find complex zeros of the polynomial function. put f in factored form.
f(x)=x^3-6x^2+21x-26
f(x)=(x-2)(x^2-4x+13)
I expect you can discover the roots from that.
To find the complex zeros of the polynomial function and factor it, you can use the Rational Root Theorem and synthetic division.
Step 1: Apply the Rational Root Theorem
The Rational Root Theorem states that if a rational number p/q is a zero of a polynomial, then p is a factor of the constant term, and q is a factor of the leading coefficient. In this case, the leading coefficient is 1, and the constant term is -26. The factors of -26 are ±1, ±2, ±13, ±26, and the factors of 1 are ±1. So, the possible rational roots are ±1, ±2, ±13, ±26.
Step 2: Use synthetic division to test the possible rational roots
Let's start with the possible root x = 1 and perform synthetic division.
1 | 1 -6 21 -26
| 1 -5 16
|----------------
1 -5 16 -10
The remainder is -10, which means that x = 1 is not a root of the polynomial.
Let's continue testing the other possible rational roots until we find one that satisfies the polynomial.
2 | 1 -6 21 -26
| 2 -8 26
|----------------
1 -4 13 0
The remainder is 0, which means that x = 2 is a root of the polynomial. Now, we have factored out a (x - 2) term from the polynomial.
Step 3: Factor the factored polynomial
Taking the result of the synthetic division, we have: (x - 2)(x^2 - 4x + 13).
Now, we need to factor the quadratic polynomial x^2 - 4x + 13. However, since the coefficients of the quadratic polynomial are not easily factorable, we can use the quadratic formula to find its complex zeros.
The quadratic formula is: x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -4, and c = 13.
x = (-(-4) ± √((-4)^2 - 4(1)(13))) / (2(1))
x = (4 ± √(16 - 52)) / 2
x = (4 ± √(-36)) / 2
x = (4 ± 6i) / 2
x = 2 ± 3i
So, the complex zeros of the polynomial function f(x) = x^3 - 6x^2 + 21x - 26 are x = 2 + 3i and x = 2 - 3i.