Factorising quadratics with a leading coefficient
Factorise quadratics such as 6x² + 11x + 3, where the x² coefficient is not 1, by splitting the middle term and grouping.
What a learner can do afterwards
- Split the middle term of 6x² + 11x + 3 using the pair that multiplies to 18 and adds to 11
- Factorise by grouping, check proposed factors by expansion or substitution, and recover a missing bracket constant
- Choose a useful factorising method by checking for a common factor or a difference-of-squares pattern
1 · Read
First check for a common factor or a special pattern. A difference of squares uses A² - B² = (A + B)(A - B); the cross terms cancel. For other quadratics, splitting the middle term is useful. Brackets (x + p)(x + q) have leading term x² because x times x is x². In any two linear brackets, multiply their x coefficients to find the leading coefficient.
For ax² + bx + c with integer coefficients and a not zero, find integers m and n with mn = ac and m + n = b. Replace bx with mx + nx. The sum condition keeps the expression equal to the original. The product condition lets the two groups reveal a shared bracket. If no signed integer pair meets both targets, there is no factorisation into two linear brackets with integer coefficients.
Factorise 2x² + 9x + 4. The product target is 2 times 4 = 8. The pair 8 and 1 adds to 9. Split and group: 2x² + 8x + x + 4 = 2x(x + 4) + 1(x + 4). Take out the shared bracket: (2x + 1)(x + 4). To check, multiply all four pairs: 2x², 8x, x and 4. These collect to 2x² + 9x + 4.
Remove a common factor from every term first. For 8x² + 24x + 16, take out 8 to get 8(x² + 3x + 2) = 8(x + 1)(x + 2). A common factor can contain x: 3x² + 6x = 3x(x + 2). Use the largest shared factor. With a negative leading term, taking out a negative factor can help: -8x² - 24x - 16 = -8(x + 1)(x + 2). Keep that outside factor in the answer.
Factorise 2x² + x - 3. The product target is -6. The pair 3 and -2 has product -6 and sum 1. Split: 2x² + 3x - 2x - 3. Group: x(2x + 3) - 1(2x + 3). The result is (x - 1)(2x + 3). Check the four products: 2x² + 3x - 2x - 3 = 2x² + x - 3.
A rectangle's area is the product of its side lengths. If its area is 3x² + 13x + 4 square units, factorising gives (3x + 1)(x + 4). If one side is 3x + 1 units, the other is x + 4 units. Use values of x that make both lengths positive. Check the side expressions by multiplying them back to the area.
Why use ac? Expanding (Ax + B)(Cx + D) gives a = AC, b = AD + BC and c = BD. The cross-term coefficients are AD and BC. Their sum is b, and their product is (AD)(BC) = (AC)(BD) = ac. This also works backward: a supplied split pair tells you b by addition.
The sum rebuilds the middle coefficient. The product links the leading and constant coefficients.
Sometimes two brackets are opposites. Since 3 - x = -(x - 3), the expression 2x(x - 3) - 5(3 - x) becomes 2x(x - 3) + 5(x - 3). Now take out x - 3 to get (2x + 5)(x - 3). Change the outside sign when you reverse every sign inside a bracket.
A full check compares all coefficients after expansion. Matching only at x = 0 checks the constant, not the whole quadratic. A mismatch at any input disproves a proposed factorisation. You can also find a missing bracket constant: (3x + 2)(x + t) has x coefficient 3t + 2. If that coefficient is 14, solve 3t + 2 = 14 to get t = 4.
Expansion can check a result or find a missing number.
Multiply a and c, find the pair that multiplies to that product and adds to b, split the middle term, group, take out the shared bracket, then check by expanding.
The practice questions and readings were written for this dataset. The referenced videos are by Khan Academy.
More to read
Finding a split that can be grouped
Try 3x² + 14x + 8. Multiplying the leading coefficient and constant gives 24. The pair 12 and 2 has product 24 and sum 14.
Replace 14x with 12x + 2x. The sum keeps the expression equal to the original: 3x² + 12x + 2x + 8.
Take a common factor from each pair: 3x(x + 4) + 2(x + 4).
Both terms now contain x + 4. Taking it out gives (3x + 2)(x + 4). Expanding produces 3x² + 12x + 2x + 8, which collects to the starting quadratic.
The two pair checks have different jobs. The sum rebuilds the middle term. The product makes this grouping method work. A split can have the right sum without having the right product.
Keeping common factors and signs
Look across every term before splitting a quadratic. For -5x² - 20x - 15, taking out -5 leaves x² + 4x + 3. The pair 1 and 3 then gives -5(x + 1)(x + 3). The outside -5 stays in the final answer.
A shared factor can include a variable. In 6x² + 15x, the largest shared factor is 3x. Dividing each term by 3x leaves 2x + 5, so the factorisation is 3x(2x + 5).
Signs also matter when grouping. Consider 2x(x - 4) - 3(4 - x). The brackets are opposites: 4 - x = -(x - 4). Therefore the second term is +3(x - 4). The shared bracket is now visible, giving (2x + 3)(x - 4).
Changing all the signs inside a bracket changes its sign. Move that minus sign outside as well to keep the expression equal.
Checking the whole quadratic
To check a factorisation, multiply every term in one bracket by every term in the other. For (4x + 3)(x + 4), the four products are 4x², 16x, 3x and 12. Collecting them gives 4x² + 19x + 12. Check the leading coefficient, middle coefficient and constant against the original expression.
A check at one input has limits. At x = 0, the expression only reveals its constant. Two different quadratics can share that constant. A mismatch at any input shows that two expressions are not equal, but one matching input does not prove equality.
Expansion can also find a missing number. In (4x + 3)(x + r), the middle coefficient is 4r + 3. If the required middle coefficient is 23, solve 4r + 3 = 23. This gives r = 5. Then check the other coefficients too.
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