SOLUTION: write the polynomial function of minimum degree in standard form that has zero -1 and 2i. Assume the leading coefficient is 1.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: write the polynomial function of minimum degree in standard form that has zero -1 and 2i. Assume the leading coefficient is 1.      Log On


   

Question 1155804: write the polynomial function of minimum degree in standard form that has zero -1 and 2i. Assume the leading coefficient is 1.
Found 3 solutions by MathLover1, Boreal, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

function of minimum degree in standard form that has zeros
x%5B1%5D=-1
+x%5B2%5D=2i-> since complex zeros come in pairs, you also have
+x%5B3%5D=-2i
then
f%28x%29=%28x-x%5B1%5D%29%28x-x%5B2%5D%29%28x-x%5B3%5D%29
f%28x%29=%28x-%28-1%29%29%28x-2i%29%28x-%28-2i%29%29
f%28x%29=%28x%2B1%29%28x-2i%29%28x%2B2i%29.....use the rule for difference of squares
f%28x%29=%28x%2B1%29%28x%5E2-%282i%29%5E2%29
f%28x%29=%28x%2B1%29%28x%5E2-4%28i%29%5E2%29
f%28x%29=%28x%2B1%29%28x%5E2-4%28-1%29%29
f%28x%29=%28x%2B1%29%28x%5E2%2B4%29
f%28x%29=x%5E3%2B4x%2Bx%5E2%2B4
f%28x%29=x%5E3%2Bx%5E2%2B4x%2B4

Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
-2i and +2i are roots, since complex roots are conjugate
-1 is other,
(x+1)
x=2i, x= -2i. (x^2+4)=0
(x+1)(x^2+4)
x^3+x^2+4x+4=0

also,
if first coefficient is 1 then
x=(1/2)(-b+/- sqrt(b^2-4ac)
the discriminate is -16 and b is 0
therefore -4ac=-16
a is 1, c=4
(x^2+4)

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
write the polynomial function of minimum degree in standard form that has zero -1 and 2i. Assume the leading coefficient is 1.
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Complex zeroes go in conjugate pairs if and only if the polynomial has real coefficients.


But the problem does not state it --- therefore both solutions of the two other tutors are IRRELEVANT.