Question

Which real numbers are zeros of the function?

Answer 1

The roots are -1/2, 0, 2, and 3.

Let's trying factoring this polynomial.

We can factor an x out of each term to start.

x(2x^3 - 9x^2 + 7x + 6)

We now know one of the roots is going to be zero.

Using the rational roots theorem, and the remainder theorem, we can try to find some more roots that way.

Factors of 6: 1, 2, 3, 6.

Factors of 2: 1

Possible rational roots: +/-1/2, +/-1, +/-2, +/-3/2, +/-3, +/-6

Using the remainder theorem, we can plug these values into the polynomial, and if we get a remainder of zero, we know it's a root.

-1/2(2(-1/2)^3 - 9(-1/2)^2 + 7(-1/2) + 6) = 0

Our first root is -1/2.

We can successfully factor a -1/2 out of this polynomial.

After diving the polynomial by -1/2, we're left with: 2x^2 - 10x + 12.

We can now try using the AC method to get our last two roots.

First, we can factor this polynomial to simplify it.

Divide all terms by 2.

2(x^2 - 5x + 6)

Now let's try using the AC method.

The digits -3 and -2 satisfy the criteria.

2(x - 3)(x - 2)

We now have all of our roots: 0, -1/2, 2, and 3.