Solving Quadratics via Factoring - If Possible

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Jason
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Joined: Sun Dec 21, 2025 8:56 pm

Quadratics can be factored if it doesn't lead to a situation with one of the following: irrational or complex roots, where factors are possible, a situation where factoring leads to 0
:!: Strategy: Try to find numbers that multiply to make A * C while add to get B.
\(ex. a\)

\(x^{2} - 6x - 8\)

\(A = 1\,\,\,C = -8\,\,\,A * C = -8\,\,\,B = -6\)

\(x^{2} + 2x + (-4x) - 8 = 0\)

\((x + 2)(x - 4) = 0\)

\(x + 2 = 0\)

\(x = -2\)

\(x - 4 = 0\)

\(x = 4\)
 

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