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Thread: Dividing polynomials

  1. #1
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    Dividing polynomials

    I'm learning how to divide polynomials by expressions like a + 1 and I'm stuck on a particular question:

    4x^4 - 3x^2 + x + 2 divided by 2x + 3

    The answer must be expressed in the form (2x+3)*Quotient+Remainder.

    Here's how I've been going about it.

    4x^4 - 3x^2 + x + 2 = (2x + 3)(ax^3 + bx + c) + R

    Then after expanding and collecting like terms, I have:

    2ax^4 + 3ax^3 + 2bx^2 + (2b+2c)x + 3c + R


    And it's here that I get stuck. The problem here is that I don't know how to go on. What for example do I do with the x^3 bit?

    Many thanks
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  2. #2
    Grand Panjandrum
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    Re: Dividing polynomials

    Quote Originally Posted by bubbletea999 View Post
    I'm learning how to divide polynomials by expressions like a + 1 and I'm stuck on a particular question:

    4x^4 - 3x^2 + x + 2 divided by 2x + 3

    The answer must be expressed in the form (2x+3)*Quotient+Remainder.

    Here's how I've been going about it.

    4x^4 - 3x^2 + x + 2 = (2x + 3)(ax^3 + bx + c) + R

    Then after expanding and collecting like terms, I have:

    2ax^4 + 3ax^3 + 2bx^2 + (2b+2c)x + 3c + R


    And it's here that I get stuck. The problem here is that I don't know how to go on. What for example do I do with the x^3 bit?

    Many thanks
    Have you done synthetic division? If so use it.

    If not try equating coeficients of like powers in each of your expressions, to get you started (assuming you have no algebraic mistakes) for the fourth power term:

    4=2a

    CB
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  3. #3
    Member sbhatnagar's Avatar
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    Re: Dividing polynomials

    You have:
    $\displaystyle 4x^4 - 3x^2 + x + 2 = (2x + 3)(ax^3 + bx^2 + {\color{red}cx}+d) + R$

    Open the brackets:

    $\displaystyle \begin{align*} 4x^4 - 3x^2 + x + 2 &=(2a)x^4+(2b)x^3+(2c)x^2+(2d)x+(3a)x^3+(3b)x^2+(3 c)x+(3d)+R \end{align*}$

    $\displaystyle \begin{align*} \implies 4x^4 - 3x^2 + x + 2 &= (2a)x^4+(2b+3a)x^3+(2c+3b)x^2+(3c+2d)x+(3d+R) \end{align*}$

    Equating the coefficients:

    $\displaystyle 2a=4$

    $\displaystyle 2b+3a=-3$

    $\displaystyle 2c+3b=0$

    $\displaystyle 3c+2d=1$

    $\displaystyle 3d+R=2$

    The correct solutions to these equations will give you the value of $\displaystyle a,b,c,d \ \textrm{and} \ R$.
    Last edited by sbhatnagar; Dec 6th 2011 at 05:06 AM.
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  4. #4
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    Re: Dividing polynomials

    Hi guys, thank you very much for your answers (and sorry for my very late reply).

    I've now learned how to do synthetic division, and in the process of doing so also learned about the general form, which ended up solving my problem with this! I was going wrong because I wasn't accounting for the "missing power" in the original equation. It jumps from x^4 to x^2. I've solved the problem by adding 0x^3 in between them -- does this sound right? It's produced the right answer, at any rate.

    Thanks again.
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  5. #5
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    Re: Dividing polynomials

    Another way to divide $\displaystyle 4x^4 - 3x^2 + x + 2$ by $\displaystyle 2x + 3$, more like "standard" numerical long division:

    $\displaystyle 2x$ divides into $\displaystyle 4x^4$ $\displaystyle 2x^3$ times. Multiplying that by $\displaystyle 2x+ 3$ gives [tex](2x+3)(2x^3)= 4x^4+ 6x^3. Subtract [tex](4x^4+ 0x^3)- (4x^4+ 6x^3)= -6x^3. That leaves us $\displaystyle -6x^3- 3x^2+ x+ 2$. Now, $\displaystyle 2x$ divides into $\displaystyle -6x^3$ $\displaystyle -3x^2$ times. Multiplying that by $\displaystyle 2x+ 3$ gives $\displaystyle -6x^3- 9x^2$. Subtract $\displaystyle (-6x^3- 3x^2)- (-6x^3- 9x^2)= 6x^2$. That leaves $\displaystyle 6x^2+ x+ 2$. $\displaystyle 2x$ divides into $\displaystyle 6x^2$ 3x times. Multiplying that by $\displaystyle 2x+ 3$ gives $\displaystyle 6x^2+ 9x$. Subtracting $\displaystyle (6x^2+ x)-(6x^2+ 9x)= -8x$. That leaves us $\displaystyle -8x+ 2$. $\displaystyle 2x$ divides into $\displaystyle -8x$ -4 times. Multiplying that by $\displaystyle 2x+ 3$ gives $\displaystyle -8x- 12$. Subtracting $\displaystyle (-8x+ 2)- (-8x- 12)= 14$.

    That is, the quotient is $\displaystyle 2x^3- 3x^2+ 3x- 4$ with remainder 14.

    (Captain Black, I was under the impression that "synthetic division" could only be used to divide by something of the form x+ a.)
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