How can a polynomial have fewer x-intercepts than its total number of roots?

A polynomial can have fewer x-intercepts than its total number of roots due to the nature of its roots, particularly when there are complex roots and repeated roots.

Let’s break this down with a specific example: consider the polynomial equation:

P(x) = (x – 1)²(x + 2)

This polynomial is of degree 3, which means it has a total of 3 roots (counting multiplicities). To find the roots, we set the equation to zero:

P(x) = 0

From this, we see that:

  • The root at x = 1 has a multiplicity of 2 because of the squared term.
  • The root at x = -2 has a multiplicity of 1.

Now let’s analyze the x-intercepts:

An x-intercept occurs where the graph of the polynomial crosses or touches the x-axis. In this case, the polynomial touches the x-axis at x = 1strong> and crosses at x = -2.

So, the x-intercepts of this polynomial are:

  • x = 1 (a repeated root, hence just one intercept)
  • x = -2 (one intercept)

This gives us a total of 2 x-intercepts. However, the polynomial has 3 roots in total:

  • One root at x = 1 (with multiplicity of 2)
  • One root at x = -2 (with multiplicity of 1)

Thus, as we can see, this polynomial has a total of 3 roots but only 2 x-intercepts. This illustrates that when roots repeat, they can lead to fewer distinct x-intercepts compared to the total number of roots.

Moreover, if a polynomial has complex roots (which occur in conjugate pairs), they do not contribute to x-intercepts, further reducing the number while still adhering to the rules of polynomial degree. Therefore, a polynomial can certainly have fewer x-intercepts than its number of roots due to the combined effects of repeated roots and the presence of complex roots.

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