how do i solve and graph these factors

y= x(x+3)^2(x+5)

y=x(x-2)(x+0.5)(2x-1)

y= x(x+3)^2(x+5)

I am not going to tell you how to graph that, but I will tell you how to sketch the graph.

for very large positive x, the thing is large positive, so in the end it will slope up to the right. So over on the right in the first quadrant draw an arrow pointing up to the right.

for very large negative x, because the highest power of this polynomial is x^4 which is even, y will again be large positive. So over in the left quadrant you have an arrow pointing up to the left.
now it is zero at x = 0 , x - -3 (twice), x = -5
Mark those 3 or really four but two are the same spots with dots on your graph.
SO
Your function comes screaming down from your arrow at the upper left and passes through (-5,0) headed down.

then it turns up and grazes the x axis again at (-3,0). I say grazes because it did not have enough oomph to get through the x axis on its way up and just touched it, heading back down. If it had gone through, we would have had two zeros close to x = -3, but they are scrunched together because the function just touched the axis there and dropped back down.
then it heads back up again to pass through the origin (0,0) on its way up to your arrow at the upper right.

Now you should be able to do the second one.

to solve, let y = 0

since the function is already factored, this is now easy
x(x+3)^2(x+5) = 0

so x=0 or x=-3 or x=-5

to graph it, realize that the solution to the corresponding equation is the x-intercept of the function.
When you have a "double root" such as you get from (x+3)^2, the graph will touch the x-axis without crossing over.

Furthermore, since the highest term is +x^4 the curve will "get lost" up in the first and second quadrants.
So looking at our solutions, the graph will cross at -5, touch at -3 and cross again at 0

I then substitute a value of x about 1/2 between x-inercepts to see approximately how high or how low the graph goes between x-intercepts.
at x=-4, y = -4
at x=-1, y = -15
at x=-2, y = -18

This gives me a pretty good idea about the behaviour of my graph.

Using the same type of logic you should be able to do the next one.

To solve and graph these factors, you can follow these steps for each equation:

Step 1: Set each factor equal to zero and solve for x.

For the first equation, y = x(x+3)^2(x+5):

1. Setting each factor to zero:
x = 0
x+3 = 0 --> x = -3
x+5 = 0 --> x = -5

For the second equation, y = x(x-2)(x+0.5)(2x-1):

1. Setting each factor to zero:
x = 0
x-2 = 0 --> x = 2
x+0.5 = 0 --> x = -0.5
2x-1 = 0 --> 2x = 1 --> x = 1/2

Step 2: Plot the points on a graph.

For the first equation:
- Plot the points (0,0), (-3,0), and (-5,0) on the x-axis.

For the second equation:
- Plot the points (0,0), (2,0), (-0.5,0), and (1/2,0) on the x-axis.

Step 3: Determine the behavior of the graph.

For the first equation:
- Since the factor (x+3)^2 is squared, it indicates a double root or zero at x = -3. This means the graph touches the x-axis at (-3,0) without crossing it.
- The factors x and x+5 indicate single roots or zeros at x = 0 and x = -5, respectively. This means the graph crosses the x-axis at (0,0) and (-5,0).

For the second equation:
- All factors are linear, so each corresponds to a single root or zero.
- The factors x-2, x+0.5, and 2x-1 indicate zeros at x = 2, x = -0.5, and x = 1/2, respectively. This means the graph crosses the x-axis at these three points.

Step 4: Determine the behavior of the graph beyond the zeros.

For both equations:
- Since there are no additional factors, the behavior of the graph is determined by the behavior of the factors and the shape of the parabolas.

For the first equation:
- The parabola shaped by the factor (x+3)^2 determines the behavior around x = -3. Since it is squared, the graph will touch and change direction at (-3,0).
- Similarly, the parabola shaped by the factor x determines the behavior around x = 0, and the parabola shaped by the factor x+5 determines the behavior around x = -5.

For the second equation:
- Each linear factor contributes a separate root, resulting in three distinct crossing points. The behavior around these points will follow the shape of the nearby parabolas.

Step 5: Connect the points to form the graph of the equation.

For both equations:
- Connect the plotted points using smooth curves or lines that reflect the behavior of the factors and parabolas to complete the graph.

Note: It is always a good idea to use additional software or graphing calculators to visualize the graphs accurately.