Factoring Quadratics

Expanding, run backward

Expanding (x + 2)(x + 3) gives $x^2+5x+6$x² + 5x + 6. Factoring a quadratic goes the other way: start from $x^2+5x+6$x² + 5x + 6 and find the two brackets it came from.

Think of $x^2+5x+6$x² + 5x + 6 as the area of a rectangle with sides x + ? and x + ?. The $x^2$x² is the square in the corner, x by x. The 6 is the small piece in the opposite corner, whose sides are the two unknown numbers, so the two numbers multiply to 6. The two strips are those numbers times x, and together they make 5x, so the two numbers add to 5.

The $x^2$x² piece is the x by x square. The two unknown numbers are the other parts of the sides.

Find the pair

List the pairs of whole numbers that multiply to 6: 1 × 6 and 2 × 3. Their sums are 1 + 6 = 7 and 2 + 3 = 5. Only 2 and 3 add to 5, so they are the numbers.

So $x^2+5x+6=\left(x+2\right)\left(x+3\right)$x² + 5x + 6 = (x + 2)(x + 3). Check by expanding: the four pieces are $x^2$x², 3x, 2x and 6, and 3x + 2x = 5x.

Drag the corner to change the two numbers. The strips add to the middle term and the corner piece is their product. Find the split that makes both 5x and 6.

A larger number at the end

Factor $x^2+7x+12$x² + 7x + 12. The two numbers multiply to 12 and add to 7. The pairs that multiply to 12 are 1 × 12, 2 × 6 and 3 × 4, whose sums are 13, 8 and 7. Only 3 + 4 = 7, so $x^2+7x+12=\left(x+3\right)\left(x+4\right)$x² + 7x + 12 = (x + 3)(x + 4).

Listing the pairs in order, starting from 1, makes sure none is missed.

When a number is negative

The same rule works with negative numbers. Factor $x^2+4x-5$x² + 4x − 5. The two numbers multiply to −5 and add to 4. A negative product means one number is positive and the other is negative. The pairs are 5 and −1, with sum 4, and −5 and 1, with sum −4. So the numbers are 5 and −1, and $x^2+4x-5=\left(x+5\right)\left(x-1\right)$x² + 4x − 5 = (x + 5)(x − 1).

When the end number is positive but the middle is negative, both numbers are negative. For $x^2-7x+12$x² − 7x + 12 the numbers multiply to 12 and add to −7, so they are −3 and −4: $x^2-7x+12=\left(x-3\right)\left(x-4\right)$x² − 7x + 12 = (x − 3)(x − 4).

Check with a number. When x = 10, $x^2-7x+12$x² − 7x + 12 is 100 − 70 + 12 = 42, and (x − 3)(x − 4) is 7 × 6 = 42.

The usual mistakes

Checking only the product. For $x^2+7x+12$x² + 7x + 12, the pair 2 and 6 multiplies to 12, but adds to 8, which belongs to $x^2+8x+12$x² + 8x + 12. The pair must pass both checks.

Swapping the two jobs. The numbers add to give the x term and multiply to give the number at the end, not the other way round.