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

(Simple small steps!)

Answer 1

To factor the polynomial \(2x^2 - 5x - 3 = 0\) and find its zeros, we'll follow these steps:

### a) Factor the Polynomial

1. **Identify the coefficients**:
The polynomial is in the standard form \(ax^2 + bx + c\). Here, \(a = 2\), \(b = -5\), and \(c = -3\).

2. **Find two numbers that multiply to \(ac\) and add to \(b\)**:
- Calculate \(ac = 2 \times -3 = -6\).
- We need to find two numbers that multiply to \(-6\) and add to \(-5\).
- The two numbers that satisfy these conditions are \(-6\) and \(1\) because:
\[
-6 \times 1 = -6 \quad \text{and} \quad -6 + 1 = -5
\]

3. **Rewrite the middle term using the two numbers**:
Rewrite \(-5x\) as \(-6x + 1x\):
\[
2x^2 - 6x + 1x - 3
\]

4. **Group the terms**:
Group the first two terms and the last two terms:
\[
(2x^2 - 6x) + (1x - 3)
\]

5. **Factor out the greatest common factor from each group**:
From the first group \(2x^2 - 6x\), factor out \(2x\):
\[
2x(x - 3)
\]
From the second group \(1x - 3\), factor out \(1\) (or just write it as is):
\[
+ 1(x - 3)
\]

6. **Combine the factored groups**:
Now we can write this as:
\[
2x(x - 3) + 1(x - 3)
\]
Now factor out the common factor \((x - 3)\):
\[
(x - 3)(2x + 1)
\]

So, the factored form of the polynomial \(2x^2 - 5x - 3\) is:
\[
\boxed{(x - 3)(2x + 1)}
\]

### b) Find the Zeros (Roots) of the Parabola

To find the zeros of the polynomial, set the factored form equal to zero:
\[
(x - 3)(2x + 1) = 0
\]

1. **Set each factor to zero**:
- For the first factor:
\[
x - 3 = 0 \implies x = 3
\]
- For the second factor:
\[
2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}
\]

2. **Write the zeros**:
The roots (zeros) of the polynomial are:
\[
\boxed{3} \quad \text{and} \quad \boxed{-\frac{1}{2}}
\]

Thus, the zeros of the polynomial \(2x^2 - 5x - 3\) are \(x = 3\) and \(x = -\frac{1}{2}\).