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p(x)=x^3+3x^2-5x-35

2007-03-30 07:57:07 · 3 answers · asked by snowboardingprojr 1 in Science & Mathematics Mathematics

3 answers

Using Descartes' rule...

1 positive zero
and
2 negative zeros

2007-03-30 08:00:36 · answer #1 · answered by Mathematica 7 · 0 0

Rational zeros are all the possible factors of d over all of the possible factors of a, for

p(x) = ax^3 + bx^2 + cx + d

All of the factors of -35 are:
+/- 1, +/- 5, +/- 7, +/- 35

All of the factors of a (which is 1) are:
+/- 1

All possible rational zeros:

+/- (1, 5, 7, 35) / +/- (1)

OR

1, -1, 5, -5, 7, -7, 35, -35

Of course, if neither of these possible rational roots work, it follows that there are irrational and/or complex roots.

2007-03-30 15:09:45 · answer #2 · answered by Puggy 7 · 0 0

x +/- 1: x+/- 5; x +/- 7; x +/- 35

2007-03-30 15:13:41 · answer #3 · answered by ironduke8159 7 · 0 0

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