SOLUTION: For the function below neatly solve for x. 𝑓(𝑥) = 6𝑥^3 + 25𝑥^2 − 24𝑥 + 5 Use the Remainder Theorem to check potential rational zeros. (Remember the techniques f

Algebra ->  Rational-functions -> SOLUTION: For the function below neatly solve for x. 𝑓(𝑥) = 6𝑥^3 + 25𝑥^2 − 24𝑥 + 5 Use the Remainder Theorem to check potential rational zeros. (Remember the techniques f      Log On


   

Question 1148165: For the function below neatly solve for x.
𝑓(𝑥) = 6𝑥^3 + 25𝑥^2 − 24𝑥 + 5
Use the Remainder Theorem to check potential rational zeros. (Remember the
techniques from the notes; you do not need to check ALL potential zeros.)
Use Synthetic Division to find a depressed polynomial once you know a zero. Repeat with additional zeros until you have the function down to a factorable function.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The possible rational roots are of the form plus-or-minus p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

You can make the list as easily as we can. p is either 1 or 5; q is either 1, 2, 3, or 6.

You can go through the list of possible rational roots and try each one using synthetic division; but that is tedious and not a good use of your time.

One way to start on the problem and make much better use of your time (still giving you practice with synthetic division) is to use a graphing calculator to find one of the roots and go from there.

Or you can do some logical reasoning, using Vieta's Theorem to see that the sum of the roots is -25/6 and the product of the roots is -5/6.

That suggests that -5 is one of the roots; and synthetic division confirms that:

    -5  |  6  25  -24   5
        |    -30   25  -5
        +-----------------
           6  -5    1   0

So

6x%5E3%2B25x%5E2-24x%2B5+=+%28x%2B5%29%286x%5E2-5x%2B1%29

Then the quadratic factors nicely: 6x%5E3%2B25x%5E2-24x%2B5+=+%28x%2B5%29%283x-1%29%282x-1%29

And so the roots are -5, 1/3, and 1/2.