How Do You Solve for m in a Polynomial Division Problem?

  • Thread starter thomasrules
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In summary, the value of m that makes the remainder equal to 3 when x^3+5x^2+6x+11 is divided by (x+m) is 4. This can be found by finding the root of x+m=0, which is -m, and then substituting it into the polynomial to get the remainder. In this case, setting x=-m gives the equation m^3-5m^2+6m-8=0, which can be solved to find m=4.
  • #1
thomasrules
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Find the value of m so that when x^3+5x^2+6x+11 is divided by (x+m) the remainder is 3.

Im having so much diffculty with this and it's so frustrating can anyoen help

Thomas
 
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  • #2
Do you know how to divide polynomials? For example if you divide ax^2+bx+c by (x+d) what would be the remainder? There's an easy way for doing this. find value of d which makes x+d=0. Then replace that value with x in the original polynomial equation, and what you got is the remainder of division!

Try doing this and let me know if there's any problem
 
  • #3
Sorry but I'm still not clear on that. I know that to find a value that's divisible in an equation you try f(x) and if that equals 0 in the equation its divisible. In my case its x^3+5x^2+6x+11 so the only things i try are the last number 11. (+-1, +-11) and that doesn't equal 0...Explain more please
 
  • #4
You divide the equation by (x+m) right? What I'm telling that find the root of this function (x+m, that is) and then place the root into the polynomial. Then you will have the remainder, which is 3. Solve for m and you got it!

The root for x+m=0 is obviously -m. now place -m in the place of x in the equation x^3+5x^2+6x+11. What I tell you is that setting x=-m in this equation gives you the remainder when you divide x^3+5x^2+6x+11 by (x+m).
 
  • #5
omg dude no I get another answer...can you tell me your answer if it's so easy lol
 
  • #6
Is it 4? If not there should be something wrong with the answer...
 
  • #7
yup,
f(x)=g(x)(x+m) +R
put x=-m
u will get
after rearranging
[tex] m^3 - 5m^2+6m-8=0 [/tex]
solving u will have
[tex](m-4)*(m^2-m+2)=0[/tex]
 

What is the Calculus-Remainder Theorem?

The Calculus-Remainder Theorem, also known as the Remainder Theorem or the Lagrange Remainder Theorem, is a fundamental theorem in calculus that provides a way to estimate the error or remainder when a polynomial is approximated by a truncated Taylor series.

What is the formula for the Calculus-Remainder Theorem?

The formula for the Calculus-Remainder Theorem is: Rn(x) = f(n+1)(c)(x-c)^(n+1)/(n+1)!, where Rn(x) is the remainder or error, n is the degree of the polynomial, c is a number between the original point and the point of approximation, and f(n+1)(c) is the (n+1)th derivative of the function evaluated at c.

How is the Calculus-Remainder Theorem used in real-world applications?

The Calculus-Remainder Theorem is used in various fields such as engineering, physics, and economics to estimate the error or uncertainty in calculations and predictions. For example, it can be used to estimate the error in numerical integration methods or in predicting future values based on a mathematical model. It is also used in computer graphics to improve the accuracy of rendered images.

What are the assumptions of the Calculus-Remainder Theorem?

The Calculus-Remainder Theorem assumes that the function is infinitely differentiable and that the Taylor series converges to the function within a given interval. It also assumes that the remainder term approaches zero as the degree of the polynomial increases.

How does the Calculus-Remainder Theorem relate to other calculus concepts?

The Calculus-Remainder Theorem is closely related to other important concepts in calculus such as derivatives and Taylor series. It provides a way to estimate the error or remainder when a polynomial is approximated by a Taylor series, which is a sum of derivatives of a function at a point. It is also used in the proof of the Taylor series formula and in the study of convergence of infinite series.

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