Find a polynomial with integer coefficients such that (sqrt3 + sqrt5) is a root of the polynomial

I thought that question was all cleared up

http://www.jiskha.com/display.cgi?id=1267933399

its sqrt 3+ sqrt5. not 3+ sqrt5

To find a polynomial with integer coefficients such that (sqrt3 + sqrt5) is a root, we can use the technique called "rationalizing the denominator."

Here's how we can do it step by step:

Step 1: Let's Assume that p(x) is the polynomial we are looking for.

Step 2: Since (sqrt3 + sqrt5) is a root, we know that (x - (sqrt3 + sqrt5)) is a factor of the polynomial. Notice that we can also write (sqrt3 + sqrt5) as ((sqrt3 + sqrt5) * (sqrt3 - sqrt5)) / (sqrt3 - sqrt5) = (3 + 2sqrt3sqrt5 - 5) / (sqrt3 - sqrt5) = -2 / (sqrt3 - sqrt5).

Step 3: Now, let's rationalize the denominator of (-2 / (sqrt3 - sqrt5)). Multiplying the numerator and denominator by the conjugate of (sqrt3 - sqrt5), which is (sqrt3 + sqrt5), gives us: (-2 * (sqrt3 + sqrt5)) / ((sqrt3 - sqrt5) * (sqrt3 + sqrt5)).

Step 4: Simplifying the denominator, we get (-2 * (sqrt3 + sqrt5)) / (3 - 5) = (-2 * (sqrt3 + sqrt5)) / (-2) = sqrt3 + sqrt5.

Step 5: Now, we know that (x - (sqrt3 + sqrt5)) is a factor of the polynomial p(x). We can multiply the polynomial by the conjugate factor to get rid of the square root: (x - (sqrt3 + sqrt5)) * (x - (sqrt3 - sqrt5)) = (x - sqrt3 - sqrt5) * (x - sqrt3 + sqrt5).

Step 6: Expanding the above expression, we get x^2 - xsqrt3 + xsqrt5 - sqrt3x + 3 - sqrt15 + sqrt3sqrt5 - sqrt3sqrt5 - 5.

Step 7: Simplifying further, we have x^2 - 2sqrt3x - 2sqrt5.

Therefore, the polynomial with integer coefficients such that (sqrt3 + sqrt5) is a root is x^2 - 2sqrt3x - 2sqrt5.