Concept explainers
To divide : using the long division

Answer to Problem 23RE
Explanation of Solution
Given information :
Concept Involved:
- Set up the long division.The divisor goes on the outside of the box. The dividend goes on the inside of the box. When you write out the dividend, make sure that you insert 0's for any missing terms.
- Divide 1st term of dividend by first term of divisor to get first term of the quotient.The quotient is written above the division box.Make sure that you line up the first term of the quotient with the term of the dividend that has the same degree.
- Take the term found in step 2 and multiply it times the divisor.Make sure that you line up all terms of this step with the term of the dividend that has the same degree.
- Subtract this from the line above.Make sure that you subtract EVERY term found in step 3, not just the first one.
- Repeat until done.You keep going until the degree of the "new" dividend is less than the degree of the divisor.Use the long division to find the other factor of the function
- Write out the answer. Your answer is the quotient that you ended up with on the top of the division box. If you have a remainder, write it over the divisor in your final answer.
The Division Algorithm : If
Calculation:
Step 1: Set up the long division. The divisor goes on the outside of the box. The dividend goes on the inside of the box. When you write out the dividend, make sure that you insert 0's for any missing terms.
Step 2: Divide 1st term of dividend by first term of divisor to get first term of the quotient. The quotient is written above the division box. Make sure that you line up the first term of the quotient with the term of the dividend that has the same degree.
Step 3: Take the term found in previous and multiply it times the divisor. Make sure that you line up all terms of this step with the term of the dividend that has the same degree.
Step 4: Subtract this from the line above.Make sure that you subtract EVERY term found in step 3, not just the first one.
Step 5: Repeat until done. You keep going until the degree of the "new" dividend is less than the degree of the divisor. Use the long division to find the other factor of the function
Step 6: Take the term found in previous and multiply it times the divisor. Make sure that you line up all terms of this step with the term of the dividend that has the same degree.
Step 7: Subtract this from the line above.Make sure that you subtract EVERY term found in step 3, not just the first one.
Conclusion:
By dividing the given polynomial
Chapter 2 Solutions
EBK PRECALCULUS W/LIMITS
- A chemical reaction involving the interaction of two substances A and B to form a new compound X is called a second order reaction. In such cases it is observed that the rate of reaction (or the rate at which the new compound is formed) is proportional to the product of the remaining amounts of the two original substances. If a molecule of A and a molecule of B combine to form a molecule of X (i.e., the reaction equation is A + B ⮕ X), then the differential equation describing this specific reaction can be expressed as: dx/dt = k(a-x)(b-x) where k is a positive constant, a and b are the initial concentrations of the reactants A and B, respectively, and x(t) is the concentration of the new compound at any time t. Assuming that no amount of compound X is present at the start, obtain a relationship for x(t). What happens when t ⮕∞?arrow_forwardConsider a body of mass m dropped from rest at t = 0. The body falls under the influence of gravity, and the air resistance FD opposing the motion is assumed to be proportional to the square of the velocity, so that FD = kV2. Call x the vertical distance and take the positive direction of the x-axis downward, with origin at the initial position of the body. Obtain relationships for the velocity and position of the body as a function of time t.arrow_forwardAssuming that the rate of change of the price P of a certain commodity is proportional to the difference between demand D and supply S at any time t, the differential equations describing the price fluctuations with respect to time can be expressed as: dP/dt = k(D - s) where k is the proportionality constant whose value depends on the specific commodity. Solve the above differential equation by expressing supply and demand as simply linear functions of price in the form S = aP - b and D = e - fParrow_forward
- Find the area of the surface obtained by rotating the circle x² + y² = r² about the line y = r.arrow_forward1) Find the equation of the tangent line to the graph y=xe at the point (1, 1).arrow_forward3) Suppose that f is differentiable on [0, 5], and f'(x) ≤ 3 over this interval. If f(0) = −1, what is the maximum possible value of f(5)?arrow_forward
- 2) Find the maximum value of f(x, y) = x - y on the circle x² + y² - 4x - 2y - 4 = 0.arrow_forwardFor the system consisting of the lines: and 71 = (-8,5,6) + t(4, −5,3) 72 = (0, −24,9) + u(−1, 6, −3) a) State whether the two lines are parallel or not and justify your answer. b) Find the point of intersection, if possible, and classify the system based on the number of points of intersection and how the lines are related. Show a complete solution process.arrow_forward3. [-/2 Points] DETAILS MY NOTES SESSCALCET2 7.4.013. Find the exact length of the curve. y = In(sec x), 0 ≤ x ≤ π/4arrow_forward
- H.w WI M Wz A Sindax Sind dy max Утах at 0.75m from A w=6KN/M L=2 W2=9 KN/m P= 10 KN B Make the solution handwritten and not artificial intelligence because I will give a bad rating if you solve it with artificial intelligencearrow_forwardSolve by DrWz WI P L B dy Sind Ⓡ de max ⑦Ymax dx Solve by Dr ③Yat 0.75m from A w=6KN/M L=2 W2=9 kN/m P= 10 KN Solve By Drarrow_forwardHow to find the radius of convergence for the series in the image below? I'm stuck on how to isolate the x in the interval of convergence.arrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning





