Question 1201673: Use the given zero to find the remaining zeros of the polynomial function
P(x) =x^4-6x^3+71x^2-146x+530; 2+7i Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! Since 2+7i is one x intercept x-2i is the other
P(x)=x^4-6x^3+71x^2-146x+530; 2+7i
x=2-7i
=x-2+7i=0
=(x-(2-7i))=0
(x-(2-7i))(x-(2+7i))=
x^2-x(2+7i)-x(2-7i)+(2+7i)(2-7i)
=x^2-2x-7xi-2x+7xi+53
=x^2-4x+53
=x^4-6x^3+71x^2-146x+530/x^2-4x+53=
x^2-2x+10
=(x^2-4x+53)(x^2-2x+10) =0
x^2-4x+53=0 or x^2-2x+10=0
x^2-4x+53=0
x^2-4x =-53
add 4 to both sides
(x^2-4x+4) = 4-53
(x-2)^2 = -49
Taking square root on both sides
x-2 = sqrt(-49)
x-2 = 7i
x=2+7i
x^2-2x+10=0
x^2-2x=-10
x^2-2x+1= 1-10
(x-1)^2 = -9
taking square root
x-1 = sqrt(-9)
x-1= 3i
x=1+3i
1-3i is the conjugate of 1+3i
The other roots are 1+3i, 1-3i
