Find all zeros (real and complex) and factor the polynomial completely. a) P(x) = x^4 - 25x^3 + 235x^2 - 995x + 1624 and x = 5 + 2i is a zero of P(x)

Answer 1

two zeros are 5+2i and 5-2i. imaginary zeros come in conjugate pairs

x=5+2i

x-5=2i

x^2 -10x +25 = -4

x^2 -10x +29 is one factor

divide it into the 4th degree polynomial

by descartes' rule of signs there are no negative real zeros, and either 0, 2 or 4 positive real zeros

since we know there are 2 imaginary zeros, that means either 0 or 2 positive real zeros and 2 or 4 imaginary zeros. but revise this if the constant term was really intended to be negative

after dividing the quadratic into the 4th degree polynomial, either factor the quotient or use the quadratic formula to find the remaining 2 zeros

(x^4 -25x^3 +235x^2 -995x +1624)/(x^2-10x+29) = a quadratic

use long division

x^2-15x -56

x^2 -10x + 29 |x^4 -25x^3 +235x^2 -995x +1624

-10x^3 + 29x^2

-15x^3 + 206x^2 -995x

150x^2-435x

56x^2 - 560x +1624

-560x -56(29)

x^2 -15x -56= the other integer factor if 1624 was a mistake and was really -56(29)= -1624

x^4 -25x^3 +235x^2 -995x -1624 = (x^2-15x-56)(x^2-15x-56)

factoring further gives you 2 imaginary factors and 2 irrational factors

then the other two zeros are

x=15/2 +/- (1/2)sqr(225 +224) = 7.5 +/-.5sqr449

irrational zeros also come in conjugate pairs

in this problem one irrational zero is positive, one is negative

but had that 2nd quadratic been x^2-15x +56

then it factors into (x-8)(x-7) with zeros 7 and 8

Answer 2

See the video.