Find all zeroes of the polynomial `3x^(3)+10x^(2)-9x-4`, if one of its zero is 1.

Answer

To find all the zeroes of the polynomial \(3x^3 + 10x^2 - 9x - 4\), given that one of its zeroes is \(x = 1\), we can follow these steps:

### Step 1: Confirm that \(x = 1\) is a zero
Since \(x = 1\) is a zero, we can use polynomial long division to divide the polynomial by \(x - 1\).

### Step 2: Perform polynomial long division
We will divide \(3x^3 + 10x^2 - 9x - 4\) by \(x - 1\).

1. Divide the leading term:
\[ \frac{3x^3}{x} = 3x^2 \]
2. Multiply \(3x^2\) by \(x - 1\):
\[ 3x^2(x - 1) = 3x^3 - 3x^2 \]
3. Subtract from the original polynomial:
\[ (3x^3 + 10x^2) - (3x^3 - 3x^2) = 10x^2 + 3x^2 = 13x^2 \]
4. Bring down the next term \(-9x\):
\[ 13x^2 - 9x \]
5. Divide the leading term again:
\[ \frac{13x^2}{x} = 13x \]
6. Multiply \(13x\) by \(x - 1\):
\[ 13x(x - 1) = 13x^2 - 13x \]
7. Subtract:
\[ (13x^2 - 9x) - (13x^2 - 13x) = -9x + 13x = 4x \]
8. Bring down the next term \(-4\):
\[ 4x - 4 \]
9. Divide the leading term:
\[ \frac{4x}{x} = 4 \]
10. Multiply \(4\) by \(x - 1\):
\[ 4(x - 1) = 4x - 4 \]
11. Subtract:
\[ (4x - 4) - (4x - 4) = 0 \]

The result of the division is:
\[ 3x^2 + 13x + 4 \]

### Step 3: Factor the quadratic polynomial
Now we need to factor \(3x^2 + 13x + 4\).

To factor, we look for two numbers that multiply to \(3 \times 4 = 12\) and add up to \(13\). The numbers are \(12\) and \(1\).

1. Rewrite the middle term:
\[ 3x^2 + 12x + 1x + 4 \]
2. Factor by grouping:
\[ (3x^2 + 12x) + (1x + 4) \]
\[ 3x(x + 4) + 1(x + 4) \]
3. Factor out the common factor:
\[ (x + 4)(3x + 1) \]

### Step 4: Find the zeroes
Now we have the complete factorization:
\[ 3x^3 + 10x^2 - 9x - 4 = (x - 1)(x + 4)(3x + 1) \]

Setting each factor to zero:
1. \(x - 1 = 0 \Rightarrow x = 1\)
2. \(x + 4 = 0 \Rightarrow x = -4\)
3. \(3x + 1 = 0 \Rightarrow 3x = -1 \Rightarrow x = -\frac{1}{3}\)

### Final Answer
The zeroes of the polynomial \(3x^3 + 10x^2 - 9x - 4\) are:
\[ x = 1, \quad x = -4, \quad x = -\frac{1}{3} \]