Find the complex zeros of the polynomial function. Write f in the factored form. f(x)=x^3-3x^2+7x-5
f(x)=
a little synthetic division shows
f(x) = (x-1)(x^2-2x+5)
And it's easy from there
First we have to find the real root, (every cubic has at least one real root)
try x = 1 , y = 1 - 3 + 7 - 5 = 0 , so x-1 is a factor
by synthetic division I got
(x-1)(x^2 - 2x + 5) = 0
So the real zero is when x=1
for complex ...
x^2 - 2x + 5= 0
x^2 - 2x = -5
x^2 - 2x + 1-5 + 1
(x-1)^2 =-4
x - 1 = ± 2i
x = 1 ± 2i
To find the complex zeros of a polynomial function, we need to factorize the polynomial function and set each factor equal to zero. Let's factorize the polynomial function f(x) = x^3 - 3x^2 + 7x - 5.
First, we need to check if any of the integers divide evenly into -5 (the constant term of the polynomial). Since -5 is a prime number, we can conclude that there are no rational zeros in this case.
Next, we can try using synthetic division to find a zero as a starting point. Let's test with x = 1:
1 | 1 - 3 + 7 - 5
| 1 - 2 + 5
|_____________________
1 - 2 + 5 0
Since the remainder is 0, it means that x = 1 is a zero of the polynomial function f(x).
Now, we have a quadratic expression as a result: x^2 - 2x + 5. We can try to find the other two zeros by factoring this quadratic expression.
The quadratic equation x^2 - 2x + 5 does not factorize further, so we need to use the quadratic formula to find the remaining zeros:
x = (-b ± √(b^2 - 4ac)) / 2a
a = 1, b = -2, and c = 5.
Plugging these values into the quadratic formula, we get:
x = (-(-2) ± √((-2)^2 - 4(1)(5))) / 2(1)
= (2 ± √(4 - 20)) / 2
= (2 ± √(-16)) / 2
= (2 ± 4i) / 2
= 1 ± 2i
Therefore, the complex zeros of the polynomial function f(x) = x^3 - 3x^2 + 7x - 5 are x = 1, x = 1 + 2i, and x = 1 - 2i.
Now, let's write the polynomial f(x) in factored form:
f(x) = (x - 1)(x - (1 + 2i))(x - (1 - 2i))
Therefore, the factored form of f(x) is f(x) = (x - 1)(x - (1 + 2i))(x - (1 - 2i)).