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**Unformatted text preview: **a zero. Descartes’ Rule of Signs told us that we may have up to three negative
real zeros, counting multiplicity, so we try −1 again, and it works once more. At this point,
we have taken f , a fourth degree polynomial, and performed two successful divisions. Our
quotient polynomial is quadratic, so we look at it to ﬁnd the remaining zeros.
−1
2 2
4 −1
3
↓ −1 − 2 −6 −3 5
4 19
8 −5
2 − 19
4 2 3 5
−8 −1
−1 2
4 −1 −6 −3
↓ −2 −2
3
3
2
2 −3 −3 0
↓ −2
0
3
2
0 −3 0 Setting the quotient polynomial equal to zero yields 2x2 − 3 = 0, so that x2 = 3 , or x = ±
2
√ √ 6
2. Descartes’ Rule of Signs tells us that the positive real zero we found, 26 has multiplicity 1.
Descartes also tells us the total multiplicity of negative real zeros is 3, which forces −1 to be
√
a zero of multiplicity 2 and − 26 to have multiplicity 1.
4
2. We know the end behavior of y = f (x) resembles its leading term, y = 2x√. This means
the graph enters the scene in Quadrant II and exits in Quadrant I. Since ± 26 are zeros of
√ odd multiplicity, we know the graph cro...

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