Question 362458: Suppose that a polynomial function of degree 5 with rational coefficients has -2-3i, -5 and square root of 7 as zeros. Find the other zeros.
a. 5, 2+3i
b. -2+3i, -square root of 7
c. 3i, -2
d. 5, -2+3i, square root of 7
e. none of the above
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! Suppose that a polynomial function of degree 5 with rational coefficients has -2-3i, -5 and square root of 7 as zeros. Find the other zeros.
a. 5, 2+3i
b. -2+3i, -square root of 7
c. 3i, -2
d. 5, -2+3i, square root of 7
e. none of the above
x + 2 - 3i = 0, x = -2 + 3i ->this should be a zero if we want i's to cancel out
x + 2 + 3i = 0, x = -2 - 3i ->know this is x is a zero
(x + 2 + 3i)(x + 2 - 3i) = x^2 + 2x - 3ix + 2x + 4 - 6i + 3ix + 6i - 9i^2
= x^2 + 2x + 2x - 3ix + 3ix - 6i + 6i + 4 + 9 = x^2 + 4x + 13
this leaves us with answers:
b. -2+3i, -square root of 7
d. 5, -2+3i, square root of 7
AND e. none of the above
were told -2-3i, -5 and square root of 7 are zeros
there should be 5 zeros
so answers are either b. -2+3i, -square root of 7
OR e. none of the above
(x + sqrt(7))(x - sqrt(7)) = x^2 - 7, square roots canceled out
so if we want i's to cancel out and square roots of 7 to cancel out
then
-2 - 3i, -5, square root of 7, -2 + 3i, and -square root of 7 are the roots
to get all rational coeffiecients
so answer is b. -2+3i, -square root of 7
and 5th degree polynomial is:
-2 - 3i, -2 + 3i, sqrt(7), -sqrt(7), -5 roots of 5th degree polynomial
(x + 2 + 3i)(x + 2 - 3i)(x - sqrt(7))(x + sqrt(7))(x + 5) = 0
(x^2 + 4x + 13)(x^2 - 7)(x + 5) = 0
(x^4 - 7x^2 + 4x^3 - 28x + 13x^2 - 91)(x + 5) = 0
(x^4 + 4x^3 + 6x^2 - 28x - 91)(x + 5) = 0
x^5 + 4x^4 + 6x^3 - 28x^2 - 91x + 5x^4 + 20x^3 + 30x^2 - 140x - 455 = 0
x^5 + 9x^4 + 26x^3 + 2x^2 - 231x - 455 = 0
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