SOLUTION: Find all real zeros of the function y=x^3 - 2 x^2 - 25 x +50.

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Question 977664: Find all real zeros of the function y=x^3 - 2 x^2 - 25 x +50.
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!

Solution 1

The polynomial  y = x%5E3+-+2x%5E2+-+25x+%2B50  has the real root  x = 2.
Indeed,  it is easy to check:  2%5E3+-2%2A2%5E2+-+25%2A2+%2B+50 = 0.

Hence,  this polynomial can be factored

x%5E3+-+2x%5E2+-+25+x+%2B50 = %28x-2%29.f%28x%29

with he linear factor  x-2,  where  f%28x%29  is a quadratic polynomial.

To find  f%28x%29,  divide the polynomial  x%5E3+-+2x%5E2+-+25x+%2B50  by  x-2  (long division).
If you perform this division,  you will get

x%5E3+-+2x%5E2+-+25x+%2B50 = %28x-2%29.%28x%5E2+-+25%29.

Hence,  two other roots of the given polynomial are the roots of the quadratic polynomial  x%5E2+-+25,  i.e.  x%5B1%5D = 5  and  x%5B2%5D = -5.

Answer.  The roots of the given polynomial  x%5E3+-+2x%5E2+-+25x+%2B50  are  2,  5  and  -5.


Solution 2

You can get factoring the polynomial  x%5E3+-+2+x%5E2+-+25+x+%2B50  by grouping its terms:

x%5E3+-+2x%5E2+-+25x+%2B50 = %28x%5E3+-+2x%5E2%29 - %2825x+-50%29 = x%5E2%28x-2%29 - 25%28x-2%29 = %28x%5E2-25%29%2A%28x-2%29.

It gives you the same roots.