Stitz-Zeager_College_Algebra_e-book

The quantity 1 is usually re labeled i the so called

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Unformatted text preview: r the trees, let us momentarily replace x2 with the variable u. In terms of u, our equation becomes u2 + u − 12 = 0. We can readily factor the left hand side of this equation as (u + 4)(u − 3) = 0, which means we have factored the left hand side of x4 + x2 − 12 = 0 as x2 − 3 x2 + 4 = 0. We get x2 = 3, which gives us √ √ x = ± 3, or x2 = −4, which admits no real solutions. Since 3 ≈ 1.73, the two zeros match what √ expected from the graph. In terms of multiplicity, the Factor Theorem guarantees we √ x − 3 and x + 3 are factors of f (x). We note that our work for finding the zeros of f shows f (x) can be factored as f (x) = x2 − 3 x2 + 4 . Since x2 + 4 has no real zeros, the √ √ quantities x − 3 and x + 3 must both be factors of x2 − 3. According to Theorem 3.7, √ x2 − 3 can have at most 2 zeros, counting multiplicity, hence each of ± 3 is a zero of f of multiplicity 1. The technique used to factor f (x) in Example 3.3.4 is called u-substitution. We shall see more of this technique in Section 5.3. In general, substitution can help us identify a...
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