It's impossible to factor x^2-6x+5 right?
mmmh, let's see, I see
(x-1)(x-5)
Ah, thank you. I was thinking that 5 had to be divided into squares but it is prime.
Actually, it is possible to factor the quadratic equation x^2 - 6x + 5. Let me explain the process step by step.
To factor a quadratic equation of the form ax^2 + bx + c, we need to find two binomial factors that multiply together to give us the original equation.
In this case, we have x^2 - 6x + 5 as our quadratic equation, so a = 1, b = -6, and c = 5.
To factor the equation, we want to find two numbers whose sum is -6 (the coefficient of x) and whose product is 5 (the constant term).
We need to find two numbers that satisfy the following equation:
m * n = ac
where m and n are the two numbers we are trying to find, and ac is the product of the coefficient of the x term and the constant term.
In this case, ac is equal to 1 * 5 = 5.
Now we need to find two numbers whose product is 5 and whose sum is -6. The numbers that satisfy this condition are -1 and -5, since -1 * -5 = 5 and -1 + -5 = -6.
Once we have found these two numbers, we can rewrite the middle term of our original equation (x^2 - 6x + 5) as the sum of these two numbers, -1x - 5x, and then factor by grouping.
So, we can rewrite the equation as:
x^2 - 6x + 5 = x^2 - x - 5x + 5.
Now, we can group the terms as follows:
(x^2 - x) + (-5x + 5).
From the first group, we can factor out the common factor x, and from the second group, we can factor out the common factor -5:
x(x - 1) - 5(x - 1).
Notice that we now have a common binomial factor (x - 1). We can combine the terms using this common factor:
(x - 1)(x - 5).
Therefore, the factored form of x^2 - 6x + 5 is (x - 1)(x - 5).
In conclusion, it is indeed possible to factor the quadratic equation x^2 - 6x + 5, using the method of factoring by grouping.