Solve for x

Question

Please help me solve the following:

 

x³+2x²-9x-18=0

Answer 1

The best approach for this problem is called factoring by grouping. 

In this method, the first two terms are one group, and the last two terms are the second group.

Looking at the first group, a factor of x² can be factored out  x²(x + 2 ).

Looking at the second group, a factor of -9 can be factored out  -9 ( x +2)

 

Now (this is the good part) it can be noticed that both of these expression have a factor of (x + 2) ,

thus whole expression can be written as

 

(x + 2) ( x² -9)

 

Also    x² - 9 = (x - 3) (x + 3)  , so the original equation is

 

(x + 2) (x - 3) (x + 3)  =0

 

From this it is easy to see the the solution set is  {-3 , -2 ,3 }

 

The factor by grouping does not always work, but when it does , it works very well and easily.

Answer 2

Grace,

          If you assume that k is one of the roots then x-k is a factor, if you then divide the cubic equation by x-k and solve for k I got k=-2 so then the equation factors into (x+2)(x²-9)=0 so the roots are x=-3,-2 and 3.

Another technique is to try a few integers to find a root by trial and error once you find one root then proceed to divide and factor the resulting quadratic.

 

Hope this helps 

 

Jim

Answer 3

x³ + 2x² - 9x -18 = 0

 

Another approach to factoring is to use the rational root theorem:

 

             multiples of (-18)  -->  ±1, ±2, ±3, ±6, ±9, ±18

Roots = --------------------         ---------------------------

              multiples of (1)   --->                ±1

 

Let's guess -2 (-2/2, or 2/-1) to start and do synthetic division.  If it divides evenly with no remainder, it's a root.

 

-2|  1  2  -9  -18

           -2   0   18

     ---------------

      1  0  -9    0

 

The remainder is 0, so -2 is a root (solution).  From the synthetic division, we're left with the quotient:

 

x² - 9 = 0

 

which factors to (x + 3)(x - 3), so 3 and -3 are also roots.

 

So the answer is x = -3, -2, 3

 

To check, plug each value of x back into the original equation and verify that it = 0.