Example

Factoring t2−11t+28t^2 - 11t + 28

Factor t2−11t+28t^2 - 11t + 28 by applying the trinomial factoring strategy. The constant term 2828 is positive and the middle coefficient −11-11 is negative, so both numbers in the factor pair must be negative. Step 1 — Set up two binomials with first terms tt: (t)(t)(t\quad)(t\quad). Step 2 — Find two negative numbers that multiply to 28 and add to −11-11. List the negative factor pairs of 28 and check their sums:

Factors of 28Sum of factors
−1,−28-1, -28−1+(−28)=−29-1 + (-28) = -29
−2,−14-2, -14−2+(−14)=−16-2 + (-14) = -16
−4,−7-4, -7−4+(−7)=−11-4 + (-7) = -11 ✓

The pair −4-4 and −7-7 has a product of 28 and a sum of −11-11. Step 3 — Use −4-4 and −7-7 as the last terms of the binomials: (t−4)(t−7)(t - 4)(t - 7). Step 4 — Check by multiplying: (t−4)(t−7)=t2−7t−4t+28=t2−11t+28(t - 4)(t - 7) = t^2 - 7t - 4t + 28 = t^2 - 11t + 28 ✓. The factored form is (t−4)(t−7)(t - 4)(t - 7). This example demonstrates that when the middle term is negative and the constant is positive, only negative factor pairs of the constant need to be tested. Both binomial factors use subtraction rather than addition.

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Updated 2026-06-25

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