First find all possible rational zeros. They will be in the set of all the real integer factors of the last term divided by the set of the real integer factors of the highest degree x's coefficient. Our possible rational zeros are
1,-1,5,-5.
(-1)^3-6(-1)-5=
-1+6-5
5-5
0
(x+1) is a factor. Now we divide and we get this:
(x+1)(x^2-x-5)
After we factor the quadratic (using the quadratic formula, we wind up with x= [-1+/- sqrt(21)]/2, x= -1
The concept i just used to factor was based on the rational zeros theorem.
Comments
if you use synthetic division, you can get (x+1)(x^2-x-5), and i think thats all you can do.
make it a good day
First find all possible rational zeros. They will be in the set of all the real integer factors of the last term divided by the set of the real integer factors of the highest degree x's coefficient. Our possible rational zeros are
1,-1,5,-5.
(-1)^3-6(-1)-5=
-1+6-5
5-5
0
(x+1) is a factor. Now we divide and we get this:
(x+1)(x^2-x-5)
After we factor the quadratic (using the quadratic formula, we wind up with x= [-1+/- sqrt(21)]/2, x= -1
The concept i just used to factor was based on the rational zeros theorem.
T5 = { 1, -1 , 5 , -5}
this are the best numbers to try
after try them you can see that x= -1 is a correct answer
and then you can dividing x³-6x-5 by (x+1)
| 1 | 0 | -6 | -5 |
| 0 | -1 | 1 | 5 |
| 1 | -1 | -5 | 0 |
solution x² -x -5
x³-6x-5 = (x+1) ( x -1/2+r5.25) ( x + 1/2 - r5.25)
(x+1)(x^2 -x - 5)
-1 is a root
x+1 is factor
divide in x+1 to original expression
getx^2-x-5
then factorise and sove this quadratic