012 if -5 and 7 are zeroes of `x^4- 6x^3- 26x^2 +138x-35` find the other zeroes

Explanation

To find the other zeroes of the polynomial \( p(x) = x^4 - 6x^3 - 26x^2 + 138x - 35 \) given that -5 and 7 are zeroes, we can follow these steps:

### Step 1: Write the known zeroes as factors
Since -5 and 7 are zeroes, we can express them as factors of the polynomial:
\[ (x + 5)(x - 7) \]

### Step 2: Expand the factors
Now, we will expand these factors:
\[ (x + 5)(x - 7) = x^2 - 7x + 5x - 35 = x^2 - 2x - 35 \]

### Step 3: Divide the polynomial by the quadratic factor
Next, we will divide the polynomial \( p(x) \) by the quadratic factor \( d(x) = x^2 - 2x - 35 \) to find the quotient \( q(x) \):
\[ p(x) = (x^2 - 2x - 35) \cdot q(x) \]

### Step 4: Perform polynomial long division
We will divide \( p(x) \) by \( d(x) \):

1. Divide the leading term \( x^4 \) by \( x^2 \) to get \( x^2 \).
2. Multiply \( x^2 \) by \( d(x) \):
\[ x^2 \cdot (x^2 - 2x - 35) = x^4 - 2x^3 - 35x^2 \]
3. Subtract this from \( p(x) \):
\[ (x^4 - 6x^3 - 26x^2 + 138x - 35) - (x^4 - 2x^3 - 35x^2) = -4x^3 + 9x^2 + 138x - 35 \]
4. Repeat the process: Divide \( -4x^3 \) by \( x^2 \) to get \( -4x \).
5. Multiply and subtract again:
\[ -4x \cdot (x^2 - 2x - 35) = -4x^3 + 8x^2 + 140x \]
\[ (-4x^3 + 9x^2 + 138x - 35) - (-4x^3 + 8x^2 + 140x) = x^2 - 2x - 35 \]
6. Finally, divide \( x^2 - 2x - 35 \) by \( x^2 - 2x - 35 \) to get 1.

### Step 5: Write the quotient
Thus, we have:
\[ p(x) = (x^2 - 2x - 35)(x^2 + 1) \]

### Step 6: Find the remaining zeroes
Now we need to find the zeroes of \( x^2 + 1 = 0 \):
\[ x^2 = -1 \implies x = i \quad \text{and} \quad x = -i \]

### Conclusion
The zeroes of the polynomial \( p(x) \) are:
\[ -5, \quad 7, \quad i, \quad -i \]