Solve : `4x^2+5x+1=0`.

Answer

To solve the quadratic equation \(4x^2 + 5x + 1 = 0\), we can use the factorization method. Here’s a step-by-step solution:

### Step 1: Write down the equation
We start with the equation:
\[ 4x^2 + 5x + 1 = 0 \]

### Step 2: Identify the coefficients
In the equation \(ax^2 + bx + c = 0\), we have:
- \(a = 4\)
- \(b = 5\)
- \(c = 1\)

### Step 3: Multiply \(a\) and \(c\)
Next, we multiply \(a\) and \(c\):
\[ 4 \times 1 = 4 \]

### Step 4: Find two numbers that multiply to \(ac\) and add to \(b\)
We need to find two numbers that multiply to \(4\) (the product from step 3) and add to \(5\) (the value of \(b\)). The numbers \(4\) and \(1\) satisfy this condition:
\[ 4 \times 1 = 4 \quad \text{and} \quad 4 + 1 = 5 \]

### Step 5: Rewrite the middle term
We can rewrite the equation using these two numbers:
\[ 4x^2 + 4x + 1x + 1 = 0 \]

### Step 6: Factor by grouping
Now, we group the terms:
\[ (4x^2 + 4x) + (1x + 1) = 0 \]
Factoring out the common terms in each group:
\[ 4x(x + 1) + 1(x + 1) = 0 \]

### Step 7: Factor out the common binomial
Now we can factor out the common binomial \((x + 1)\):
\[ (4x + 1)(x + 1) = 0 \]

### Step 8: Set each factor to zero
Now, we set each factor equal to zero:
1. \(4x + 1 = 0\)
2. \(x + 1 = 0\)

### Step 9: Solve for \(x\)
For the first equation:
\[ 4x + 1 = 0 \implies 4x = -1 \implies x = -\frac{1}{4} \]
For the second equation:
\[ x + 1 = 0 \implies x = -1 \]

### Final Solution
The solutions to the equation \(4x^2 + 5x + 1 = 0\) are:
\[ x = -1 \quad \text{and} \quad x = -\frac{1}{4} \]