Answer:
-1 and 3
Step-by-step explanation:
First, note that there is an evident root in x=-1. You can see this because the coefficients of x3 and x sum the same number than the ones of x2 and the independent term (they sum 4: 1+3=4 and -5+9=4).
We can use this root to apply the Ruffini's rool with all the coefficients:
1 -5 3 9
-1 -1 6 -9
1 -6 9 0
This gives as that we can write the equation x3 – 5x² + 3x+9=0 as:
(x+1)*(x2-6x+9) = 0
The first term is due to the root x=-1 and the following is a product of Ruffini's rool. Now we need to find the 2 (at maximum) remaining roots, which are the roots of (x2-6x+9)=0
For this we could use Baskara's formula (you can try it) but lets do something simpler. We know that with the 2 roots, call them a and b, we can write the equation as: (x-a)(x-b)=0
If we apply distributive to (x-a)(x-b) we must get x2-6x+9. So, (-a)(-b) MUST be equal to 9. Working with integers there are not many options. Lets suppose a=b=3:
(x-3)*(x-3)=x2 -3x -3x + 9 = x2 - 6x + 9, and BINGO!
If we didnt get the results we could have tried with a=1 and b=9, for example, and so on. (please, compare with Baskara!).
So, we can write:
x3 – 5x² + 3x+9= (x+1)*(x-3)*(x-3) = (x-1)*(x-3)^2
And the roots are the values that make this equation zero:
x=-1 and x=3 where the last one is a double root.
