You use rational root theorem.
Let p = constant term
Let q = numerical coefficient of highest degree of x
x = +/- p/q
In here, p = -20 and q = 1. Thus,
x = (+/-) 1 , 2 , 4 , 5 , 10, 20
You have to substitute each factor (for instance x = -1) to the expression, and if you get a zero answer, that value of x is a factor. Well, it's kind of like trial and error, but at least you know which ones are to be substituted.
Hope this helps :3
Is it possible to factor x^3 - 9x^2 + 24x - 20 and solve for x by hand?
I tried using the grouping method and got x^2(x-9) + 4(6x-5) but how can I continue solving for x?
If I use a calculator I get x= 2 and 5 but is it possible to find this by hand? thanks
3 answers
*I mean you get all the factors of x = +/- p/q. In the problem, x = +/- 20, and its factors are written there above. The factors are the ones you have to check by substituting them to the expression. :)
After showing that both f(1) and f(-1) ≠ 0
try f(2)
= 8 -36+48-20 = 0
so (x-2) is a factor
Now use synthetic division and
x^3 - 9x^2 + 24x - 20
= (x-2)(x^2 - 7x + 10)
= (x-2)(x-2)(x-5)
so x^3 - 9x^2 + 24x - 20
= (x-5)(x-2)^2
For a cubic, once you find one root, you can reduce your cubic to a quadratic, by either long or synthetic division.
Of course you could just feel lucky and keep going to find other roots from the list that Jai gave you, and you would hit x = 5
And of course, as soon as you found 2 factors of a cubic you can "reason" out what the third factor would be.
try f(2)
= 8 -36+48-20 = 0
so (x-2) is a factor
Now use synthetic division and
x^3 - 9x^2 + 24x - 20
= (x-2)(x^2 - 7x + 10)
= (x-2)(x-2)(x-5)
so x^3 - 9x^2 + 24x - 20
= (x-5)(x-2)^2
For a cubic, once you find one root, you can reduce your cubic to a quadratic, by either long or synthetic division.
Of course you could just feel lucky and keep going to find other roots from the list that Jai gave you, and you would hit x = 5
And of course, as soon as you found 2 factors of a cubic you can "reason" out what the third factor would be.
4 answers