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Thread: 2 Polynomial Problems

  1. #1
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    2 Polynomial Problems

    Problem 147. The numbers 1 and 2 are roots of polynomial P. Prove that P is divisible by (x-1)(x-2).

    I have the solution but I don't understand one of the steps.

    Solution. P is divisible by (x-1) because 1 is a root of P. Therefore $\displaystyle P = (x-1)\cdot{Q}$ for some polynomial Q. Substituting 2 for x in this equality we find that 2 is a root of Q, so Q is divisible by (x-2), that is $\displaystyle Q = (x-2)\cdot{R}$ for some polynomial R. So $\displaystyle P = (x-1)(x-2)\cdot{R}$

    Why is it, that you can just substitute 2 into that? Am I correct in saying it shows that P = Q therefore (x-2) is a root of Q, because it's a root of P? I don't understand the reasoning behind that step. Could someone explain?



    A second problem - to which I've no solution - has me stumped.

    Problem 154. Assume that $\displaystyle x^3+ax^2+x+b$ (where a and b are some numbers) is divisible by $\displaystyle x^2-3x+2$. Find a and b.

    Thanks in advance.
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by jerry View Post
    Why is it, that you can just substitute 2 into that? Am I correct in saying it shows that P = Q therefore (x-2) is a root of Q, because it's a root of P? I don't understand the reasoning behind that step. Could someone explain?
    2 is a root of $\displaystyle P$ so $\displaystyle P(2)=0$
    In $\displaystyle P(x)=(x-1)Q(x)$, we substitute $\displaystyle x=2$ and get $\displaystyle 0=P(2)=1\cdot Q(2)=Q(2)$ : 2 is also a root of $\displaystyle Q$

    Problem 154. Assume that $\displaystyle x^3+ax^2+x+b$ (where a and b are some numbers) is divisible by $\displaystyle x^2-3x+2$. Find a and b.
    The first polynomial is cubic, the second quadratic so there exists $\displaystyle c,\,d\in \mathbb{R}$ such that $\displaystyle x^3+ax^2+x+b=(cx+d)(x^2-3x+2)$
    Then, you can develop the RHS and identify the coefficients of the two polynomials which yields 4 equations with 4 unknowns.
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  3. #3
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    Problem 147

    A polynomial $\displaystyle P(x)$ has a root when $\displaystyle P(x)=0$, for some given value of $\displaystyle x$. If $\displaystyle 1$ is a root of $\displaystyle P(x)$ then $\displaystyle P(1)=0$, hence $\displaystyle P(x)$ is divisible by $\displaystyle (x-1)$ because when $\displaystyle x=1$, ($\displaystyle x-1)=0$. The same can be said for $\displaystyle x=2$.

    Problem 154

    If $\displaystyle x^3+ax^2+x+b$ is divisible by $\displaystyle x^2-3x+2$ then the roots of the latter must also be roots for the former for a given $\displaystyle x$ value.

    Let $\displaystyle x^2-3x+2~=~0$

    $\displaystyle (x-2)(x-1)~=~0$ therefore $\displaystyle x=2~,~x=1$ are roots of the equation $\displaystyle x^3+ax^2+x+b$

    $\displaystyle x=1~,~~2+a+b~=~0$
    $\displaystyle x=2~,~~10+4a+b~=~0$

    Now solve simultaneously to find a and b
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  4. #4
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    Thanks Flyingsquirrel: your explanation for Problem 147 helped. I understand the solution now.

    Sean, thanks for your help on 154. I understand everything you did but I've run into a problem. Each time I solve simultaneously, I get $\displaystyle a=\frac{-8}{3} , b=\frac{2}{3}$ which works fine for $\displaystyle 2+a+b=0$ but yields a very strange, nonzero value indeed for $\displaystyle 10+4a+b$. I've tried a few times and get the same values every time.

    Here's my working:

    $\displaystyle 2+a+b=0$

    $\displaystyle a=-2-b$

    $\displaystyle 10+4(-2-b)+b=0$

    $\displaystyle
    b=\frac{2}{3}$ and $\displaystyle a=-2-\frac{2}{3}=\frac{-8}{3}$

    Does this happen for you as well? Or am I erring?
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  5. #5
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    Quote Originally Posted by jerry View Post
    Thanks Flyingsquirrel: your explanation for Problem 147 helped. I understand the solution now.

    Sean, thanks for your help on 154. I understand everything you did but I've run into a problem. Each time I solve simultaneously, I get $\displaystyle a=\frac{-8}{3} , b=\frac{2}{3}$ which works fine for $\displaystyle 2+a+b=0$ but yields a very strange, nonzero value indeed for $\displaystyle 10+4a+b$. I've tried a few times and get the same values every time.

    Here's my working:

    $\displaystyle 2+a+b=0$

    $\displaystyle a=-2-b$

    $\displaystyle 10+4(-2-b)+b=0$

    $\displaystyle
    b=\frac{2}{3}$ and $\displaystyle a=-2-\frac{2}{3}=\frac{-8}{3}$

    Does this happen for you as well? Or am I erring?
    The solution works fine!

    $\displaystyle 10+4a+b$:

    $\displaystyle 10 + 4 \left( \frac{-8}{3}\right) + \frac{2}{3} = 10 - \frac{32}{3} + \frac{2}{3} = 10 - \frac{30}{3} = 10 - 10 = 0$, as it should .....!!
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  6. #6
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    It appears we're done here. Thanks everybody!
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