An SAT Math micro-topic under Factoring quadratic and polynomial expressions (Advanced Math). Free to read — no account needed.
Factoring ax2+bx+c by grouping begins by naming the coefficients: a is the number in front of x2, b is the number in front of x, and c is the constant term. The one search that drives the method is to find two numbers whose product is a×c and whose sum is b.
Those two numbers let you split the middle term bx into two terms, turning the three-term quadratic into four terms you can group into pairs. Factor each pair, and the two pairs will share a common bracket that you factor out.
With a=1, factor x2+2x−8: here a=1, b=2, c=−8, so a×c=−8. Two numbers that multiply to −8 and add to 2 are 4 and −2. Split the middle term: x2+4x−2x−8. Group in pairs: x(x+4)−2(x+4). Both share (x+4), so the factors are (x+4)(x−2).
With a=1 and a negative c, factor 3x2+2x−5: here a=3, b=2, c=−5, so a×c=−15. Two numbers that multiply to −15 and add to 2 are 5 and −3. Split: 3x2−3x+5x−5. Group: 3x(x−1)+5(x−1). Both share (x−1), so the factors are (x−1)(3x+5).
If every term shares a common factor, pull it out first to keep the numbers small. 6x2+4x−10=2(3x2+2x−5), and then you factor the bracket by grouping as above.
Worked examples
Factor x2+7x+12. Here a=1, b=7, c=12. Two numbers that multiply to a×c=12 and add to 7 are 3 and 4. Split the middle term: x2+3x+4x+12. Group and factor: x(x+3)+4(x+3)=(x+3)(x+4).
Factor 6x2+13x+6. Here a=6, b=13, c=6. Two numbers that multiply to a×c=36 and add to 13 are 4 and 9. Split the middle term: 6x2+4x+9x+6. Group and factor: 2x(3x+2)+3(3x+2)=(3x+2)(2x+3).
Factor 2x2+x−1. Here a=2, b=1, c=−1. Two numbers that multiply to a×c=−2 and add to 1 are −1 and 2. Split the middle term: 2x2−x+2x−1. Group and factor: x(2x−1)+1(2x−1)=(2x−1)(x+1).
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