find a quadratic equation that has the given two zeros have

find a quadratic equation that has the given two zeros, have your answer in expanded form. (y = ax^2 + bx + c). 2 - 3 i and 2 + 3 i 3 - Squareroot 7 and 3 + Squareroot 7 - 8 i and 8 i 1 - 2 Squareroot 3 and 1 + 2 Squareroot 3 find all the zeros of the given polynomial functions. Zeros found on the calculator must be verified. No Decimal answers!! Show your work. f(x) = x^4 - 38 x^2 + 80 x + 21 f(x) = x^4 - 3x^3 - 50 x^2 + 276 x - 224 f(x) = 3x^4 - 8x^3 - 77 x^2 + 432 x - 252 f(x) = 4 x^4 - 28 x^3 - 130 x^2+ 532 x - 9 f(x) = x^4 - 16 x^3 + 55 x^2 - 106 x - 1326 f(x) = x^5 - 5x^4 - 22 x^3 + 86 x^2 - 235 x + 175 f(x) = x^4 - 6x^3 + 29 x^2 - 96 x + 208 Given: 4i is a zero f(x) = x^4 - 18 x^3 + 72 x^2 + 126 x - 553 Given: Squareroot 7 is a zero f(x) = x^4 - 22 x^3 + 183 x^2 - 718 x + 1150 Given: 7 + Squareroot 3 is a zero f(x) = 9x^4 - 78 x^3 + 160 x^2 + 362 x + 31 Given: 5 - i Squareroot 6 is a zero

Solution

Hi, You have posted 14 different questions. I will be answering the first one.

Whenever roots of quadratic are given, You can find the equation using this method. If a and b are the zeros or roots , equation is (x-a)*(x-b)=0

therefore, for 1st one it is(x-(2-3i))(x-(2+3i))=0 , this can also be written as ((x-2)^2 - (3i)^2)=x^2 + 4 -4x + 9=

x^2-4x+13