Factoring Trinomials

Algebra 1 · Axiom Academy

Run FOIL in reverse: turn x^2+bx+c back into (x+p)(x+q) by finding two numbers with the right product and sum. Multiply out (x+p)(x+q) and you always land on x^2+(p+q)x+pq : the constant is the two numbers multiplied , the middle coefficient is those same two numbers added . So to factor x^2+bx+c , run that backward — hunt for two numbers whose product is c and whose sum is b . The area model makes it visible: a rectangle of area x^2+7x+12 has side lengths (x+3) and (x+4) . Multiply out — the pattern to reverse The two conditions to search for 2. Find the Pair, Then Fix the Signs Turn the idea into a procedure: list every pair of integers that multiplies to c , then keep the one pair that also adds to b . For x^2+7x+12 the factor pairs of 12 are , , and — and only 3+4 hits 7 . Watch the search run. Constant positive, middle positive. x^2+7x+12=(x+3)(x+4) . Constant positive, middle negative. x^2-8x+15=(x-3)(x-5) . Constant negative. x^2-5x-14=(x-7)(x+2) . A positive product means the two numbers share a sign; a negative product means they differ. The sum then tells you which sign dominates. Check every answer by expanding Factor x^2+3x-10 . Among the pairs that multiply to -10 , the pair 5 and -2 also adds to 3 : Multiply it back out to be sure nothing slipped: 3. When the Leading Coefficient Isn't 1

This is the written version of the interactive lesson above. See the full Algebra 1 course.