find x-intts, critical pts of 3x^5 + 5x^4 - 11x^3 - 21x^2 -

[Guest]

Hi, I need help with these questions:

1) Find the x-intercept(s) and the critical point(s).

x-intercepts:

. . .y = 3x^5 + 5x^4 - 11x^3 - 21x^2 - 4x + 4

. . .0 = 3x^5 + 5x^4 - 11x^3 - 21x^2 - 4x + 4

Now what?

Critical points:

. . .y' = 15x^4 - 20x^3 - 33x^2 - 42x - 4

. . .0 = 15x^4 - 20x^3 - 33x^2 - 42x - 4

. . .4 = x( 15x^3 - 20x^2 - 33x - 42)

Now what?

I don't know how to factor these types of problems, I thought of factering by grouping but there is nothing in common between all of the terms..so I don't know what to do.

Thanks.

 

[galactus]

Re: find the x intercept and the critical points

Hi, I need help with these questions

find the x intercept and the critical points

X- Intercepts
y= 3x^5 + 5x^4 -11x^3 - 21x^2 - 4x +4
o= 3x^5 + 5x^4 -11x^3 - 21x^2 - 4x +4

now what?

critical point

y'= 15x^4 + 20x^3 - 33x^2 - 42x - 4
0= 15x^4 - 20x^3 - 33x^2 - 42x - 4
4= x( 15x^3 - 20x^2 - 33x - 42)
now what?

I don't know how to factor these types of problems, I thought of factering by grouping but there is nothing in common between all of the terms..so I don't know what to do.

Thanks.

You have a sign error in your derivative.

It factors to:

\(\displaystyle \L\\(x+1)(15x^{3}+5x^{2}-38x-4)\)

To factor f(x), you can use the remainder theorem

Using the potential factors of \(\displaystyle \pm{4}, \;\ \pm\frac{4}{3}, \;\ \pm{2}, \;\ \pm\frac{2}{3}, \;\ \pm{1}, \;\ \pm\frac{1}{3}\)

Dividing by (x-2) we find that we have:

\(\displaystyle \L\\3x^{4}+11x^{3}+11x^{2}+x-2\)

Therefore, 2 is a factor.

Try some of the others and we whittle it down to:

\(\displaystyle \L\\(x-2)(x+1)^{2}(x+2)(3x-1)\)

Those are the x-intercepts

Try graphing it to find the critical values. Solve it with technology.