The key piece of information for these questions is the Fundamental Theorem of Algebra, which states that a degree n polynomial has n complex roots. A complex root can be either real or imaginary.
First question, regarding the polynomial y = x^3 - 3x^2 + 16x - 48:
We know there is one real root, the x-intercept.
Since it's a third degree polynomial, there are three complex roots in total.
Therefore, there is one real root and two imaginary roots.
Answer is B
Second question:
You probably can guess the answer, now that you know the Fundamental Theorem of Alegebra:
There are 3 real zeros, each with multiplicity one, meaning each root only happens once. It's a 5th degree polynomial, so there are a total of 5 roots, implying 2 imaginary roots.
Answer is C) 3 real and 2 imaginary zeroes.
Answer:
The correct answer is B, the second option
Step-by-step explanation:
(-2, 5/3)
Solve for the first variable in one of the equations, then substitute the result into the other equation.
Answer: C. 5/6
Steps:
Use the formula x1-x2/y1-y2 to find the slope
(5,11), (-5,-1)
-5-5/-1-11
= -10/-12
=5/6
Answer:
See below.
Step-by-step explanation:
(a) Because the solution led to a square root of a negative number:
x^2 -10x+40=0
x^2 - 10x = -40 Completing the square:
(x - 5)^2 - 25 = -40
(x - 5)^2 = -15
x = 5 +/-√(-15)
There is no real square root of -15.
(b) A solution was found by introducing the operator i which stands for the square root of -1.
So the solution is
= 5 +/- √(15) i.
These are called complex roots.
(c) Substituting in the original equation:
x^2 - 10 + 40:
((5 + √(-15)i)^2 - 10(5 + √(-15)i) + 40
= 25 + 10√(-15)i - 15 - 50 - 10√(-15)i + 40
= 25 - 15 - 50 + 40
= 0. So this checks out.
Now substitute 5 - √(-15)i
= 25 - 10√(-15)i - 15 - 50 + 10√(-15)i + 40
= 25 - 15 - 50 + 40
= 0. This checks out also.
