Write a polynomial function of minimum degree in standard form with real coefficients whose zeros and their multiplicities include those listed.3 (multiplicity 2) 5+ i ​(multiplicity 1)

Answer 1

x=3, x=3, x=5+i and 5-i, as imaginary roots come in conjugate pairs, every time, so 4 roots

(x-3)(x-3) are two roots' corresponding factors = x^2 -6x +9

then the other two: (x-5-i)(x-5+i) = x^2 -10x +25+1 = x^2 -10x +26

multiply them together

(x^2-6x+9)(x^2 -10x +26)

=x^4 -10x^3 +26x^2 -6x^3 +60x^2 -156x +9x^2 -90x + 234

= x^4 -16x^3 +95x^2 -246x +234

or any multiple (except times zero) such as double

2x^4 -32x^3 +190x^2 -492x +468

check the solution, graph it and there is just one x intercept 3 which = the only real solution

the curve is tangent to the x axis at (3,0) a tangency means multiplicity for 3.

Answer 2

The polynomial you want is found by multiplying all of the factors together.

Each factor is found by converting the zeros to factored form. For example

if 3 is a multiplicity then (x - 3) is a factor.

Another note is that if (x-3) is a factor with multiplicity then it is raised to that power

So (x-3)² is the factor you want.

Additionally when (5 + i) is a factor, so is its conjugate (5 - i)

So the solution is found by multiplying out (x - 3)² ( x -(5 + i)) (x - (5 - i))

Multiply the complex ones first ( x -(5 + i)) (x - (5 - i))

First simplify within the parenthesis. (x - 5 - i) ( x - 5 + i) to get x² -5x + xi - 5x + 25 -5i - xi + 5i - i²

Set i² = -1 and collect like terms to get x² - 10x + 25 - (-1) or x² -10x + 26.

Now multiply out (x - 3)² to get x² - 6x + 9

Now multiply both of these together for the final answer

(x² -10x + 26)(x² - 6x + 9) = (x⁴ - 6x³ + 9x²) + (- 10x³ + 60x² - 90x) + (26x² - 156x + 234) or

x⁴ - 16x³ + 95x² - 246x + 234

Answer 3

5 - i is zero also. So, f(x) = a(x- 3)²(x - 5 - i)(x - 5 + i) = a(x- 3)²(x² - 10x + 26), where a ≠ 0 any real number