SOLUTION: f(x)= 2x^2-16x+30
g(x)= x*(x-3)^2*(x+1)^4
Find the zeros of g(x) and f(x).
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Question 1009152: f(x)= 2x^2-16x+30
g(x)= x*(x-3)^2*(x+1)^4
Find the zeros of g(x) and f(x).
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
set f(x) equal to 0 and you get:
2x^2 - 16x + 30 = 0
factor out a 2 and you get:
2 * (x^2 - 8x + 15) = 0
factor x^2 - 8x + 15 to get:
2 * (x-3) * (x-5) = 0
set each of these factors equal to 0 to get:
x-3 = 0 and x-5 = 0
solve for x to get x = 3 and x = 5.
the zeroes of y = 2x^2 - 16x + 30 are at x = 3 and x = 5.
here's the graph:
your second equation is:
g(x)= x(x-3)^2 (x+1)^4
set g(x) equal to 0 and you get:
x * (x-3)^2 * (x+1)^4 = 0
set each of these factors equal to 0 and you get:
x = 0 and (x-3)^2 = 0 and (x+1)^4 = 0
solve for x in each of these equations to get:
x = 0
(x-3)^2 = 0
take square root of both sides of this equation to get:
(x-3) = 0
solve for x to get x = 3
(x+1)^4 = 0
take the fourth root of both sides of this equation to get:
x + 1 = 0
solve for x to get x = -1
your zeroes are at x = -1, x = 0, and x = 3
here's the graph:
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