SOLUTION: could someone help me with this.... Use the given zero to find the remaining zeros of the function. ​h(x)=2x^4+3x^3+48x^2+75x-50; zero: −5i The remaining zeros of h a

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: could someone help me with this.... Use the given zero to find the remaining zeros of the function. ​h(x)=2x^4+3x^3+48x^2+75x-50; zero: −5i The remaining zeros of h a      Log On


   

Question 1157330: could someone help me with this.... Use the given zero to find the remaining zeros of the function.
​h(x)=2x^4+3x^3+48x^2+75x-50; zero: −5i


The remaining zeros of h are_______
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

h%28x%29=2x%5E4%2B3x%5E3%2B48x%5E2%2B75x-50
if given zero x%5B1%5D=+-5i, then you also have x%5B2%5D=+5i
so, find a product %28x-x%5B1%5D%29%28x-x%5B2%5D%29 and divide given equation using long division
%28x-%28-5i%29%29%28x-5i%29
%28x%2B5i%29%28x-5i%29
x%5E2-cross%285i%2Ax%29%2Bcross%285i%2Ax%29-%285i%29%5E2
x%5E2-25%28i%29%5E2
x%5E2-25%28-1%29
x%5E2%2B25

.......................(2x%5E2%2B3x-2
x%5E2%2B25|2x%5E4%2B3x%5E3%2B48x%5E2%2B75x-50
.....................2x%5E4%2B0%2Ax%5E3%2B50x%5E2..........subtract
....................................3x%5E3-2x%5E2......bring down 75x
....................................3x%5E3-2x%5E2%2B75x
....................................3x%5E3%2B0%2Ax%5E2%2B75x .........subtract
................................................-2x%5E2......bring down -50
................................................-2x%5E2-50.
................................................-2x%5E2-50.......subtract
.............................................................0
so 2x%5E4%2B3x%5E3%2B48x%5E2%2B75x-50= %28x%5E2%2B25%29%282x%5E2%2B3x-2%29
now factor 2x%5E2%2B3x-2=0
2x%5E2-x%2B4x-2=0
%282x%5E2-x%29%2B%284x-2%29=0
x%282x-1%29%2B2%282x-1%29=0
%28x+%2B+2%29+%282+x+-+1%29+=+0
remaining zeros are:
x=-2
x=1%2F2