To find the zeros of the function, we need to set it equal to zero and solve for x:
x(x-1)(2x+4)^2 = 0
Since we have a product of factors equal to zero, one or more of the factors must be equal to zero.
Setting x = 0:
(0)(0-1)(2(0)+4)^2 = 0
This simplifies to:
0 * (-1) * 4^2 = 0
0 * -1 * 16 = 0
0 = 0
So one zero is x = 0.
Setting x - 1 = 0:
(x-1)(0)(2x+4)^2 = 0
This simplifies to:
(x-1) * 0 * 4^2 = 0
0 * 0 * 16 = 0
0 = 0
So another zero is x = 1.
Setting 2x + 4 = 0:
x(x-1)(0)^2 = 0
This simplifies to:
x(x-1) * 0 = 0
0 * 0 = 0
0 = 0
Since this equation is always true, the factor 2x + 4 does not contribute any additional zeros.
Therefore, the function f(x) = x(x-1)(2x+4)^2 has 2 zeros: x = 0 and x = 1.
The answer is 4.
How many zeros, including repeated zeros, does the function f(x)=x(x−1)(2x+4)^2 have?(1 point)
1. The degree cannot be determined from the information provided.
2. 4
3. 3
4. 2
1 answer
1 answer