Factoring cubics can be tough, especially since most of them can't be factored. But some certainly can. Usually the strategy is to factor a common term from the original terms, but two at a time. For example, if you have 5x^3 + 5x^2 + 4x + 4, this is 5x^2(x+1) + 4(x+1), which then factors to (5x^2 + 1)(x+1).
Here we have three terms, but seeing the "13" and the "12" makes me think that we can more or less change the 13 to 12+1, and then group and factor from there:
many methods, you can use the cubic function but itll take too long, so just use long division
find a factor of x3 - 13x + 12
*a factor is a number that will make the function = 0
*if leading coeffecient is 1, then try numbers that that multiply to 12 (number without the x in it) (+/-3, +/- 4, +/- 6, +/- 2, +/- 12, +/- 1)
alright so...
ill try 1 in place of x
=1^3 - 13 (1) + 12
=0
so 1 works, so your factor would be (x-1)
now, divide the function by (x-1) using long division method
*i added a - 0x2 in the equation because it has to be in order( i.e x^3, x^2, x^1, x^0) these are called fake terms that have no affect on the question, u dont have to add them, i just do to help
-------------x^2 +x-12
___________________
x-1 divide x^3 - 0x^2 - 13x + 12
-------------x^3 - x^2
------------------------------------
x^2 -13x
x^2 - x
---------------------------------------
-12x + 12
-12x + 12
---------------------------------------------
0
now, you have x^2 + x -12, factor that quadratic even more
(find out 2 numbers that multiply to -12 and add to +1)
ok, you do it like this: first u write -13x as -9x-4x and you get
x^3 -9x -4x+12
now first two go together and the last two together; --> x(x^2 -9) - 4(x-3)
and x^2 -9 is (x-3)(x+3)--> x(x+3)(x-3) -4(x-3) and now you "expose" /sorry don't know the correct english expression) (x-3) to get --> (x-3)(x(x+3)-4)=(x-3)(x^2+3x-4) and the second bracket can be further factorized; x^2+3x-4=(x+4)(x-1)-->and then you get (x-3)(x+4)(x-1)
Comments
Factoring cubics can be tough, especially since most of them can't be factored. But some certainly can. Usually the strategy is to factor a common term from the original terms, but two at a time. For example, if you have 5x^3 + 5x^2 + 4x + 4, this is 5x^2(x+1) + 4(x+1), which then factors to (5x^2 + 1)(x+1).
Here we have three terms, but seeing the "13" and the "12" makes me think that we can more or less change the 13 to 12+1, and then group and factor from there:
x^3 - 13x + 12
x^3 - 12x - x + 12
x^2 (x-12) - x + 12
x^2 (x-12) - (x-12)
(x^2 - 1)(x - 12)
(x+1)(x-1)(x-12)
many methods, you can use the cubic function but itll take too long, so just use long division
find a factor of x3 - 13x + 12
*a factor is a number that will make the function = 0
*if leading coeffecient is 1, then try numbers that that multiply to 12 (number without the x in it) (+/-3, +/- 4, +/- 6, +/- 2, +/- 12, +/- 1)
alright so...
ill try 1 in place of x
=1^3 - 13 (1) + 12
=0
so 1 works, so your factor would be (x-1)
now, divide the function by (x-1) using long division method
*i added a - 0x2 in the equation because it has to be in order( i.e x^3, x^2, x^1, x^0) these are called fake terms that have no affect on the question, u dont have to add them, i just do to help
-------------x^2 +x-12
___________________
x-1 divide x^3 - 0x^2 - 13x + 12
-------------x^3 - x^2
------------------------------------
x^2 -13x
x^2 - x
---------------------------------------
-12x + 12
-12x + 12
---------------------------------------------
0
now, you have x^2 + x -12, factor that quadratic even more
(find out 2 numbers that multiply to -12 and add to +1)
numbers are : -3 and 4
so (x+3), (x-4), (x-1) are your factors
helpful websites: http://www.purplemath.com/modules/polydiv2.htm
http://calc101.com/webMathematica/long-divide.jsp#...
rational root theorem suggests trying ±1, ±2, and so on (factors of 12) as roots.
synthetic division with x = 1 gives us
(x - 1)(x² + x - 12) =
(x - 1)(x + 4)(x - 3)
ok, you do it like this: first u write -13x as -9x-4x and you get
x^3 -9x -4x+12
now first two go together and the last two together; --> x(x^2 -9) - 4(x-3)
and x^2 -9 is (x-3)(x+3)--> x(x+3)(x-3) -4(x-3) and now you "expose" /sorry don't know the correct english expression) (x-3) to get --> (x-3)(x(x+3)-4)=(x-3)(x^2+3x-4) and the second bracket can be further factorized; x^2+3x-4=(x+4)(x-1)-->and then you get (x-3)(x+4)(x-1)
x^(3)-13x+12
Factor the polynomial using the rational roots theorem.
(x-1)(x-3)(x+4)