SOLUTION: Find the real-number solutions of the equation. {{{ x^3+2x^2-25x-50 }}} = 0

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Question 625496: Find the real-number solutions of the equation.
+x%5E3%2B2x%5E2-25x-50+ = 0
Found 2 solutions by EdwinParker, solver91311:
Answer by EdwinParker(16) About Me  (Show Source):
You can put this solution on YOUR website!
         x³ + 2x² - 25x - 50 = 0

Factor the first two terms on the left side by taking out GCF x²

        x²(x + 2) - 25x - 50 = 0

Now factor the last two terms on the left side by taking out GCF -25

       x²(x + 2) - 25(x + 2) = 0

Now factor out the GCF (x + 2) leaving x² and -25 in parentheses:

            (x + 2)(x² - 25) = 0

Factor the second parentheses as the difference of squares:

       (x + 2)(x - 5)(x + 5) = 0

Use the zero-factor principle by setting each factor = 0:

x + 2 = 0;  x - 5 = 0;  x + 5 =  0
    x =-2;      x = 5;      x = -5

The real-number solutions are -5, -2, and 5.

Edwin

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


First use the rational roots theorem.

If a polynomial equation of the form:



has any rational roots, they will be of the form where and

Since your lead coefficient is 1, all of the possible rational zeros of your equation are the integer factors of your constant term, namely


Use synthetic division to start testing the possibilities.

Click on: Purple Math Synthetic Division if you need a refresher on synthetic division.

Write back once you have found the first of the three rational roots.

John

My calculator said it, I believe it, that settles it
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