how do i factorize?

Question

x^3 + 4X^2-2X-5=0

I want to factorize it .
tell the general procedure for this . How do i start ?

please tell me a procedure which i could apply for other similar kind of problems.

it would be a very bad idea , if we try guessing roots of the given equation.

what would be the best way to attack this problem ?

Answer 1

you may not like it, but guessing the roots is one of the best things you can do
of course, there are some shortcuts - for example, the rational root must be r = u/v where u is a factor of the free coefficient and v is a factor of the leading coefficient

Answer 2

this is a cubic function. In order to factorise it you first have to identify the first factor by guessing method, normally one has to try different values of x that will make the expression =0
try numbers closer to 0. example(0,1,-1,2,-2) and so on . the first number that will make it zero will be the solution.
for this we try: 0
then: o^3+4(0)^2 -2(0)-5=-5 so 0 is not a solution.
we try: 1
then :1^3+4(-1) ^2-2(1)-5=-2 it is not also a solution
we try:-1
then :-1^3+4(-1) ^2-2(-1)-5=0 and we got it
since -1 is a solution , then (x+1) is a factor. because if x=-1 then x+1=0 that is why (x+1) is a factor.
we now do the long division to find the remainder:

____ x^2+3x-5
x+1 ┌──────────────
x^3+4x^2-2x-5
- x^3+x^2
-------------
0+3x^2-2x-5
- 3x^2+3x
----------------------
0-5x-5
- 5x-5
---------------------
0

and we now have: (x+1) (x^2+3x-5)
when we can factorise the quadratic one, things would look better , but in this case, i think we can't so we leave it as it is.

hope this is helpful

Answer 3

x³ + 4x² - 2x - 5 = 0
(x + 1)(x² + 3x - 5) = 0

I'm sure you can solve the quadratic portion yourself.

If there are rational roots they will be some combination of the terms for the highest power of x and the constant. Since the leading coefficient is 1 and the constant is a prime, then the only possibililties are ±1 and ±5. If those don't work, any solution that exists is irrational.

Answer 4

factor using the roots (x ints of the graph)
if you look at the graph it crosses at -1, therefore (x+1) is a factor
so factor that out we get
x^2+3x-5, then use quadratic formula, you get
x=-4.19, x=1.19

Answer 5

x³ + 4x² - 2x - 5=0
Factoring:
x(x² + 4x - 2 - 5) = 0
Baskara => ax² + bx + c = 0
x² + 4x - 7 = 0
delta = b² - 4ac
delta = 4² - 4.1.-7
d = 16 + 28
d = 44

x = (-4 +/- \/44) : 2
x' = (-4 + 2\/11) : 2
x' = -2 + \/11
x" = -2 - \/11
x'" = 0
::==::