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One Factor Is Enough

Part of the Factoring & Quadratics section of Coddy's Algebra journey. Lesson 31 of 61.

A product is zero if at least one of its factors is zero. In (x−4)(x+6)=0\left(x-4\right)\left(x+6\right) = 0, substitute x=4x = 4. The first factor is 0 and the second factor is 10, so the product is 0⋅10=00\cdot 10 = 0. The second factor does not need to be zero as well.

Each case is solved on its own. In the first case, the first factor is zero, which gives x=4x = 4. In the second case, the second factor is zero, which gives x=−6x = -6. Each of the answers, 4 and -6, makes the original equation true by itself. You can check this by substituting each one.

Solve each factor equation separately, and join the solutions with the word or. For (x−4)(x+6)=0\left(x-4\right)\left(x+6\right) = 0, the solutions are x=4x = 4 or x=−6x = -6. A solution needs to make at least one factor zero. It does not need to make both factors zero.

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Challenge

Easy

Solve (x−1)(x+8)=0\left(x-1\right)\left(x+8\right) = 0.

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(x−1)(x+8)=0\left(x-1\right)\left(x+8\right) = 0

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