How many x-intercepts and how many local extrema does the polynomial P(x)= x(sqrt)3-4x have?

How many x-intercepts and how many local extrema does the polynomial Q(x)=x^3-4x have?

and If a>0, how many x-intercepts and how many local extrema does each of the polynomials P(x)=x^3-ax and Q(x)=x^3+ax? Explain your answer.

If P(x) = x^2-4x, then since

P(x) = x(x-4), it has two x-intercepts.

Q(x) = x(x^2-4) = x(x-2)(x+2)
so Q has 3 roots.

x^2-ax = x(x^2-a) = x(x-√a)(x+√a)
so there are 3 roots
With 3 roots, there must be 2 extrema.

x^2+ax = x(x^2+a)
since a>0, x^2+a can never be zero, so there is only 1 x-intercept, at x=0.
There are no extrema, since the derivative is never zero.

To find the number of x-intercepts of a polynomial, we need to determine the values of x where the polynomial equals zero. Similarly, to find the number of local extrema, we need to find the values of x where the derivative of the polynomial equals zero.

1. Polynomial P(x) = x√3 - 4x:
- X-intercepts: To find the x-intercepts, we set P(x) equal to zero and solve for x:
x√3 - 4x = 0
Factor out x: x(√3 - 4) = 0
Two possibilities: either x = 0 or √3 - 4 = 0
The second equation, √3 - 4 = 0, does not have a real solution. Therefore, there is only one x-intercept at x = 0.

- Local extrema: To find the local extrema, we need to find the derivative of P(x), set it equal to zero, and solve for x:
P'(x) = (√3 - 4) = 0
Since (√3 - 4) is a constant, it never equals zero. Thus, there are no local extrema for this polynomial.

2. Polynomial Q(x) = x^3 - 4x:
- X-intercepts: To find the x-intercepts, we set Q(x) equal to zero and solve for x:
x^3 - 4x = 0
Factor out x: x(x^2 - 4) = 0
Two possibilities: either x = 0 or x^2 - 4 = 0
Solving x^2 - 4 = 0, we find two solutions: x = -2 and x = 2.
Therefore, there are three x-intercepts at x = 0, x = -2, and x = 2.

- Local extrema: To find the local extrema, we need to find the derivative of Q(x), set it equal to zero, and solve for x:
Q'(x) = 3x^2 - 4 = 0
Solving 3x^2 - 4 = 0, we find two solutions: x = -√(4/3) and x = √(4/3).
Therefore, there are two local extrema for this polynomial at x = -√(4/3) and x = √(4/3).

3. Polynomial P(x) = x^3 - ax and Q(x) = x^3 + ax (assuming a > 0):
- X-intercepts: To find the x-intercepts, we set P(x) and Q(x) equal to zero and solve for x:
For P(x): x^3 - ax = 0
Factor out x: x(x^2 - a) = 0
Two possibilities: either x = 0 or x^2 - a = 0
Solving x^2 - a = 0, we find two solutions: x = -√a and x = √a.
Therefore, there are three x-intercepts for P(x) at x = 0, x = -√a, and x = √a.

For Q(x): x^3 + ax = 0
Factor out x: x(x^2 + a) = 0
Two possibilities: either x = 0 or x^2 + a = 0
The second equation, x^2 + a = 0, does not have a real solution. Therefore, there is only one x-intercept at x = 0.

- Local extrema: To find the local extrema, we need to find the derivative of P(x) and Q(x), set it equal to zero, and solve for x:
For P(x): P'(x) = 3x^2 - a = 0
Solving 3x^2 - a = 0, we find two solutions: x = -√(a/3) and x = √(a/3).
Therefore, there are two local extrema for P(x) at x = -√(a/3) and x = √(a/3).

For Q(x): Q'(x) = 3x^2 + a = 0
Similarly, solving 3x^2 + a = 0 does not have any real solutions. Therefore, there are no local extrema for Q(x).

In summary:
- Polynomial P(x) = x√3 - 4x has one x-intercept (at x = 0) and no local extrema.
- Polynomial Q(x) = x^3 - 4x has three x-intercepts (at x = 0, x = -2, and x = 2) and two local extrema (at x = -√(4/3) and x = √(4/3)).
- Polynomial P(x) = x^3 - ax has three x-intercepts (at x = 0, x = -√a, and x = √a) and two local extrema (at x = -√(a/3) and x = √(a/3)) when a > 0.
- Polynomial Q(x) = x^3 + ax has one x-intercept (at x = 0) and no local extrema when a > 0.