L7 · Two-variable factoring forms

Learning goal

“I can factor two-variable trinomials ax^2 + bxy + cy^2 with the same pair and ac methods - treating the y-parts as part of every term - plus two-variable common factors and differences of squares, verifying by expanding.”

Two-variable polynomial — a polynomial in two variables (e.g. x^2 + 5xy + 6y^2); its terms carry x-parts, y-parts, or both.

xy-term — the middle term of ax^2 + bxy + cy^2; in the factored form it is built from BOTH cross partials, exactly as in L2.

y-multiple constant — in (x + 3y)(x + 2y) the "constants" are 3y and 2y: they multiply to 6y^2 (the c-term) and their cross partials with x build the xy-term.

Factored form — (review) product form; in this lesson the binomials mix x and y.

Worked example

Factor x^2 + 5xy + 6y^2. Step 1: Set y = 1 mentally: the skeleton is x^2 + 5x + 6, whose pair is 2 and 3 (L5). Step 2: Restore the y's: the constants become 2y and 3y. Check both conditions in the two-variable world: (2y)(3y) = 6y^2 = the c-term; cross partials: x*3y + x*2y = 3xy + 2xy = 5xy = the middle. Step 3: x^2 + 5xy + 6y^2 = (x + 2y)(x + 3y). Step 4: Verify by expanding fully: x^2 + 3xy + 2xy + 6y^2.

Check yourself — explain it

Why do the constants have to carry y (2y, 3y) rather than being plain 2 and 3?

Paired problem — example still visible

Factor x^2 - 7xy + 12y^2 -> (x - 3y)(x - 4y) (pair -3, -4 with y attached: product +12y^2, cross sum -7xy)

Faded completion — last step is yours

"x^2 + 4xy - 12y^2 = (x + 6y)(x - ___)" -> 2y ((6y)(-2y) = -12y^2; cross: -2xy + 6xy = +4xy)

Ready to practise?

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

L7 · Two-variable factoring forms