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Math Help - 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 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 Q = (x-2)\cdot{R} for some polynomial R. So 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 x^3+ax^2+x+b (where a and b are some numbers) is divisible by 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 P so P(2)=0
    In P(x)=(x-1)Q(x), we substitute x=2 and get 0=P(2)=1\cdot Q(2)=Q(2) : 2 is also a root of Q

    Problem 154. Assume that x^3+ax^2+x+b (where a and b are some numbers) is divisible by x^2-3x+2. Find a and b.
    The first polynomial is cubic, the second quadratic so there exists c,\,d\in \mathbb{R} such that 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 P(x) has a root when P(x)=0, for some given value of x. If 1 is a root of P(x) then P(1)=0, hence P(x) is divisible by (x-1) because when x=1, ( x-1)=0. The same can be said for x=2.

    Problem 154

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

    Let x^2-3x+2~=~0

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

    x=1~,~~2+a+b~=~0
    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 a=\frac{-8}{3}          , b=\frac{2}{3} which works fine for 2+a+b=0 but yields a very strange, nonzero value indeed for 10+4a+b. I've tried a few times and get the same values every time.

    Here's my working:

    2+a+b=0

    a=-2-b

    10+4(-2-b)+b=0

    <br />
b=\frac{2}{3} and 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 a=\frac{-8}{3} , b=\frac{2}{3} which works fine for 2+a+b=0 but yields a very strange, nonzero value indeed for 10+4a+b. I've tried a few times and get the same values every time.

    Here's my working:

    2+a+b=0

    a=-2-b

    10+4(-2-b)+b=0

    <br />
b=\frac{2}{3} and a=-2-\frac{2}{3}=\frac{-8}{3}

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

    10+4a+b:

    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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