How do we factorize a cubic equation???

Question

I dont want to use hit and trial method.

Please give sensible answers.

Answer 1

x^3 + x^2 + x + 1 could be factored by spliting the terms
that is
(x^3 + x^2) + (x + 1) then factor out x^2 from the first term
(x^2)(x + 1) + (x + 1) now factore out the (x + 1)
(x+1)(x^2 + 1) and it is factored

the trick is to split it, this however, only works with some for other equations you must use long division. which is fairly a long process to explain, but if you look in a pre-calculus book you will find it easily and it will be explained fairly well and if you still have questions feel free to im me or email me... good luck!

Answer 2

It is too complex to give a details here.
Please refer to below for the method

However an equation of the form if it has at least one rational toot is easy as below

ax^3+bx^2+cx+ d = 0

where a b c and d if it has got rational roots then

the zero shall be of the form s/t where s is a factor of d and t a factor if a
then one can check with all the cases and find it


for example

f(x) = x^3 -6x^2 + 11x - 6= 0

if it has got a rational root has to be factor of 6

1/-1/2/-2/3-3

by checking f(1) = 1 - 6+ 11 - 6 = 0

so x-1 is a factor.

by taking x-1 a factor we get

f(x) = (x-1)(x^2-5x+6)

2nd one being quadratic it can be factored using standard methods for quadratic

Answer 3

ax^3+bx^2+cx+ d = 0
where a b c and d if it has got rational roots then
the zero shall be of the form s/t where s is a factor of d and t a factor if a
then one can check with all the cases and find it

for example
f(x) = x^3 -6x^2 + 11x - 6= 0
if it has got a rational root has to be factor of 6
1/-1/2/-2/3-3
by checking f(1) = 1 - 6+ 11 - 6 = 0
so x-1 is a factor.
by taking x-1 a factor we get
f(x) = (x-1)(x^2-5x+6)
2nd one being quadratic it can be factored using standard methods for quadratic equations

or
by synthetic division which is the above method

Answer 4

There is a formula you could use, but it's so bulky and ugly that only computers use it.

It depends on the problem you are asked to factor. Usually, you can determine a zero (or it's given in the problem) and factor that out using long division or synthetic division. From there it's a matter of factoring a quadratic.

Second, you can use a computer algebra system such as Mathematica, Maple, of the Voyage 200 calculator.

Unfortunately, there is no straight forward method. I wish there were!

Answer 5

It would have been better if you would have given some expression to explain this.
To facterise you must know certain rules
A cubic expression contains 3 +1 = 4 terms
A square expression contains 2 +1 = 3 terms
(a+b)^n contains n + 1 terms
In cubic expression index of1st term is 3, of 2nd term is 2, of 3rd term is 1 and of 4th term is 0
The index of "a" goes on decreasing from 3 to 0 and the index of "b" goes on increasing from 0 to 3
you will understand better from the following example
(a+b)^3 = a^3b^0 + 3a^2b^1 + 3a^1b^2 +a^0b^3
(a+b)^3 = a^3 + 3a^2b + 3ab^2 +b^3 --------(1)
(a–b)^3 = a^3 – 3a^2b + 3ab^2 – b^3 sign of terms + and – are alternate.---------(2)
a^3 + b^3 = (a + b)(a^2 – ab +b^2) ----------(3)
a^3 – b^3 = (a – b)(a^2 + ab +b^2) ---------(4)
After knowing these rules you can solve any expression depending upon the expression.
If the expression contains 2 terms of type e.g. x^3a^3 + y^3b^3 then you can use above rule 3
If the expresion contains 4 terms then arrange the terms in the expression as per the decreasing index
e.g. a^3 + 3b^3 + ab^2 + 3a^2b
arrange the terms in decreasing order of index of "a"
a^3 + 3a^2b – ab^2 – 3b^3
make groups of 2 terms first group a^3 + 3a^2b & second group ab^2 + 3b^3 (generally marked by undrerling )
Then from first group a^2 is common and from second group – b^2 is common. & write the remaining terms in bracket.
a^3 + 3a^2b – ab^2 – b^3
= a^2(a + 3b) – b^2(a + 3b)
= (a + 3b)(a^2 – b^2)
now (a^2 –b^2) can again be facterised as (a+b)(a –b)
a^3 + 3a^2b – ab^2 – b^3
= (a + 3b)(a+b)(a –b)
Is OK ?

Answer 6

Sorry there is no other way.There exists
a formula for the solution(as there is one for the 2nd degree equation) but it is too complicate for normal use.
So hit and try searching possible rational roots ,study the derivative and find out intervals where roots are etc

Answer 7

YOU CAN BREAK CUBIC EQUATION
AND NOW
YOU CAN SOLVE THIS QUESTION