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Question

x^4+5x^3-27x-135=0

Answer 1

Hi Brian,

 

x⁴+5x³-27x-135 = 0

 

Let's try grouping

 

(x⁴+5x³)-(27x+135) = 0

 

Try factoring each group

 

x³(x+5)-27(x+5) = 0

 

This becomes

(x³-27) (x+5) = 0

 

For the first term the difference between two cubes can be factored as with the (x+5) appended

(x-3)(x²+3x+9) (x+5) = 0

 

x-3 = 0     x=3

x+5 = 0   x=-5

x²+3x+9 = 0

 

This quadratic cannot be factored.  We can solve it though using the quadratic formula

 

-b±√(b²-4ac)

---------------

       2a

 

-3±√[3²-(4)(1)(9)]

---------------------

        2(1)

 

3±√(9-36)

------------

      2

 

3±√-27

--------

    2

 

3±3√-3

---------

    2

 

3±3i√3

--------

    2

 

So we have 4 solutions:

2 real : 3, -5

2 imaginary:  3±3i√3

                         --------

                             2