What is a polynomial function in standard form with zeros 1, 2, -2 and -3
2 answers:
The zeros of the polynomial are <span>1, 2, -2 and -3, so this polynomial must have at least one of each of these factors:
(x-1), (x-2), (x-(-2)), and (x-(-3)); rewriting: </span>(x-1), (x-2), (x+2), and (x+3).
Thus, any such polynomial must have a factor (x-1)(x-2)(x+2)(x+3).
The simplest such polynomial we can think of, is p(x)=(x-1)(x-2)(x+2)(x+3).
To write in standard form, lets first multiply the factors two by two as follows:
by the difference of squares formula,
.
Next, we multiply our results:
.
Answer:
(x-1), (x-2), (x-(-2)), and (x-(-3)); rewriting: </span>(x-1), (x-2), (x+2), and (x+3).
Thus, any such polynomial must have a factor (x-1)(x-2)(x+2)(x+3).
The simplest such polynomial we can think of, is p(x)=(x-1)(x-2)(x+2)(x+3).
To write in standard form, lets first multiply the factors two by two as follows:
Next, we multiply our results:
Answer:
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Answer:
2π/3, 4π/3
Step-by-step explanation:
I interpret your question as asking us to solve the equation
2cos(x) + 1 = 0
2cos(x) = -1
cos(x) = -½x
x = arccos(-½), 0 ≤ x ≤2π
Since the cosine is negative, x must be in the second or third quadrant.
Referring to the unit circle, we find that
x = 2π/3 (120°) or x = 4π/3 (240°)
