L6 · Factoring ax^2 + bx + c (a != 1) and difference of squares

Learning goal

“I can factor ax^2 + bx + c by splitting the middle term with a pair that multiplies to ac and adds to b, and I can recognize and factor a difference of squares - explaining why it is the b = 0 case of trinomial factoring.”

Leading coefficient — the coefficient of the x^2 term (the a in ax^2 + bx + c).

Split the middle term — rewrite bx as two terms whose coefficients multiply to ac and add to b, so the four terms can be grouped.

Difference of squares — a^2 - b^2 form: (a - b)(a + b). As a trinomial it is ax^2 + 0x - c: the pair that works is +k and -k, summing to 0.

ac product — the product of the leading coefficient and the constant; the split pair multiplies to ac (not just to c - this is the one change from L5).

Worked example

Factor 2x^2 + 7x + 3. Step 1: GCF? None. a = 2, c = 3, so ac = 6. Need a pair multiplying to 6 and adding to b = 7: 6 and 1. Step 2: Split: 2x^2 + 6x + x + 3. Step 3: Group: 2x(x + 3) + 1(x + 3). The brackets MATCH - that is the signal the split was right. Step 4: 2x^2 + 7x + 3 = (2x + 1)(x + 3). Step 5: Verify by expanding: 2x^2 + 6x + x + 3.

Check yourself — explain it

Why must the two grouped brackets be identical?

Paired problem — example still visible

Factor 3x^2 + 10x + 8 -> (3x + 4)(x + 2) (ac = 24; pair 6, 4: 3x^2 + 6x + 4x + 8 = 3x(x + 2) + 4(x + 2))

Faded completion — last step is yours

"5x^2 + 13x + 6: ac = 30; the pair is 10 and 3; split: 5x^2 ___ + 3x + 6; group: 5x(x + 2) + 3(x + 2)" -> + 10x

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10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.

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Grade 10 — Mathematics 10C › C10U06 Polynomials: Multiply & Factor

L6 · Factoring ax^2 + bx + c (a != 1) and difference of squares