Math test tomorrow and I need help on just one thing...Finding Complex zeros!?

Question

is someone could walk me through how to solve this sample problem I will love you forever...okay the question is...

find the complex zeros of x^3-4x^2+25x-100

please walk me through it or give me a good website that will walk me through it

Answer 1

Well, the worst part of your problem is finding your first positive root (guaranteed to have at least one positive root because of Descartes' rule of signs).

Using synthetic division, when you test +4, your bottom line translates into x^2 + 25 = 0 (because your bottom line is
1, 0, 25.

so your non real answers come from x = + or - sqrt(-25) which is plus or minus (5i).

Answer 2

The zeros of this polynomial are the solutons of the equation
x^3 - 4x^2 + 25x - 100 = 0
The expression factors using integer factoring methods:
(x^3 - 4x^2) + (25x - 100) = 0
x^2(x - 4) + 25(x - 4) = 0
(x - 4)(x^2 + 25) = 0
Since you have only linear and quadratic factors, you can solve this equation using the Zero-Product Property.
(x - 4)(x^2 + 25) = 0
x - 4 = 0 or x^2 + 25 = 0
x = 4 or x^2 = -25
x = 4 or x = (+/-) 5i

So the complex zeros of x^3 - 4x^2 + 25x - 100 are
4, 5i, and -5i.

Answer 3

first consider the equation x^3-4x^2 +25x-100 = 0.By trial and error we see that left hand side is zero when x=4.so it is a zero.now we divide the polynomial x^3-4x^2 +25x-100 by 4 (by long division or synthetic division) :
we get x^3-4x^2 +25x-100 = (x-4)(x^2+25)
Now x^2+25 = 0 gives x^2= - 25 which gives x = square root of -25 which is + or -5i as square root of a negative number is imaginary and i is the imaginary unit which stands for square root of - 1.so the complex roots are +5i or -5i.

Answer 4

I assume you mean to set the expression equal to zero and solve for x.

Any cubic equation will always have at least one real solution. So first try to factor it. Since the coefficient of x³ is one, so if the expression has a rational factor, it will be a factor of 100, either + or -.

x³ - 4x² + 25x - 100 = 0
(x - 4)(x² + 25) = 0

We have just factored out (x - 4). Now just use the formula for quadrilaterals to solve the rest.

x = 4, 5i, -5i

Complex zeros always come in pairs.

Answer 5

Factorize the polynomial
x^3-4x^2 +25x-100 = (x-4)(x^2+25)
So one root is 4. The other two zeroes occur when x^2 = -25
So x^2 = (5i)^2
So x = + 5i and -5i

Answer 6

x^3 - 4x² + 25x - 100 = 0
x²(x - 4) + 25(x - 4) = 0
(x² + 25)(x - 4) = 0
x - 4 = 0
x = 4
x² + 25 = 0
x² = -25
x = ±√-25
x = ±5i

Answer 7

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