Find a polynomial P(x) with real coefficients which has degree 5, leading coefficient of -2, and zeros of 0, 5i and 1-3i

Answer 1

zeros of 0, 5i & 1-3i. with degree 5.

any imaginary zero has a conjugate pair. so if 5i is a zero, so is -5i. if 1-3i is a zero, so is 1+3i.

that's 5 zeros. Take each zero and change its sign. then stick an x in front of it, to get the factor. Multiply all the 5 factors together to get the 5 degree polynomial.

for 0, the factor is just x.

for 5i and -5i, the factor is found by setting x = 5i, squaring both sides, x^2 = -25, x^2 + 25 = 0

with luck I multiplied this correctly. No guarantees. still, it's the basic method, short of any arithmetic mistakes

that'2 3 factors: x and x^2 + 25 or x^3 + 25x

then find the remaining 2 factors from the zero 1 - 3i

and the conjugate pair 1+3i

which gives x^2 -2x +10

multiply them all together. (x^2 -2x +10)(x^3 + 25x)

= x^5 - -2x^4 + 35x^3 - 50x^2 +250x

a 5th degree polynomial

but you want a leading coefficient of -2

so multiply by -2

to get

-2x^5 + 4x^4 -20x^3 -50x^2 +100x