check/ solve algebraically and state exact roots

[kathy's]

I have two questions. Can someone check my work for me?

Solve the equations algebraically and state the exact roots.

1) X^3 - 4X = 3X^2 - 12

-X^3 + 3X^2 - 4X - 12 = 0

I find the factors of 12 and when I plug it back into the equation it gives me zero. I found 2 works so (x - 2) is a factor. Then I use synthetic division like so:

2 | -1  3  4 -12
  |    -2  2  12
  --------------
    -1 -1  6   0

-X^2 + X + 6

(X + 3)(X + 2)

The exact roots are X = 2, -2, and -3. Is this right?

The second one has the same instructions but different equation:

2) m^3 - 5m^2 - 5m +1 = 0

I find the factors of 1 and when I plug it back into the equation it gives me zero. So I found -1 works so (x + 1) is a factor. Then I use synthetic division like so:

-1 | 1 -5 -5  1
   |   -1  6 -1
   -------------
     1 -6  1  0

m^2 - 6m +1 = 0

(m - 2)(m + 3)

So the exact roots are m = 2, -3, and -1. Is this one correct as well?

Thanks for your help!

 

[tkhunny]

You have the idea, I think, but you are slopping around a bit.

The first is almost magic. You have a coefficient of "-4", but managed the correct "+4" when you set up the synthetic division. 3-1 = +1. You wrote down "-1", but on multiplication by '2', still managed the correct "+2" in the table. Also, why did you manipulate to a negative leading coefficient. That's OK, but a bit unusual.

On the second, the synthetic division looks OK, but subsequent factoring is rather confused. All you had when you started was factors of "1". How did you manage 2 and 3 in the final answer? I think you got your '1' and '6' mixed up.