Question 715593
One can find the maximum number of positive and negative zeros pretty quickly using Descartes' Rule of Signs:<ul><li>The maximum number of positive zeros will be the number of changes in sign in the polynomial f(x). The actual number of positive zeros will be either this maximum or some multiple of 2 less than the maximum.</li><li>The maximum number of negative zeros will be the number of changes in sign in the polynomial f(-x). The actual number of positive zeros will be either this maximum or some multiple of 2 less than the maximum.</li></ul>Using this rule on your polynomial...
There are three sign changes (from positive 5x^4 to negative 2x^3 (one change) to positive 3x (two changes) to negative 3 (three changes)). So there will be a maximum of 3 positive zeros. The actual number will be 3 or 1.<br>
f(-x) = 5(-x)^4-2(-x)^3+3(-x)-3 = 5x^4+2x^3-3x-3
f(-x) has just 1 sign change (from positive 2x^3 to negative 3x) so there will be a maximum of 1 negative zero to f(x). The actual number of negative zeros will be 1 (since all multiples of 2 less than 1 are negative).<br
So f(x) has exactly 1 negative zero and either 3 or 1 positive zeros. Note: A 4th degree polynomial with real coefficients, like this one, should have 4 zeros. But that does not mean that there must be 3 positive zeros (to go with the one negative zero. The "missing" zeros could be complex (which always come in conjugate pairs of polynomials with real coefficients).<br>
If you need to know exactly how many positive zeros there are (we already know there is exactly one negative zero) then you have to either:<ul><li>Use a graphing calculator to graph f(x) and see how many times it intersects the positive part of the x-axis (where the y-coordinates are zero).</li><li>Actually find the zeros by factoring, if possible.</li></ul>Unfortunately f(x) does not seem to factor. So its zeros must be irrational or complex. All we have left is to look at the graph:
{{{graph(400, 400, -3, 3, -3, 3, 5x^4-2x^3+3x-3)}}}
As we can see, the graph intersects the negative part of the x-axis once (near -1), as expected, and the positive part of the x-axis once (between 0 and 1). The other two zeros must be complex.