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Math Help - P(x) leaves two remainder when divided by two polynomials , find r(x) when divided...

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    P(x) leaves two remainder when divided by two polynomials , find r(x) when divided...

    Let $p(x)$ be a polynomial. When $p(x)$ is divided by $x-1$, it leaves the remainder $4$. When it is divided by $x-2$ it leaves the remainder $5$. Then find the remainder when $p(x)$ is divided by $x^2-3x+2$.


    $p(1)=4$

    $p(2)=5 $

    $\Rightarrow p(x)=x+3.$

    [/TEX]

    I don't think I am right on guessing p(x)=x+3, If yes how can I divide it by a quadratic polynomial.
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Hint: $\displaystyle \begin{align*} x^2 - 3x + 2 = (x - 1)(x - 2) \end{align*}$...
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Quote Originally Posted by Prove It View Post
    Hint: $\displaystyle \begin{align*} x^2 - 3x + 2 = (x - 1)(x - 2) \end{align*}$...
    Yes I saw that already and the divisors of p(x) coinciding with the factors of the third divisor.

    But I don't know what does it mean? Should I multiply, divide, add or subtract 4 and 5?????


    $p(x)=(x-1)q_1(x)+4=(x-2)q_2(x)+5=(x-1)(x-2)q_3(x)+ax+b$

    What to do with their quotients. I am struggling
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Why is no one giving me hint? I think all this is because of spammers and advertisers who push the suitable posts behind.
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Or it's because the problem is so simple it's hard to give a "hint" without telling you exactly how to solve it!

    First, since p leaves a constant remainder when divided by a linear term, p must be quadratic: that is, p(x)= x^2+ ax+ b
    Further, since dividing p(x) by x- 1 leaves remainder 4, p(1)= 1+ a+ b= 4. Since dividing p(x) by x- 2 leaves remainder 5, p(2)= 4+ 2a+ b= 5.
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Quote Originally Posted by HallsofIvy View Post
    Or it's because the problem is so simple it's hard to give a "hint" without telling you exactly how to solve it!

    First, since p leaves a constant remainder when divided by a linear term, p must be quadratic: that is, p(x)= x^2+ ax+ b
    Further, since dividing p(x) by x- 1 leaves remainder 4, p(1)= 1+ a+ b= 4. Since dividing p(x) by x- 2 leaves remainder 5, p(2)= 4+ 2a+ b= 5.
    Thanks. I will work it out.

    How do you say if the remainder is constant and dicisor being linear, then the dividend is quadratic. As far as I know, if the divisor is linear then the remainder is constant. Is there any dividend theorem?
    Last edited by NameIsHidden; June 28th 2014 at 06:13 AM.
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Quote Originally Posted by NameIsHidden View Post
    Thanks. I will work it out.

    How do you say if the remainder is constant and dicisor being linear, then the dividend is quadratic. As far as I know, if the divisor is linear then the remainder is constant. Is there any dividend theorem?
    Anyway, $a+b=3 \Rightarrow b=3-a$ and $2a+b=1=2a+3-a=1 =\Rightarrow a=-2 \text{ and } b=5$.

    Then $p(x)=x^2-2x+5$. I can take it from now. But I am still unclear Why p(x) should be quadratic and the leading co-efficient should be 1?
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Adding to my confusion, $(x^4-x^3+4) \text{ leaves the remainder (4) when divided by }
    \left(x-1 \right) \text{ and quotient } x^3$.

    Here it is quartic expression.
    Last edited by NameIsHidden; June 28th 2014 at 06:47 AM.
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Note that we don't actually have to find $p(x)$, which is good (there are too many possibilities).

    Here is what we know:

    $p(x) = q_1(x)(x - 1) + 4 = q_2(x)(x - 2) + 5$.

    We want to find $r(x)$ such that $p(x) = q(x)(x^2 - 3x + 2) + r(x)$.

    Note that $r(x) = ax + b$, since $\text{deg}(x^2 - 3x + 2) = 2$.

    Let's just play with this a bit.

    $r(x) = p(x) - q(x)(x^2 - 3x + 2)$, so:

    $r(x) - 4 = p(x) - 4 - q(x)(x^2 - 3x + 2) = q_1(x)(x - 1) + q(x)(x - 2)(x - 1) = [q_1(x) - q(x)(x - 2)](x - 1)$, so $r(x) - 4$ is divisible by $(x - 1)$. This mean that $r(1) - 4 = 0$, that is:

    $a + b = 4$.

    Similarly,

    $r(x) - 5 = p(x) - 5 - q(x)(x^2 - 3x + 2) = q_2(x)(x - 2) + q(x)(x - 2)(x - 1) = [q_2(x) - q(x)(x - 1)](x - 2)$, so $r(x) - 5$ is divisible by $(x - 2)$. What does this say about what $r(2) - 5$ is?
    Thanks from NameIsHidden
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Quote Originally Posted by Deveno View Post
    Note that we don't actually have to find $p(x)$, which is good (there are too many possibilities).

    Here is what we know:

    $p(x) = q_1(x)(x - 1) + 4 = q_2(x)(x - 2) + 5$.

    We want to find $r(x)$ such that $p(x) = q(x)(x^2 - 3x + 2) + r(x)$.

    Note that $r(x) = ax + b$, since $\text{deg}(x^2 - 3x + 2) = 2$.

    Let's just play with this a bit.

    $r(x) = p(x) - q(x)(x^2 - 3x + 2)$, so:

    $r(x) - 4 = p(x) - 4 - q(x)(x^2 - 3x + 2) = q_1(x)(x - 1) + q(x)(x - 2)(x - 1) = [q_1(x) - q(x)(x - 2)](x - 1)$, so $r(x) - 4$ is divisible by $(x - 1)$. This mean that $r(1) - 4 = 0$, that is:

    $a + b = 4$.

    Similarly,

    $r(x) - 5 = p(x) - 5 - q(x)(x^2 - 3x + 2) = q_2(x)(x - 2) + q(x)(x - 2)(x - 1) = [q_2(x) - q(x)(x - 1)](x - 2)$, so $r(x) - 5$ is divisible by $(x - 2)$. What does this say about what $r(2) - 5$ is?
    $r(2)-5=0$

    Thank you very much. HallsofIvy confused me.

    $2a-b=5$

    $2a-4+a=5=3a-4\Rightarrow a=3, b=1$

    Then $r(x)=3x+1$
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Shouldn't $r(2) = 2a + b$?
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    Re: P(x) leaves two remainder when divided by two polynomials , find r(x) when divide

    Quote Originally Posted by Deveno View Post
    Shouldn't $r(2) = 2a + b$?
    Sorry a small typo resulting in a very huge deviation from correct answer!

    $r(x)=x+3$
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