Factoring quadratic and polynomial expressions

How the SAT tests this, with worked examples — plus the 6 skills it covers. Free to read, no account needed.

Factoring rewrites a sum as a product, which matters because a product equalling zero tells you something a sum never can: at least one factor must be zero.
For x2+bx+cx^2 + bx + c, you need two numbers that multiply to cc and add to bb. The area model shows why.
et_factoring_area_model.png
The two inner rectangles are the xx terms, and they add to give the middle term, while the corner rectangle is the constant. So (x+2)(x+3)(x+2)(x+3) gives x2+5x+6x^2 + 5x + 6 because 2×3=62 \times 3 = 6 and 2+3=52 + 3 = 5.
Always check for a common factor first. 2x2+10x+122x^2 + 10x + 12 looks awkward until you take out the 22, leaving 2(x2+5x+6)2(x^2 + 5x + 6), which factors easily.
Two patterns are worth recognising on sight. The difference of squares, a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b), and the perfect square, a2+2ab+b2=(a+b)2a^2 + 2ab + b^2 = (a+b)^2. Both appear far more often than random quadratics.
When the leading coefficient is not 11, factoring by grouping works: split the middle term into two parts that multiply to acac and add to bb, then factor in pairs.

Worked examples

Factor x2+7x+12x^2 + 7x + 12.
Find two numbers multiplying to 1212 and adding to 77: those are 33 and 44
x2+7x+12=(x+3)(x+4){x^2 + 7x + 12 = (x+3)(x+4)}
Factor x2−49x^2 - 49.
  • A(x−7)(x+7)(x-7)(x+7)
  • B(x−7)2(x-7)^2
This is a difference of squares, since 49=7249 = 7^2, so (A) is correct. (B) would expand to x2−14x+49x^2 - 14x + 49.
Solve x2+5x+6=0x^2 + 5x + 6 = 0.
Factor first, then use the fact that a product is zero only if a factor is
(x+2)(x+3)=0  ⇒  x=−2 or x=−3{(x+2)(x+3) = 0 \;\Rightarrow\; x = -2 \text{ or } x = -3}
This is why factoring is worth doing: it converts one quadratic into two trivial linear equations.

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All the skills in Factoring quadratic and polynomial expressions