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krok68 [10]
3 years ago
8

Use synthetic division to solve (x3 + 1) ÷ (x – 1). What is the quotient?

Mathematics
2 answers:
kolbaska11 [484]3 years ago
6 0

The quotient is of the division of {x^3} + 1 and x - 1 \:is \:\boxed{{x^2} + x + 1} and the remainder is \boxed2.

Further explanation:

The numerator of the division is {x^3} + 1 and the denominator isx - 1.

Solve the given polynomial P\left( x \right) = {x^3} + 1 to obtain the quotient by the use of synthetic division.

Now obtain the value of x from the denominator.

\begin{aligned} x - 1&= 0 \\x &= 1\\\end{aligned}

Divide the coefficients of the polynomial by  , to check whether   is a zero of the polynomial.

\begin{aligned}1\left| \!{\nderline {\,{1\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,1\,} \,}} \right.\hfill \\\,\,\,\,\underline {\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,1}\hfill \\\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,2 \hfill\\\end{aligned}

The last entry of the synthetic division told about remainder and the last entry of the synthetic division is 2. Therefore, the remainder of the synthetic division is  .

The entry in the last line of synthetic division gives us the coefficients of the quotient.

The new numbers are the coefficients of the quotient.

The quotient is of the division of {x^3} + 1 \:and \:x - 1 \:is\: \boxed{{x^2} + x + 1} and the remainder is \boxed2.

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.

2. Learn more about equation of circle brainly.com/question/1506955.

3. Learn more about range and domain of the function brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Synthetic Division

Keywords: division, synthetic division, long division method, coefficients, quotients, remainders, numerator, denominator, polynomial, zeros, degree.

mamaluj [8]3 years ago
4 0
Steps:

Use the divisor and find x :
x - 1 = 0  **add 1
x= 1
Now we will use the 1 in dividing:

take the coefficients from in front of all terms
   ** make sure you include 0's for x^2 and x since you have to have all terms

set it up with a 1 in a box:

1|  1   0   0  1   **bring the first number down
____________
     1                    **multiply the boxed number by the first number and add it                                 to the second number

1|  1   0   0  1   
____+1_____       **repeat with the rest of the terms
     1   1   

1|  1   0    0    1
  ___+1_+1_+1   
     1   1    1    2                              
**when you're done, use the new numbers to write an equation starting with a term with a degree one less than the previous equation.
**since there is a remainder, rewrite it divided by the original divisor

final answer:
x^2 + x + 1 + (2/ x -1)
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Read 2 more answers
I already have the first question done! I just need help with the other two.
Rainbow [258]

Problem 2

Part 1

I'll replace h with y and replace t with x.

The equation turns into y = -5x^2 + 40x + 45

It's of the form y = ax^2+bx+c where

  • a = -5
  • b = 40
  • c = 45

Use the 'a' and b values to find the value of h, which is the x coordinate of the vertex

h = -b/(2a)

h = -40/(2(-5))

h = -40/(-10)

h = 4

At the four second mark is when the rocket will reach its peak height.

Plug this into the original equation to find its paired y value

y = -5x^2 + 40x + 45

y = -5(4)^2 + 40(4) + 45

y = 125

The vertex is at (h,k) = (4, 125).

<h3>The highest the rocket goes is 125 feet.</h3>

---------------

Part 2

Plug in y = 0 and solve for x to find when the rocket hits the ground. I'll use the quadratic formula.

x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(40)\pm\sqrt{(40)^2-4(-5)(45)}}{2(-5)}\\\\x = \frac{-40\pm\sqrt{2500}}{-10}\\\\x = \frac{-40\pm50}{-10}\\\\x = \frac{-40+50}{-10} \ \text{ or } \ x = \frac{-40-50}{-10}\\\\x = \frac{10}{-10} \ \text{ or } \ x = \frac{-90}{-10}\\\\x = -1 \ \text{ or } \ x = 9\\\\

Ignore the negative x value. We cannot have negative time values.

The only practical root is that x = 9, meaning the rocket hits the ground at the 9 second mark.

<h3>The rocket is in the air for 9 seconds.</h3>

=====================================================

Problem 3

You have a product that sells for $10 and 1000 people buy per month.

The revenue you pull in based on those figures is 10*1000 = 10,000 dollars per month.

If you raised the price by $1, then the $10 jumps to $11. The downside is that the 1000 people drops to 900 people (you lose 100 customers). The revenue would be 11*900 = 9900. So far, it seems like a bad idea to raise prices. But we'll increase the price once more to see what happens.

If the price goes to $12, then you lose another 100 customers and you now have 800 customers. So that's 12*800 = 9600 dollars in revenue per month. It seems the trend is getting worse.

---------------

Let's generalize what's going on.

x = number of times you raise the price by $1

The old price is $10 per item. It jumps up to 10+x per item. At the same time, the customer count goes from 1000 to 1000-100x. Each time x goes up by 1, the expression 1000-100x goes down by 100.

Multiply the price and customer count to get the amount earned.

revenue = (price)*(number of customers)

revenue = (10+x)*(1000-100x)

revenue = 10(1000 - 100x) + x(1000-100x)

revenue = 10,000 - 1000x + 1000x - 100x^2

revenue = 10,000 - 100x^2

revenue = -100x^2 + 10,000

If you were to use the methods done in the previous problem, you should find the vertex is at (0,10000). This means that the max revenue was already reached when x = 0 price increases were done. This reinforces the previous results we got earlier before we started generalizing in terms of x.

You should <u>not</u> raise the price, or else you'll just continue to lose customers until you go to 0. No amount of price raising will get your revenue up, which in turn means the profits will suffer as well. With the product at $10, you are already at the max revenue point.

<h3>Conclusion: Keep the price at $10</h3>
3 0
3 years ago
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