Question

Solve 4x
2
+12x+5=0 by grouping.
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Answer 1

To solve the quadratic equation \( 4x^2 + 12x + 5 = 0 \) by factoring by grouping, we will follow these steps:

1. **Identify the equation**:
\[
4x^2 + 12x + 5 = 0
\]

2. **Multiply \(a\) and \(c\)** (the coefficients of \(x^2\) and the constant term):
- Here, \( a = 4 \) and \( c = 5 \).
- Therefore, \( ac = 4 \times 5 = 20 \).

3. **Find two numbers that multiply to \(ac\) and add up to \(b\) (the coefficient of \(x\))**:
- We need two numbers that multiply to 20 and add up to 12.
- These numbers are 10 and 2 because \( 10 \times 2 = 20 \) and \( 10 + 2 = 12 \).

4. **Rewrite the middle term using these numbers**:
\[
4x^2 + 10x + 2x + 5 = 0
\]

5. **Group the terms**:
\[
(4x^2 + 10x) + (2x + 5) = 0
\]

6. **Factor out common factors from each group**:
- From the first group \(4x^2 + 10x\), we can factor out \(2x\):
\[
2x(2x + 5)
\]
- From the second group \(2x + 5\), there is no common factor:
\[
1(2x + 5)
\]
- So now we have:
\[
2x(2x + 5) + 1(2x + 5) = 0
\]

7. **Factor out the common binomial factor**:
\[
(2x + 5)(2x + 1) = 0
\]

8. **Set each factor to zero**:
\[
2x + 5 = 0 \quad \text{or} \quad 2x + 1 = 0
\]

9. **Solve for \(x\)**:
- From \(2x + 5 = 0\):
\[
2x = -5 \implies x = -\frac{5}{2}
\]
- From \(2x + 1 = 0\):
\[
2x = -1 \implies x = -\frac{1}{2}
\]

10. **Final solutions**:
\[
x = -\frac{5}{2} \quad \text{and} \quad x = -\frac{1}{2}
\]

Thus, the solutions to the equation \( 4x^2 + 12x + 5 = 0 \) are \( x = -\frac{5}{2} \) and \( x = -\frac{1}{2} \).