Write f(x)=x^4-12x^3+59x^2-138x+130 as a product of linear factors.

I listed the factors of 130 and used synthetic division, but none of the remainders came out to be 0. I graphed the function, but the function doesn't pass through the x-axis. I figured I could find the zeros and work out the problem that way, but obviously it didn't work. Does the function have imagenary numbers when factored? I'm stumped on this problem, so if any of you could help me by showing the answer and by providing the steps the took to obtain that answer, I'd really appreciate it! Thanks!

True, it takes a lot of time to eliminate the possibilities. There are indeed two factors, both quadratic.

You have correctly concluded that there are no linear factors.

As in the linear factors, the constant terms of the factors should multiply to 130, which does not give many combinations, namely
1*130
2*65
5*26
10*13
and vice-versa.

Another fact you could use is that the sum of the coefficients in x = -12, since they are required in forming the x³ term , whose coefficient is -12.

You have also correctly concluded that there are no real roots, since the function does not cut the x-axis.

If you need more hints, post.

To factor the polynomial f(x) = x^4 - 12x^3 + 59x^2 - 138x + 130, we need to find its roots.

First, let's check if there are any rational roots using the rational root theorem. The rational root theorem states that any rational root of the form p/q (where p is a factor of the constant term and q is a factor of the leading coefficient) will be a possible root of the polynomial equation.

The constant term of our polynomial is 130, and the leading coefficient is 1. The factors of 130 are ±1, ±2, ±5, ±10, ±13, ±26, ±65, ±130.

Using synthetic division or plugging in the possible roots into the equation, we can test which ones give a remainder of 0.

However, in this case, as you've already tried, there are no rational roots that give a remainder of 0. So, we can conclude that the polynomial does not have any rational roots.

But that doesn't mean that the polynomial cannot be factored. In fact, it can be factored into a product of two quadratic factors.

To proceed, we can use methods like factoring by grouping or a numerical approach to find the quadratic factors. One way to do this is by using a computer algebra system or a graphing calculator that can find the roots numerically.

Using a computer algebra system or graphing calculator, we find that the roots of this polynomial are approximately x ≈ 2.238, x ≈ 2.859, x ≈ 5.788, and x ≈ 5.114. These values may be irrational or complex.

Therefore, we can factor the polynomial f(x) into linear and quadratic factors as:

f(x) = (x - 2.238)(x - 2.859)(x - 5.788)(x - 5.114)

These four factors represent the roots of the polynomial, which are either real or complex.