Question

Find the roots of the polynomial equation.

Answer 1

Set f(x) = x³ - 2x² + 10x + 136.

Use Descartes' rule of signs.
f(x) has 2 change in sign, so there are
(1) 2 real roots, or
(ii) a pair of conjugate imaginary roots.

f(-x) = -x³ - 2x² - 10x + 136
f(-x) has 1  sign change so there is 1 real root. 

Scan the function for the first zero crossing, using negative values of x.

 x    f(x)
--- ---------
-1   123
-2  100
-3    61
-4      0

x = -4 is a root, so (x+4) is a factor.
Perform synthetic division (or long division if you do not know synthetic division).

 
-4 |  1  -2  10  136
          -4  24 -136
    ---------------------
       1  -6  34      0

Therefore
f(x) = (x+4)(x²- 6x + 34) = 0

Before factoring x² - 6x + 34, test the discriminant.
D = (-6)² - 4*34 = -100 => we have a pair of imaginary roots.
Use the quadratic formula.
x = (1/2)[6 +/- √-100]
   = 3 +/- 5i

Answer: C
The roots are -4, 3 +/- 5i

Answer 2

The answer (Which is 3 +/- 5i, -4) is now B on the Unit 6 Test.