find a polynomial function of lowest degree with rational coefficients that has the given numbers as zeros: 2-rad3, 1+i

Answer 1

F√ind a polynomial function of lowest degree with rational coefficients that has the given numbers as zeros:

(2-√3), (1+i)

***Important*** Complex numbers always, I repeat always come in conjugate pairs!

Recall complex number form as a + bi and its conjugate a - bi, this is very important to rewrite/convert complex zeros back into their factored form, based on this information, I know that the polynomial will be at minimal a degree three function (cubic equation).

Recall the the general pattern for and degree polynomial, such as:

y = a (x - r_,)ⁿ (x - r₂)ⁿ, where n represents the multiplicity of the factor.

Step 1) We convert and rewrite zeroes into the factored form and we will start with the easier of the two:

2-√3, setting it equalled to x as follows:

x = 2 - √3 → Using algebra we manipulate it to set it zero (x - 2 +√3)=0

Step 2) Find the conjugate of the given complex zero: 1 + i, which is 1 - i

Step 3) Set both complex zeros equal to zero and rewrite/convert back to factored form:

x = 1+i and x = 1 - i → Using algebra to manipulate we get: (x - 1 - i) = 0 and (x - 1 + i) = 0

Step 4) We multiply together these two complex factors using FOIL/distribution to eliminate the i generating a quadratic expressions as follows:

(x - 1 - i)(x - 1 + i) → After FOIL and eliminating i, we get: x²- 2x + 2

Step 5) Combine factors and plug them into the general polynomial factor form as:

y = a (x - 2 +√3) (x²- 2x + 2), we need a point (x,y) that lies on this polynomial in order to find a, the leading coefficient of the function.