Find the zeroes of the polynomial `p(x)=4x^2-7x+3` by factorising it and verify the relationship between the zeroes and coefficients of p(x).

Answer

To find the zeroes of the polynomial \( p(x) = 4x^2 - 7x + 3 \) by factorizing it, we will follow these steps:

### Step 1: Write down the polynomial
We start with the polynomial:
\[ p(x) = 4x^2 - 7x + 3 \]

### Step 2: Identify the coefficients
From the polynomial, we identify the coefficients:
- \( a = 4 \) (coefficient of \( x^2 \))
- \( b = -7 \) (coefficient of \( x \))
- \( c = 3 \) (constant term)

### Step 3: Find two numbers that multiply to \( ac \) and add to \( b \)
We need to find two numbers that multiply to \( ac = 4 \times 3 = 12 \) and add to \( b = -7 \).

The two numbers that satisfy these conditions are \( -3 \) and \( -4 \) because:
- \( -3 \times -4 = 12 \)
- \( -3 + (-4) = -7 \)

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

### Step 5: Factor by grouping
Now, we can factor by grouping:
1. Group the first two terms and the last two terms:
\[ p(x) = (4x^2 - 3x) + (-4x + 3) \]
2. Factor out the common factors:
\[ p(x) = x(4x - 3) - 1(4x - 3) \]
3. Now, factor out \( (4x - 3) \):
\[ p(x) = (4x - 3)(x - 1) \]

### Step 6: Set the factors to zero
To find the zeroes, we set each factor to zero:
1. \( 4x - 3 = 0 \)
\[ 4x = 3 \]
\[ x = \frac{3}{4} \]
2. \( x - 1 = 0 \)
\[ x = 1 \]

### Step 7: State the zeroes
The zeroes of the polynomial \( p(x) = 4x^2 - 7x + 3 \) are:
\[ x = \frac{3}{4} \quad \text{and} \quad x = 1 \]

### Step 8: Verify the relationship between the zeroes and coefficients
The relationship between the zeroes \( \alpha \) and \( \beta \) and the coefficients \( a, b, c \) of the polynomial \( ax^2 + bx + c \) is given by:
- Sum of the zeroes \( \alpha + \beta = -\frac{b}{a} \)
- Product of the zeroes \( \alpha \beta = \frac{c}{a} \)

Calculating the sum and product:
1. Sum of the zeroes:
\[ \alpha + \beta = \frac{3}{4} + 1 = \frac{3}{4} + \frac{4}{4} = \frac{7}{4} \]
\[ -\frac{b}{a} = -\frac{-7}{4} = \frac{7}{4} \]
(This matches!)

2. Product of the zeroes:
\[ \alpha \beta = \frac{3}{4} \times 1 = \frac{3}{4} \]
\[ \frac{c}{a} = \frac{3}{4} \]
(This also matches!)

### Conclusion
Thus, we have verified the relationship between the zeroes and the coefficients of the polynomial.