1Cademy - Factoring $$6b^2

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Factoring Trinomials of the Form $ax^2 + bx + c$ Using Trial and Error

Example

Factoring $6b^2 - 13b + 5$

Factor $6b^2 - 13b + 5$ completely using the trial and error method. This trinomial has a leading coefficient with multiple factor pairs, which increases the number of combinations to test.

Step 1 — Write in descending order. The trinomial $6b^2 - 13b + 5$ is already in descending order.

Step 2 — Find factor pairs of the first term. The term $6b^2$ can be factored into first-degree terms in two ways: $b \cdot 6b$ or $2b \cdot 3b$ .

Step 3 — Find factor pairs of the last term and consider signs. The last term $5$ is positive, so its factors must have the same sign — both positive or both negative. Since the middle coefficient $-13$ is negative, both factors must be negative: $-1$ and $-5$ .

Step 4 — Test all combinations. With two factor pairs for the first term and one factor pair for the last term (in two possible arrangements each), there are four combinations to test:

Possible factors	Product
$(b - 1)(6b - 5)$	$6b^2 - 11b + 5$
$(b - 5)(6b - 1)$	$6b^2 - 31b + 5$
$(2b - 1)(3b - 5)$	$6b^2 - 13b + 5$ ✓
$(2b - 5)(3b - 1)$	$6b^2 - 17b + 5$

The combination $(2b - 1)(3b - 5)$ produces the correct middle term $-13b$ .

Step 5 — Check by multiplying: $(2b - 1)(3b - 5) = 6b^2 - 10b - 3b + 5 = 6b^2 - 13b + 5$ ✓

The factored form is $(2b - 1)(3b - 5)$ . Unlike previous examples where the leading coefficient (such as 3) had only one factor pair, $6b^2$ can be split as either $b \cdot 6b$ or $2b \cdot 3b$ , doubling the number of trial factorizations. This demonstrates that as the leading coefficient gains more factor pairs, the trial and error process requires testing more combinations.

Updated 2026-04-21

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OpenStax Elementary Algebra

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