Which of these is a zero of the polynomials p(y) = 3y³- 16y-8!
Answers
Answer:
-2
Step-by-step explanation:
substitute each y term with -2,u will get +32 and -32=0 thus -2 is zero of polynomial
Given : p(y) = 3y³ - 16y - 8
To Find : Zeros of the polynomial
Solution:
p(y) = 3y³ - 16y - 8
p(1) = 3(1)³ - 16(1) - 8 = -21
p(2) = 3(2)³ - 16(2) - 8 = -16
p(3) = 3(3)³ - 16(3) - 8 = 25
p(-1) = 3(-1)³ - 16(-1) - 8 = 5
p(-2) = 3(-2)³ - 16(-2) - 8 = 0
Using factor theorem
(y - (-2)) = y + 2 is a factor
and - 2 is a zero
3y² -6y -4
y + 2 3y³ - 16y - 8
3y³ + 6y²
___________
-6y² - 16y - 8
-6y² -12y
______________
-4y - 8
-4y - 8
______________
0
3y³ - 16y - 8 = (y + 2)(3y² -6y -4)
3y² -6y -4 = 0
=> y = (6 ± √(36 + 48))/(2* 3)
=> y = ( 3± √21 )/3
= y = 1 ± √7/√3
zeroes are - 2 , 1 + √7/√3 , 1 - √7/√3
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