Find the polynomial with rational coefficients and whose roots are
` a+b,a-b,-a+b,-a-b`

Answer

To find the polynomial with rational coefficients whose roots are \( a+b, a-b, -a+b, -a-b \), we will follow these steps:

### Step 1: Identify the roots
The roots given are:
- \( \alpha = a + b \)
- \( \beta = a - b \)
- \( \gamma = -a + b \)
- \( \delta = -a - b \)

### Step 2: Write the polynomial in factored form
The polynomial can be expressed as:
\[ P(x) = (x - \alpha)(x - \beta)(x - \gamma)(x - \delta) \]
Substituting the roots into the polynomial:
\[ P(x) = (x - (a+b))(x - (a-b))(x - (-a+b))(x - (-a-b)) \]
This simplifies to:
\[ P(x) = (x - (a+b))(x - (a-b))(x + (a-b))(x + (a+b)) \]

### Step 3: Group the factors
We can group the factors:
\[ P(x) = [(x - (a+b))(x + (a+b))][(x - (a-b))(x + (a-b))] \]
Using the difference of squares:
\[ P(x) = (x^2 - (a+b)^2)(x^2 - (a-b)^2) \]

### Step 4: Expand the squares
Now, we expand each square:
\[ P(x) = (x^2 - (a^2 + 2ab + b^2))(x^2 - (a^2 - 2ab + b^2)) \]
This simplifies to:
\[ P(x) = (x^2 - (a^2 + b^2 + 2ab))(x^2 - (a^2 + b^2 - 2ab)) \]

### Step 5: Let \( y = x^2 \)
Let \( y = x^2 \), then we have:
\[ P(y) = (y - (a^2 + b^2 + 2ab))(y - (a^2 + b^2 - 2ab)) \]

### Step 6: Expand the polynomial
Now we expand:
\[ P(y) = y^2 - (2a^2 + 2b^2)y + ((a^2 + b^2)^2 - (2ab)^2) \]
This simplifies to:
\[ P(y) = y^2 - 2(a^2 + b^2)y + (a^2 - b^2)^2 \]

### Step 7: Substitute back \( y = x^2 \)
Finally, substituting back \( y = x^2 \):
\[ P(x) = x^4 - 2(a^2 + b^2)x^2 + (a^2 - b^2)^2 \]

### Final Polynomial
Thus, the polynomial with rational coefficients whose roots are \( a+b, a-b, -a+b, -a-b \) is:
\[ P(x) = x^4 - 2(a^2 + b^2)x^2 + (a^2 - b^2)^2 \]