Example

Factoring 4y2+8y+34y^2 + 8y + 3 Using the ac Method

Factor the trinomial 4y2+8y+34y^2 + 8y + 3 using the ac method.

Step 1 — Factor any GCF. The terms 4y24y^2, 8y8y, and 33 share no common factor.

Step 2 — Find the product acac. Here a=4a = 4 and c=3c = 3, so ac=43=12ac = 4 \cdot 3 = 12.

Step 3 — Find two numbers mm and nn that multiply to 12 and add to 8. Since both the product and sum are positive, both numbers must be positive. The pair m=2m = 2 and n=6n = 6 works: 26=122 \cdot 6 = 12 and 2+6=82 + 6 = 8.

Step 4 — Split the middle term 8y8y into 2y+6y2y + 6y: 4y2+8y+3=4y2+2y+6y+34y^2 + 8y + 3 = 4y^2 + 2y + 6y + 3

Step 5 — Factor by grouping. Group into two pairs and factor the GCF from each: 2y(2y+1)+3(2y+1)2y(2y + 1) + 3(2y + 1) Both groups share the common binomial (2y+1)(2y + 1). Factor it out: (2y+1)(2y+3)(2y + 1)(2y + 3)

Step 6 — Check by multiplying: (2y+1)(2y+3)=4y2+6y+2y+3=4y2+8y+3(2y + 1)(2y + 3) = 4y^2 + 6y + 2y + 3 = 4y^2 + 8y + 3

The factored form is (2y+1)(2y+3)(2y + 1)(2y + 3).

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Updated 2026-04-29

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