How to I factor the following polynomial

x^3 - 5x^2 + x - 5


SHOW ALL Work please! I don't want the answer but the guidelines to solving this. I have the answer in this textbook in front of me. (Test tomorrow)

x³ - 5x² + x - 5 = x³ + x - 5(x²+1) = x(x²+1) - 5(x²+1) = (x-5)(x²+1)

If you divide the polynomial into two sides, you'll note that (x-5) is common....it yields

(x-5)*x^2 + (x-5)*1

This means that (x-5) is common to "both sides".

Hence, one level of factorization is (x-5)(x^2+1).

This means that the polynomial only has one real root x^2 + 1 = 0 does not have real roots.

You can verify this by applying Descartes test to the polynomial.


Another approach is to use Vieté:

If you assume the roots are a, b, and c, then (x-a)(x-b)(x-c) = x^3 - 5x^2 + x - 5.

This means

(1) a + b + c = 5
(2) ab + bc + ac = 1
(3) abc = 5

Now you can substitute and solve algebraically. However, having 2 non-real roots will prove challenging in solving this system of non-linear equations.

I hope this helps.

The first two terms have a common factor in x^2. The last two terms have a common factor in 1. Factor those terms out.
x^2(x-5) + 1(x-5)
Now you have a binomial. Each term has a factor of (x - 5). Factor that out for the final answer.
Answer: (x^2 + 1) (x - 5)


hope this helped! :)

Factor by grouping:

(x^3-5x^2) + (x-5)
x^2(x-5) + 1(x-5)
take out common factor x-5 to get (x-5)(x^2+1)

taking x^2 common
x^2(x-5)+1(x-5) {since x-5 has only 1 as factor}
(x^2+1)(x-5)


either
x=5(equating with zero for x)
pr,
x=i

x^3 - 5x^2 + x - 5 =
x^2(x-5)+1(x-5) =
(x-5)(x^2+1)

x^3-5x^2+x-5=(x^2+1)(x-5)


checking:

(x^2+1)(x-5)

= x^3-5x^2
x-5
_________
x^3-5x^2+x-5





^__^

x^3 - 5x^2 + x - 5= x^2(x-5)+(x-5)=(X-5)(x^2+1)