Which of the following sets contains all roots of the polynomial f(x)=2x^3+3x^2-3x-2?
1 answer:
Answer:
C
Step-by-step explanation:
Given
f(x) = 2x³ + 3x² - 3x - 2
Note that
f(1) = 2 + 3 - 3 - 2 = 0 , thus
(x - 1) is a factor
Dividing f(x) by (x - 1) gives
f(x) = (x - 1)(2x² + 5x + 2) = (x - 1)(x + 2)(2x + 1)
To find the roots equate f(x) to zero, that is
(x - 1)(x + 2)(2x + 1) = 0
Equate each factor to zero and solve for x
x - 1 = 0 ⇒ x = 1
x + 2 = 0 ⇒ x = - 2
2x + 1 = 0 ⇒ 2x = - 1 ⇒ x = -
The solution set is therefore
{ - 2, - , 1 } → C
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Answer:
B. Y = 6/4 X
Step-by-step explanation:
Well to find its parallel line we need to put,
2y - 3x = 4 into slope-intercept.
+3x to both sides
2y = 3x + 4
Now we divide everything by 2,
y = 3/2x + 2
So a line that is parallel to the given line will have the same slope but different y intercept, meaning we can cross out choices A and C.
To check look at the image below ↓
<em>Thus,</em>
<em>answer choice B. Y = 6/4 X is correct.</em>
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<em>Hope this helps :)</em>
