1Cademy - Factoring $$-10y^4 - 55y^3

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

Example

Factoring $-10y^4 - 55y^3 - 60y^2$

To factor $-10y^4 - 55y^3 - 60y^2$ completely, first check for a greatest common factor (GCF). The three terms share a common numerical factor of $5$ and a variable factor of $y^2$ . Because the leading coefficient is negative, factor out the negative GCF $-5y^2$ , which yields $-5y^2(2y^2 + 11y + 12)$ . Next, factor the resulting trinomial $2y^2 + 11y + 12$ using the trial and error method. The only factor pair for $2y^2$ is $y \cdot 2y$ , and the positive factor pairs for $12$ are $1 \cdot 12$ , $2 \cdot 6$ , and $3 \cdot 4$ . Testing the combinations shows that $(y + 4)(2y + 3)$ produces the correct middle term $11y$ (since $3y + 8y = 11y$ ). Combine this with the GCF to write the completely factored form: $-5y^2(y + 4)(2y + 3)$ . Finally, verify by multiplying: $-5y^2(y + 4)(2y + 3) = -5y^2(2y^2 + 3y + 8y + 12) = -5y^2(2y^2 + 11y + 12) = -10y^4 - 55y^3 - 60y^2$ .

Updated 2026-05-26

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OpenStax Intermediate Algebra

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