The polynomial can be factorised to make it simpler to solve:
m(9m3-6m2-32m+32).
So m=0 is a zero of the polynomial. We need only look for one zero of the cubic which would then reduce it to a quadratic which can be solved by simple factoring or using the quadratic formula.
We can try rational zeroes by looking at the factors of the lead coefficient 9 and the constant 32.
Factors of 9 are 1, 3, 9 and the factors of 32 are 1, 2, 4, 8, 16, 32.
The rational zeroes are found by dividing each of the factors of 32 by those of 9 (omitting duplicates):
|
1 |
2 |
4 |
8 |
16 |
32 |
| 1 |
1 |
2 |
4 |
8 |
16 |
32 |
| 3 |
⅓ |
⅔ |
1⅓ |
2⅔ |
5⅓ |
10⅔ |
| 9 |
1/9 |
2/9 |
4/9 |
8/9 |
16/9 |
32/9 |
Each of these can be in the positive or negative form. There are therefore 36 values to plug in turn into the cubic. That's a lot of computing! What we're looking for is a value of m which makes the cubic evaluate to zero.
Let's start with whole numbers. It's clear that neither 1 or -1 is a zero. So let's try m=2: 72-24-64+32=16, so m=2 is not a zero. But if m=-2: -72-24+64+32=0 so m=-2 is a zero.
Using synthetic division divide by this zero:
-2 | 9 -6 -32 32
9 -18 48 | -32
9 -24 16 | 0 = 9m2-24m+16 = (3m-4)2, giving us the duplicate zero m=4/3 or 1⅓.
So the zeroes are 0, -2, 1⅓.
