero. If you divide an equation without checking that condition, you may remove a valid solution.
For 4x² = 20x, factorising gives 4x(x - 5) = 0. The solutions are 0 and 5. Dividing by 4x would retain 5 but exclude 0.
Not every quadratic equation has zero as a solution. You can handle the zero case separately and then divide in the nonzero case. Factorising often keeps both cases visible.
Using roots to build an equation
A factor x - r becomes zero at x = r. To make a quadratic with roots -3 and 7, use (x + 3)(x - 7) = 0. Expanding gives x² - 4x - 21 = 0.
The reverse check is to solve the two zero-factor cases. They give -3 and 7 again.
If both roots are the same value r, the factors repeat: (x - r)² = 0. There are two identical factors but only one distinct solution.
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