A polynomial of degree `nge1` can have at most n real zeroes. A quadratic polynomial can have at most two real zeroes .
find the zeroes of the polynomial `p(x)=3x^(2)+7x+2` .

Options

A. `1,-1//3`
B. `-1//3, -2`
C. `2//3,1`
D. `1//3,2`

Answer

To find the zeroes of the polynomial \( p(x) = 3x^2 + 7x + 2 \), we can use the method of splitting the middle term. Here’s a step-by-step solution:

### Step 1: Identify the polynomial
We have the polynomial:
\[ p(x) = 3x^2 + 7x + 2 \]

### Step 2: Multiply the coefficient of \( x^2 \) term with the constant term
We need to find two numbers that multiply to \( 3 \times 2 = 6 \) and add up to \( 7 \) (the coefficient of \( x \)).

### Step 3: Find the two numbers
The two numbers that satisfy these conditions are \( 6 \) and \( 1 \) because:
\[ 6 \times 1 = 6 \quad \text{and} \quad 6 + 1 = 7 \]

### Step 4: Rewrite the polynomial
We can rewrite the polynomial by splitting the middle term:
\[ p(x) = 3x^2 + 6x + 1x + 2 \]

### Step 5: Factor by grouping
Now, we will group the terms:
\[ = (3x^2 + 6x) + (1x + 2) \]
Now, factor out the common terms from each group:
\[ = 3x(x + 2) + 1(x + 2) \]

### Step 6: Factor out the common binomial
Now, we can factor out the common binomial \( (x + 2) \):
\[ = (3x + 1)(x + 2) \]

### Step 7: Set each factor to zero
To find the zeroes of the polynomial, we set each factor equal to zero:
1. \( 3x + 1 = 0 \)
2. \( x + 2 = 0 \)

### Step 8: Solve for \( x \)
For the first equation:
\[ 3x + 1 = 0 \implies 3x = -1 \implies x = -\frac{1}{3} \]

For the second equation:
\[ x + 2 = 0 \implies x = -2 \]

### Conclusion
The zeroes of the polynomial \( p(x) = 3x^2 + 7x + 2 \) are:
\[ x = -\frac{1}{3} \quad \text{and} \quad x = -2 \]