Find all real-number solutions: x^3-5x^2+8x-4=0

Question

find all real-number solutions 

Answer 1

The Rational Root test yields the candidates ±1, ±2, ±4.

Testing all of them yields 1 and 2 are both roots.

Since the leading coefficient is 1, the constant coefficient equals negative of the product of the roots, thus the product of the roots is 4.

So 1*2*r = 4 so r = 2 is the last root.

Result: 1 is a single root and 2 is a double root.

Answer 2

Let f(x) = x^3-5x^2+8x-4

Since f(1) = 1-5+8-4 = 0, by factor theorem, x-1 is a factor. Now you can use synthetic division to get the other factors.

1 | 1 -5 8 -4

........1 -4 4

-------------------

....1 -4 4 | 0

So, you have another factor x^2-4x+4 = (x-2)^2

x^3-5x^2+8x-4 = (x-1)(x-2)^2 = 0

Answer: x = 1, 2