finding an equation given a degree and zeros

Question

degree 5; leading coefficient 3; zeros: 5 (multiplicity 2),1,4-i

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Answer 1

3(x-1)(x-5)^2(x^2 -8x+17) multiplied out is a 5th degree polynomial with leading coefficient 3 and 5 zeros of 1, 5, 5, 4-i and 4+i. 5 repeats. The polynomial when graphed crosses the x axis once at (1,0) and is tangent to the x axis at (5,0) The degree of a polynomial is equal to the number of zeros if you count the multiplicity and imaginary zeros.

x=4-i

x-4= -i

square both sides

x^2 -8x +16 = -1

x^2 -8x +17 is another factor with two imaginary zeros, a conjugate pair: 4-i and 4+i. Imaginary zeros always come in conjugate pairs

(4-i)(4+i) = 16+1 = 17

x^2 -8 +16 = -1

x-4 = + or - sqr(-1) = + or - i

x =4+i or 4-i are two imaginary zeros

3(x-1)(x-5)(x-5)(x^2-8x+17)

=3(x-1)(x^2-10x+25)(x^2-8x+17)

=3(x^3 -11x^2 +45x -25)(x^2 -8x+17)

=(3x^3 -33x^2 +135x -75)(x^2-8x+17)

=3x^5 -42x^4 +450x^3 -1716x^2 +2895x -1275 is degree 5, leading coefficient 3, with zeros 1,5,5,4+i