Factoring Quadratics with a Leading Coefficient

A number in front of $x^2$x²

To factor $x^2+7x+12$x² + 7x + 12, you look for two numbers that multiply to 12 and add to 7. They are 3 and 4, so $x^2+7x+12=\left(x+3\right)\left(x+4\right)$x² + 7x + 12 = (x + 3)(x + 4).

Now try $2x^2+7x+3$2x² + 7x + 3. The 2 in front of $x^2$x² is the leading coefficient, and it changes the search. The pair you want must still add to 7, the coefficient of x, but it must multiply to 2 × 3 = 6, the leading coefficient times the constant term. For $a{x}^2+bx+c$ax² + bx + c the pair multiplies to a × c and adds to b.

The rectangle (2x + 1) by (x + 3). The two middle pieces, 6x and x, add to 7x. They also multiply to $6x\times x=6x^2$6x × x = 6x², the same as the two corner pieces, $2x^2\times 3=6x^2$2x² × 3 = 6x².

Why a times c

Look at the rectangle whose area is $2x^2+7x+3$2x² + 7x + 3. Its four pieces are $2x^2$2x², 6x, x and 3. The two corner pieces on one diagonal, $2x^2$2x² and 3, multiply to $6x^2$6x². The two middle pieces on the other diagonal, 6x and x, multiply to $6x^2$6x² as well. That is true of every such rectangle: each piece is one side piece times another, so each diagonal multiplies all four side pieces, 2x, 1, x and 3, together.

So the two numbers in front of the middle pieces multiply to 2 × 3 = 6, and they add to 7 because the middle pieces make up the 7x. The pairs that multiply to 6 are 1 and 6, and 2 and 3. Only 1 + 6 = 7, so the pair is 1 and 6.

Split the middle term, then group

Use the pair to split the middle term: 7x = x + 6x. The expression has not changed, since x + 6x is still 7x, but now it has four terms: $2x^2+7x+3=2x^2+x+6x+3$2x² + 7x + 3 = 2x² + x + 6x + 3.

Factor the four terms in two pairs, as in Factoring by Grouping. The first pair, $2x^2+x$2x² + x, has a common factor of x, so it is x(2x + 1). The second pair, 6x + 3, has a common factor of 3, so it is 3(2x + 1). Now $2x^2+x+6x+3=x\left(2x+1\right)+3\left(2x+1\right)$2x² + x + 6x + 3 = x(2x + 1) + 3(2x + 1).

The same bracket, 2x + 1, appears in both parts. Take it out as a common factor, and x and 3 are left in the other bracket: $2x^2+7x+3=\left(2x+1\right)\left(x+3\right)$2x² + 7x + 3 = (2x + 1)(x + 3).

If the two brackets do not match, the split or one of the common factors is wrong, so check them before going on.

Check by expanding

Multiply the brackets back out: $\left(2x+1\right)\left(x+3\right)=2x^2+6x+x+3=2x^2+7x+3$(2x + 1)(x + 3) = 2x² + 6x + x + 3 = 2x² + 7x + 3. The four products are the four pieces of the rectangle, and they add back to the expression you started with.

When the signs are negative

The method works for any signs. For $3x^2-10x+8$3x² − 10x + 8, the pair must multiply to 3 × 8 = 24 and add to −10. Two negative numbers multiply to a positive number, so try negative pairs: −4 and −6 multiply to 24 and add to −10.

Split the middle term: $3x^2-4x-6x+8$3x² − 4x − 6x + 8. The first pair gives x(3x − 4). For the second pair, take out −2, not 2, so that the same bracket appears: −6x + 8 = −2(3x − 4). Then $3x^2-10x+8=x\left(3x-4\right)-2\left(3x-4\right)=\left(3x-4\right)\left(x-2\right)$3x² − 10x + 8 = x(3x − 4) − 2(3x − 4) = (3x − 4)(x − 2).

The usual mistakes

Looking for a pair that multiplies to c alone. For $2x^2+7x+3$2x² + 7x + 3 that would be a pair for 3, but the 2 in front of $x^2$x² changes the target to 2 × 3 = 6.

Putting the numbers in the wrong brackets. (2x + 3)(x + 1) also uses 2x, x, 3 and 1, but it expands to $2x^2+5x+3$2x² + 5x + 3, so its middle term is wrong. Expanding your answer shows at once which bracket each number belongs in.