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finlep [7]
2 years ago
6

What is the possible number of positive and negative real roots or zeroes in f(x)=x^3+x^2-4

Mathematics
1 answer:
Tanya [424]2 years ago
3 0

This is a third degree polynomial, so we know that the maximum amount of roots will be 3.

We also know that imaginary roots come in pairs, so there is either 0 or 2 imaginary roots.

We can tell the amount of positive / negative roots by applying Descartes' rule of signs.

Let's examine the polynomial: x^3 + x^2 - 4

Note the sign changes.

From left to right, there is one sign change.

We now know there is only one positive real root.

Now, we can replace all values of x with -1 and then simplify. The number of sign changes will tell us the amount of negative roots.

(-1)^3 + (-1)^2 - 4

-1 + 1 - 4

From left to right, there are two sign changes, which tells us there will be 2 or 0 negative roots.

If there are 2, they are not imaginary.

If there are 0, they are imaginary.

<h3>So, our guaranteed root is:</h3><h3>1 positive, real root.</h3><h3>And our possible roots are:</h3><h3>2 negative, real roots.</h3><h3>0 negative, real roots, but 2 imaginary roots.</h3>
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