What are all the roots of P(x) = (x – 1)(x + 3)(x + 2)?

The roots are all the x-values that make the polynomial equal 0. If any chunk in parentheses equals 0, then the whole thing will. That gives us three choices:

x – 1 = 0, x + 3 = 0, or x + 2 = 0

Solving these equations gives us our roots: x = 1, -3, and -2.

What are the roots of P(x) = (x + 3)(x – 4)(x – 2)?

If everything is factored to start out, then finding the roots is straightforward.

x + 3 = 0, x – 4 = 0, or x – 2 = 0

The roots are x = -3, 4, and 2.

Write a polynomial with roots of -1, 1, and 2.

It's easy to create a set of factors that have these roots—just multiply some binomials together that will equal 0 if you plug in one of those values.

P(x) = (x + 1)(x – 1)(x – 2)

This is the polynomial, but we'd like to see it expanded fully.

P(x) = (x² – 1)(x – 2)

P(x) = x³ – 2x² – x + 2

If you plug in -1, 1, or 2 into this equation, it will still equal 0, showing that they are still the roots.

Write a polynomial with roots of -2, 2, and 5.

The factors for this polynomial will be:

(x + 2), (x – 2), (x – 5)

The polynomial then equals:

P(x) = (x + 2)(x – 2)(x – 5)

P(x) = (x² – 4)(x – 5)

P(x) = x³ – 5x² – 4x + 20

Find all roots of x³ – 4x² – x + 4 given that one root is 4.

We know that one root is 4, so that means x – 4 is a factor.

The other factors can be found using synthetic division.

Multiply the numbers on the bottom by 4, then add the result to the next column. In this case, there must be no remainder, because we've been told that 4 is a factor of the polynomial.

Now factor x² – 1:

(x + 1)(x – 1)

The full polynomials factors to:

P(x) = (x – 4)(x + 1)(x – 1)

The roots are x = 4, -1, and 1.