1Cademy - Showing $$u^2 - 9uv - 12v^2$$ Is a Prime Trinomial

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Factoring Trinomials of the Form $x^2 + bxy + cy^2$

Example

Showing $u^2 - 9uv - 12v^2$ Is a Prime Trinomial

Attempt to factor $u^2 - 9uv - 12v^2$ by applying the two-variable trinomial factoring strategy. Because the first term is $u^2$ , each binomial factor begins with $u$ . Because the last term contains $v^2$ , the second term of each binomial must include $v$ . The last term of the trinomial is negative ( $-12v^2$ ), so the factors must have opposite signs.

Step 1 — Set up two binomials: $(u\_v)(u\_v)$ , where the blanks will be filled with coefficients of $v$ and the signs will be opposite.

Step 2 — Find two numbers that multiply to $-12$ and add to $-9$ . List all factor pairs of $-12$ and check their sums:

Factors of $-12$	Sum of factors
$1, -12$	$1 + (-12) = -11$
$-1, 12$	$-1 + 12 = 11$
$2, -6$	$2 + (-6) = -4$
$-2, 6$	$-2 + 6 = 4$
$3, -4$	$3 + (-4) = -1$
$-3, 4$	$-3 + 4 = 1$

None of the factor pairs produce a sum of $-9$ .

Since no pair of integers has a product of $-12$ and a sum of $-9$ , the trinomial $u^2 - 9uv - 12v^2$ cannot be factored — it is a prime trinomial. This example shows that two-variable trinomials of the form $x^2 + bxy + cy^2$ can also be prime: the same exhaustive factor-pair check used for single-variable trinomials applies, and when no pair works, the trinomial is prime regardless of how many variables it contains.

Updated 2026-04-29

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