if a polynomial has roots at -3 and 4+5i, what is the polynomial in standard form?

Question

show all of your work so i can understand it better please !

Answer 1

the standard way to solve this is change the sign of each zero, include the conjugate of the imaginary zero, put an x in front of each, and multiply them altogether

Another method is use the solution for the imaginary zeroes is use the quadratic formula

x = -b/2a + or - (1/2a)sqr(b^2-4ac) with -b/21 = 4 and and the rest = 5i, solve for a,b & c. let a=1

-b/2a = 4

-b/2 = 4

b = -8

x = 4 + or - sqr((b^2-4ac/4a^2) = 4 + or - sqr(b^2/4a -c/a) =4 + or - sqr(16-c/a)

where sqr(16-c/a) = 5i = sqr(16-c) =5i

16-c = (5i)^2 = -25

c= -25-16 = -41

x^2 -8x -41

one factor is ax^2 + bx + c = x^2 -8x -41.

with root of -3, the other factor is x+3

x =-3 + 0 where 0=1/2asqr(0) the discriminant b^2-4ac=0

x+3 = 0

multiply (x+3)(x^2-8x-41)= x^3 -5x +17x - 123 is the polynomial in standard form

the other method is easier, but this way shows where the quadratic formula involves the roots or zeroes, including imaginary ones.

Answer 2

If 4+5i is a root, then its conjugate, 4-5i is also a root.

p(x) = (x+3)(x - 4-5i)(x-4+5i) = (x+3)(x²-8x+41)

= x³-8x²+41x +3x²-24x+123

= x³-5x²+17x+123