In determining the factorization, it is a good trick to know that when the coefficient of the x^2 is indeed one (as it is in this case), solving for the two factors is much easier.

Basically, what two numbers, when multiplied, equal positive 5, but when added together, equal negative 6?

The only possibility is -1 and -5. Therefore (x-1), (x-5)

To check your answer, you can use the FOIL method of multiplying your two factors (first, outside, inside, last), to arrive back to the original equation:

First = x times x equals x^2.

Outside = x times -5 equals -5x.

Inside = -1 times x equals -x.

Last = -1 times -5 equals 5.

Combined then you have x^2 -5x -x +5, simplified to x^2-6x+5.

The solution of a quadratic equation ax^2 + bx + c = 0 is given by `(-b+-sqrt(b^2-4ac))/(2*a)`

In the problem, the solution of the equation x^2 - 6x + 5 = 0 has to be determined. Using the formula given earlier, a = 1, b = -6 and c = 5.

The roots of the equation are `(6 +-sqrt(36-20))/(2*1) = (6+-4)/2`

The roots are 5 and 1.

It is also possible to factorize the expression x^2 - 6x + 5 to determine the roots of the equation.

x^2 - 6x + 5 = 0

=> x^2 - 5x - x + 5 = 0

=> x(x - 5) - 1(x - 5) = 0

=> (x - 1)(x - 5) = 0

=> x = 1 and x = 5

Again, the same roots are obtained.

**The roots of the equation x^2 - 6x + 5 = 0 are 1 and 5**

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