find the zero's for the following polynomial and give the multiplicity for each zero. For each zero, state whether the graph crosses the x-axis or touches the x-axis and turns around.
f(x)=5(x-2)(x+4)^3
I am having trouble with this. Can you please show me how you got the answer. Thanks.
x = 2 , multiplicity: 1
x = -4 , multiplicity : 3
crosses at both x=2 and x=-4 (odd multiplicities )
What is the value of the zero with the smaller multiplicity from that same equation?
To find the zeros of the given polynomial, we need to set f(x) equal to zero and solve for x. So, we have:
5(x-2)(x+4)^3 = 0
Setting each factor equal to zero, we get:
x-2 = 0 (1st factor)
x+4 = 0 (2nd factor)
Solving equation 1 for x, we have:
x = 2
Solving equation 2 for x, we have:
x = -4
So, the zeros of the polynomial are x = 2 and x = -4.
Now, let's determine the multiplicity of each zero. The multiplicity of a zero refers to the number of times the factor appears in the polynomial.
For the first factor (x-2), it appears once as (x-2). So, the multiplicity of x = 2 is 1.
For the second factor (x+4), it appears three times as (x+4)^3. So, the multiplicity of x = -4 is 3.
To determine whether the graph crosses the x-axis or touches the x-axis and turns around at each zero, we can use the fact that the multiplicity of a zero determines the behavior of the graph near that zero.
For x = 2 (with multiplicity = 1), the graph crosses the x-axis.
For x = -4 (with multiplicity = 3), the graph touches the x-axis and turns around.
I hope this helps! Let me know if you have any further questions.
To find the zeros of a polynomial, we need to set the polynomial equal to zero and solve for the variable. In this case, we have the polynomial:
f(x) = 5(x-2)(x+4)^3
To find the zeros, we set f(x) equal to zero:
0 = 5(x-2)(x+4)^3
Now we solve for x. To do this, we can set each factor equal to zero and solve individually.
Setting the first factor equal to zero:
x-2 = 0
x = 2
This gives us one zero at x = 2.
Next, we set the second factor equal to zero:
x+4 = 0
x = -4
This gives us a second zero at x = -4.
Lastly, we set the third factor equal to zero:
(x+4)^3 = 0
For this factor, we can take the cube root of both sides to find the values of x:
x+4 = 0
x = -4
Since we have been given that the exponent of this factor is 3 (multiplicity), we know that x = -4 has a multiplicity of 3. This means that x = -4 is a zero that repeats 3 times.
For each zero, we need to determine if the graph crosses the x-axis or touches the x-axis and turns around. This is determined by the multiplicity of the zero.
For x = 2, since the multiplicity is 1, the graph crosses the x-axis at x = 2.
For x = -4, since the multiplicity is 3, the graph touches the x-axis and turns around at x = -4.
To summarize:
The polynomial f(x) = 5(x-2)(x+4)^3 has the following zeros and multiplicities:
Zero: x = 2 (Multiplicity: 1)
Zero: x = -4 (Multiplicity: 3)
At x = 2, the graph crosses the x-axis. At x = -4, the graph touches the x-axis and turns around.