nth degree polynomial has atmost n real roots

Algebra Level 3

$\large \displaystyle 3x^5 + 15x - 8 = 0$

Find the number of real roots of the equation above.

Try my set

4 2 5 0 1 3

2 solutions

Vishwak Srinivasan
Jun 3, 2015

I did it the calculus way. Any other methods are welcome.

Let $f(x) = 3x^5 + 15x - 8$

$f'(x) = 15x^4 + 15 = 15(x^2 +1) > 0 \text{ for all } x$

This means it is an increasing function for all $x$ , and hence crosses the x - axis only $\boxed{\text{once}}$

Moderator note:

Can you solve this without using calculus?

Root lies between 0 and 1.

Shubhendra Singh - 6 years ago

That doesn't mean that the graph can pass the x-axis only once.

Vishwak Srinivasan - 6 years ago

We can see that f'(x) = 0 has imaginary roots which means that the graph is always increasing. That's why it'll cut x - axis only once between 0 and 1

Shubhendra Singh - 6 years ago

@Shubhendra Singh – That was pretty much my solution.

Vishwak Srinivasan - 6 years ago

Soutrik Bandyopadhyay
Jun 3, 2015

By Descartes rule of Signs - number of positive solutions of f(x) is equal to or less than number of sign changes in polynomial by even number. So we see the signs in polynomial are ++- so 1 positive root. Now for negative roots we do the same thing for f(-x) . We get signs --- thus no negative roots. The conclusion - 1 real root

Moderator note:

Good standard approach.

nth degree polynomial has atmost n real roots

Try my set

2 solutions

Moderator note:

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