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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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