X^3-6x^2-15x+35=0, how do you find the x intercept for this problem. i keep looking this up but everywhere just gives me the answer and doesn't show how they factor out the roots.

Explanation:

There are 3 irrational x-intercepts for this cubic.

In general, solving a cubic is not easy. There are formulas for it, and even the formulas for the real roots (which this has 3 of) involve irrational complex numbers. (Actually, there is a trigonometric formula I like for three real roots.)

I like to use a graphing calculator for roots of higher-degree polynomials. A good one will find the roots to calculator precision quickly and easily. Some will even show the radical form of the roots.

The first attachment shows a graph of the equation and the approximate roots. (This same calculator makes Newton's method iteration easy, so refining the roots is a snap.)

The second attachment shows input and output from a graphing calculator that will actually solve this cubic. It displays the root values to calculator precision with no further effort.

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You can start by trying the roots suggested by the Rational Root theorem. Here, those suggestions are ±1, ±5, and ±7. Finding the function values for these x-values will tell you there are real roots in the intervals (-5, -1), (1, 5), and (7, 35). The latter root being close to 7, but a little bit more.

You can try use any of several iterative methods to refine these intervals. One commonly seen is the bisection method.

We assume we're looking for roots of g(x) = 0.

Bisection Method

For this method, you need values of x that bracket the root of interest. (Possible starting values are the intervals listed above.) For each iteration, you compute the midpoint of the interval, then find the value of the function at that midpoint. The x-value of the midpoint replaces the x-value of the interval whose function value has the same sign.

For example, g(7) = -21; g(8) = 43. The midpoint of the interval (7, 8) is 7.5, so we find g(7.5) = 6.875. This has the same sign as g(8), so our new interval is (7, 7.5). On the next iteration, we evaluate g(7.25).

As the interval is cut in half at each iteration, it takes about 10 iterations to gain 3 decimal places in accuracy.

Secant Method

The secant method is similar to the bisection method, except that the next x-value used is not the midpoint of the interval, but is one calculated based on the slope of the secant line between the points at the ends of the interval.

In the above example, g(7) = -21 and g(8) = 43, so the next choice of x for evaluation would be x = 7 +(0 -(-21))/(43-(-21)) = 7 21/64. That function value is ...

  g(7 21/64) ≈ -3.59968

Now, the interval of interest is (7 21/64, 8) and the x-intercept of the next secant line is computed based on these endpoints and function values.

Convergence can be more rapid than with the bisection method, but depends on the function being well-behaved between the ends of the intervals.

Newton's Method

Another iteration method that works well with well-behaved polynomial functions is Newton's method. The iterator here is ...

  x' = x - g(x)/g'(x)

where x' is the next "guess" and g'(x) is the derivative of the polynomial function g(x). This method converges quadratically, so nearly doubles the number of good decimal places at each iteration.

Other Iteration Methods

Sometimes, you can create an iterator from the function itself, solving for x:

  x = ∛(6x^2 +15x -35) . . . . . using the cubed term

  x = (x^3 -6x^2 +35)/15 . . . . using the linear term

These tend to converge slowly, if at all. The closer you are to start with, the better the results will be.

goog may have it for you

a) one student is chosen from classroom a, then a student is chosen from classroom b.  

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b.   this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes.   this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes.   this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes.   this makes them dependent events.