L5 · Factoring out a common factor
“I can factor a polynomial by removing its greatest common factor - coefficients and variable powers together - and I can prove the factoring is right, and complete, by expanding it back.”
Factor 6x^2 + 9x. Step 1: GCF of the coefficients: GCF(6, 9) = 3. Step 2: Variable part: x^2 and x share the lowest power present in both, which is x^1 = x. (Not x^2 - the 9x term does not have an x^2 to give.) Step 3: The GCF is 3x. Divide each term by it: 6x^2 / 3x = 2x; 9x / 3x = 3. Step 4: Write the product: 3x(2x + 3). Step 5 - the non-negotiable check: expand back. 3x · 2x = 6x^2; 3x · 3 = 9x. The original is recovered, and the bracket (2x + 3) has no common factor left: fully factored.
Why is the variable part of the GCF x, and not x^2?
Factor 8x^3 - 4x^2, and check by expanding. -> 4x^2(2x - 1) (GCF: 4 from the coefficients; x^2 is the lowest power in both terms. Check: 4x^2 · 2x = 8x^3; 4x^2 · (-1) = -4x^2)
"10x^2 - 15x = 5x(___ - 3). What is the missing term in the bracket?" -> 2x (10x^2 / 5x = 2x; and -15x / 5x = -3)
10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.