a |
a³+3a²-13a-38 |
-4 |
-2 |
-3 |
1 |
-2 |
-8 |
-1 |
-23 |
0 |
-38 |
1 |
-47 |
2 |
-44 |
3 |
-23 |
4 |
22 |
The table shows a change of sign between a=-4 and a=-3, between -3 and a=-2, between a=3 and a=4, so there are three real roots.
To find them (to factorise) we can use Newton’s method, which will quickly find the zeroes, but it requires calculus.
If we differentiate the function of a we get 3a²+6a-13.
We use the iterative relationship: a=a-(a³+3a²-13a-38)/(3a²+6a-13). We start with a value of a close to a zero and substitute it into the expression after the equals sign. This generates a revised value for the a on the left of the equals sign. This value is then inserted into the right-hand side expression, which gives another value for a. The process is repeated until the value of a ceases to change.
a=-4: after a few iterations, we get a=-3.779310253.
a=-3: a=-2.805118087.
a=4: a=3.58442834.
So the factors are (a+2.805118087)(a-3.58442834)(a+3.779310253).