Please solve it for me 3x^4 + 6x^3 - 123x^2 - 126x + 1080 = 0?

Answer 1

devide by 3

x^4+2x^3-41x^2-42x + 360 = 0
now if has a real solution it is factore of 360.
1/-1 2/-2 can be tried out

f(3) is zero so x-3 is a factor
x^4-3x^3+5x^3-15x^2-26x^2+78x - 120 x + 360 =0

or(x-3)(x^3+5x^2-26x-120) = 0

x = 3
or say g(x) x^3+5x^2-26x-120 =0

g(5) = 125+125-130-120 = 0
so x(-5) is a factor

g(x) = (x-5)(x^2+10x+24)

x= 5 or x^2+10x+24 =0

x = -6 or -4

so x = 3 or -4 or 5 or -6

Answer 2

given: 3x^4 + 6x^3 - 123x^2 - 126x + 1080 = 0 <=>
x^4 + 2x^3 - 41x^2 - 42x + 360=0 (¤¤¤¤)
I have only divided with 3:

in general the equation can be written
A*x^4 +B*x^3 + C*x^2 + Dx+E = 0...(####)
in our case A=1, D= 360

if p/q is a rational root in (####) then
p is divisor in A and q is divisor in E =>

p belongs to {-1,1}

look at 360 = 1 * 2 * 3 * 3 * 4 * 5, I have resolved 360 in its prime factor, therefore we have that q consist of a combination of these prime numbers:

there is a lot of combinations but we can make an educated guess lets say 3 ,-4 ,5 , -6 (use your calculator)

now we need to control our result, and we can do in two different ways, by setting our guess into (¤¤¤¤) or by calculating the product:
(x-3) * (x+4) * (x-5) *(x+6) = 0 <=>
(x^2+x -12) * ( x^2+x-30) = 0 <=>
x^4 + 2x^3 - 41x^2 - 42x + 360 = 0, but this equation is equal to (¤¤¤¤)

so we have show that the roots = { x in R I x = 3 or x = - 4 or x = 5 or x = - 6 }
===================================

Answer 3

3x^4 + 6x^3 - 123x^2 - 126x + 1080 = 0
=>3(x^4 + 2x^3 - 41x^2 - 42x + 360) = 0
=>x^4 + 2x^3 - 41x^2 - 42x + 360 = 0
=>x^4 + (x^3+x^3) -(12x^2+30x^2+x^2) - (36x +12x)+ (30*12) = 0
=>(x^2)(x^2) + (x^3+x^3) -(12x^2+30x^2+x^2) - (36x +12x)+ (30*12) = 0
=>(x^2)*[x^2 + x - 30] + (x)*[x^2 + x - 30] - 12*(x^2)[x^2 + x - 30]=0
=>[x^2 +x -12]*[x^2 + x - 30]=0
=>[ x -3 ]*[ x + 4 ][ x - 5 ][ x + 6 ]=0
=> x = 3 or x = -4 or x = 5 or x = -6

Answer 4

3x^4 + 6x^3 - 123x^2 - 126x + 1080 = 0
3(x + 6)(x + 4)(x - 3)(x - 5) = 0
x = -6 or -4 or 3 or 5