1Cademy - Factoring $$4u^3 + 16u^2

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Strategy to Factor Polynomials Completely

Example

Factoring $4u^3 + 16u^2 - 20u$

Factor $4u^3 + 16u^2 - 20u$ completely by first extracting a monomial GCF that includes a variable, and then factoring the resulting trinomial.

Step 1 — Check for a GCF: The three terms $4u^3$ , $16u^2$ , and $-20u$ share a numerical factor of $4$ and a variable factor of $u$ (the lowest power of $u$ among all terms). The GCF is $4u$ . Factor it out: $4u(u^2 + 4u - 5)$ .

Step 2 — Classify the expression inside the parentheses: The expression $u^2 + 4u - 5$ is a trinomial with a leading coefficient of 1, so the "undo FOIL" method applies. The constant term $-5$ is negative, so the two factor-pair numbers must have opposite signs. Set up two binomials: $(u\quad)(u\quad)$ .

Step 3 — Find two numbers whose product is $-5$ and whose sum is 4. List the factor pairs of $-5$ and check their sums:

Factors of $-5$	Sum of factors
$-1, 5$	$-1 + 5 = 4$ ✓
$1, -5$	$1 + (-5) = -4$

The pair $-1$ and $5$ works.

Step 4 — Write the fully factored form: $4u(u - 1)(u + 5)$ .

Step 5 — Check by multiplying: $4u(u - 1)(u + 5) = 4u(u^2 + 5u - u - 5) = 4u(u^2 + 4u - 5) = 4u^3 + 16u^2 - 20u$ ✓.

The completely factored form is $4u(u - 1)(u + 5)$ . Unlike previous examples where the GCF was purely numerical, here the GCF is $4u$ — a monomial containing both a number and a variable. When the original polynomial has degree 3 or higher, the GCF may include a variable factor, and after extracting it the remaining expression can still be a trinomial that factors further. The result is a product of three factors: one monomial and two binomials.

Updated 2026-04-21

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OpenStax Elementary Algebra

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