Question

Consider the polynomial p(x) = x^3 + 4x^2 + 6x − 36.
a. Graph y = x^3 + 4x^2 + 6x − 36, and find the real zero of polynomial p.
b. Write p as a product of linear factors.
c. What are the solutions to the equation p(x) = 0?

Answer 1

Answer:

a. attached graph; zero real: 2

b. p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)

c. the solutions are 2, -3-3i and -3+3i

Step-by-step explanation:

p(x) = x³ + 4x² + 6x - 36

a. Through the graph, we can see that 2 is a real zero of the polynomial p. We can also use the Rational Roots Test.

p(2) = 2³ + 4.2² + 6.2 - 36 = 8 + 16 + 12 - 36 = 0

b. Now, we can use Briott-Ruffini to find the other roots and write p as a product of linear factors.

2 |  1     4     6    -36

     1      6    18     0

x² + 6x + 18 = 0

Δ = 6² - 4.1.18 = 36 - 72 = -36 = 36i²

√Δ = 6i

x = -6±6i/2 = 2(-3±3i)/2

x' = -3-3i

x" = -3+3i

p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)

c. the solutions are 2, -3-3i and -3+3i