#42 02.03 Real Zeros of Polynomial Functions

Mathematics
Tutor: None Selected Time limit: 1 Day

May 12th, 2015

p(x) = x^3 + 0x^2 - 28x - 48.

-4 |  1 |  0  | -28  |  -48
   |    | -4  |  16  |  -48
-----------------------------
     1 -4 -12 0

I.e. p(x) = (x+4)*(x^2 - 4x -12). To find remaining zeros we have to solve
x^2 - 4x - 12 = 0,  a quadratic equation. Use the simplified formula for even b:

x2,3 = 2 +- sqrt(4+12) = 2 +- 4,
so x2 = -2, x3 = 6.

Zeros: -4, -2, 6.

May 12th, 2015

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