Question

Find the roots of the polynomial equation.

x3 - 2x2 + 10x + 136 = 0

A. –3 ± i, 4
B. 3 ± i, 4
C. 3 ± 5i, –4
D. –3 ± 5i, –4

Answer is C, Just got it correct on a test.

Answer 1

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

Answer 2

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