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
A. Categorical
b. I think walking group (I’m not sure)
c. Bar graph
8,7,9,3,4,9,2
This is correct because I’m in middle school
Answer:
1. if x + 5 = 12, then x = 7
:Subtraction property of equality
2. If x + y = 20, and y = 5, then x + 5 = 20.
<u>:Substitution property of equality</u>
3. If 2x3 = 11, then 11 = 2x - 3.
<u>:Symmetric property of equality</u>
In that statement of a function, 'a', 'b', and 'c' all represent
plain old ordinary constant numbers.