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What are the quotient and remainder of (x^5-x^3+-5) divided by (x-2)

pishuonlain [190]

Answer:

x⁴+2x³ +3x²+6x +12 r. $\dfrac{19}{x-2}$

Step-by-step explanation:

You can do this through synthetic division. It is a shorthand way of dividing polynomials.

First you need to make your divisor equal to zero so you can solve for what goes into the division box:

x - 2 = 0 → x = 2

The next step is to arrange your polynomials in descending powers. All missing terms, you will put in a zero.

x⁵ - x³ + (-5) → x⁵ + 0x⁴- x³ + 0x² + 0x + (-5)

Now you can proceed to synthetic division. Make an upside down division box with the divisor outside and the coefficients of the dividend listed, along with their sign. Leave a space below the divident

+2 | +1 0 -1 0 0 -5

|

Next you bring down the first coefficient:

+2 | +1 0 -1 0 0 -5

|

+1

Then you multiply it by the divisor and write the product under the next coefficient:

+2 | +1 0 -1 0 0 -5

| +2

+1

Next add the column and put the sum below it:

+2 | +1 0 -1 0 0 -5

| +2

+1 +2

Then multiply again and repeat until you reach the last coefficient:

+2 | +1 0 -1 0 0 -5

| +2 +4 +6 +12 +24

+1 +2 +3 +6 +12 +19

Now that you have your results, add in the x and their powers. The powers will start with the highest power but 1 less than the dividend. Since the dividend's highest power is 5, then the quotient's highest power will be 4. Then write it in descending order :

+1x⁴ +2x³ +3x² +6x +12 +19

Now the last coefficient is your remainder. So your results will be:

x⁴+2x³ +3x²+6x +12 r. $\dfrac{19}{x-2}$

7 0

3 years ago

Read 2 more answers

F(x)=x^3+3

lawyer [7]

Answer:

(a) -3/4

(b) -0.75

(c) -0.75

Step-by-step explanation:

It's a bit hard to tell what constitutes an "iteration" when using the bisection method to approximate a polynomial root. For the purpose here, we'll say one iteration consists of ...

evaluating the function at the midpoint of the bracketing interval
choosing a smaller bracketing interval
identifying the x-value known to be closest to the solution

Thus, the result of the iteration consists of a bracketing interval and the choice of one of the interval's ends as the solution approximation.

__

(a) We observe that the graphs intersect in the interval (-1, 0). For the first iteration, we evaluate f(x)-g(x) at x=-1/2. This tells us the solution is in the interval (-1, -1/2). The x-value closest to the root is x=-1/2.

For the second iteration, we evaluate the function f(x)-g(x) at x=-3/4. This tells us the solution is in the interval (-1, -3/4). The x-value closest to the root is x=-3/4.

For the third iteration, we evaluate the function f(x)-g(x) at x=-7/8. This tells us the solution is in the interval (-7/8, -3/4). The x-value closest to the root is x=-3/4.

__

(b) The graph tells us the solution is approximately 0.7549. Rounded to 2 decimal places, the solution is approximately 0.75.

__

(c) The above solution found after 3 iterations rounded to 2 decimal places is exactly 0.75.

__

See the attached table for function values.

_____

Comment on bisection iteration

Since you cut the interval containing the root in half with each iteration, you gain approximately one decimal place for each 3 iterations. When the function value is very nearly zero at one of the interval endpoints, it can take many more iterations to achieve a better result.

Here, it takes 4 more iterations before an x-value becomes closer to the solution (x≈-97/128). And it takes one more iteration to move the end of the interval away from -3/4. After these 5 more iterations (8 total), the solution is known to lie in the interval (-97/128, -193/256). The corresponding solution approximation is -193/256. It is still only correct to 2 decimal places.