Asked by Marissa
How many real roots does f(x) = x^3+5x+1 have?
x^3 = odd polynomial = at least 1 real root
Can I follow Descartes' rule for this? No sign change, so there aren't any positive zeros.
when I sub in -x for x
(-x)^3+ 5(-x) + 1
= -x^3-5x+1
Then there is one sign change, so 1 negative root.
So my conclusion is that there is one real root. Is this right?
x^3 = odd polynomial = at least 1 real root
Can I follow Descartes' rule for this? No sign change, so there aren't any positive zeros.
when I sub in -x for x
(-x)^3+ 5(-x) + 1
= -x^3-5x+1
Then there is one sign change, so 1 negative root.
So my conclusion is that there is one real root. Is this right?
Answers
Answered by
drwls
I agree. The Descartes rule says that there cannot be more than one real root in this case, and that it will be negative. There is another rule that says there must be at least one real root. Therefore the number of real roots is one.
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