Question

Find the zeros of the quadratic equation y=6x2+13x+5 . use the following method factoring,

Answer 1

To find the zeros of the quadratic equation \( y = 6x^2 + 13x + 5 \) by factoring, we need to express the quadratic in the form \( (ax + b)(cx + d) = 0 \).

1. **Identify \(a\), \(b\), and \(c\)**: In our case, the coefficients are:
- \( A = 6 \)
- \( B = 13 \)
- \( C = 5 \)

2. **Multiply \( A \) and \( C \)**: We need to find two numbers that multiply to \( A \times C = 6 \times 5 = 30 \) and add to \( B = 13 \).

The pairs of factors of \( 30 \) that add up to \( 13 \) are:
- \( 3 \) and \( 10 \) (because \( 3 + 10 = 13 \))

3. **Rewrite the middle term**: We can rewrite the quadratic by splitting the middle term using the two numbers found:
\[
6x^2 + 3x + 10x + 5
\]

4. **Factor by grouping**:
- Group the first two terms and the last two terms:
\[
(6x^2 + 3x) + (10x + 5)
\]
- Factor out the greatest common factor from each group:
\[
3x(2x + 1) + 5(2x + 1)
\]
- Now, factor out the common binomial \((2x + 1)\):
\[
(3x + 5)(2x + 1) = 0
\]

5. **Set each factor to zero**:
- \( 3x + 5 = 0 \) gives:
\[
3x = -5 \quad \Rightarrow \quad x = -\frac{5}{3}
\]
- \( 2x + 1 = 0 \) gives:
\[
2x = -1 \quad \Rightarrow \quad x = -\frac{1}{2}
\]

6. **Conclusion**: The zeros of the quadratic equation \( y = 6x^2 + 13x + 5 \) are:
\[
x = -\frac{5}{3} \quad \text{and} \quad x = -\frac{1}{2}
\]