Question

a) Factor the polynomial using any method. All work must be shown to earn full credit.

b) Find the zeros (roots of parabola). All work must be shown to earn full credit.

2x2 - 5x - 3 = 0

Answer 1

Let's complete the two parts of the problem: factoring the quadratic polynomial \(2x^2 - 5x - 3\) and finding its zeros.

### Part a: Factor the polynomial

The polynomial we need to factor is:

\[ 2x^2 - 5x - 3 \]

To factor this polynomial, we can use the method of splitting the middle term. First, we look for two numbers that multiply to give the product of the leading coefficient (2) and the constant term (-3). That is:

\[
2 \times -3 = -6
\]

We need two numbers that multiply to -6 and add up to -5. The numbers that fit this requirement are -6 and +1, because:

\[
-6 + 1 = -5 \quad \text{and} \quad -6 \times 1 = -6
\]

Now, we can rewrite the middle term (-5x) using these two numbers:

\[
2x^2 - 6x + x - 3
\]

Next, we group the terms:

\[
(2x^2 - 6x) + (x - 3)
\]

Factoring out the common factors in each group gives:

\[
2x(x - 3) + 1(x - 3)
\]

Now we can factor out the common binomial factor \((x - 3)\):

\[
(2x + 1)(x - 3)
\]

So, the factored form of the polynomial \(2x^2 - 5x - 3\) is:

\[
\boxed{(2x + 1)(x - 3)}
\]

### Part b: Find the zeros (roots of the parabola)

To find the zeros of the polynomial, we set it equal to zero:

\[
(2x + 1)(x - 3) = 0
\]

Using the zero product property, we set each factor equal to zero:

1. \(2x + 1 = 0\)
2. \(x - 3 = 0\)

**For the first equation**:

\[
2x + 1 = 0
\]
\[
2x = -1
\]
\[
x = -\frac{1}{2}
\]

**For the second equation**:

\[
x - 3 = 0
\]
\[
x = 3
\]

Thus, the zeros (roots) of the parabola are:

\[
\boxed{-\frac{1}{2}} \quad \text{and} \quad \boxed{3}
\]

These results indicate that the polynomial \(2x^2 - 5x - 3\) crosses the x-axis at the points \(x = -\frac{1}{2}\) and \(x = 3\).