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Math Help - Remainder theorem

  1. #1
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    Remainder theorem

    given that f(x)= x^3+kx^2-2x+1 and when f(x) is divided by (x-k) the remainder is k. find the possible values for k.

    i have k^3+k(k)^2-kx+1= k
    2k^3 - 2k+1 = k

    am stuck here. i know that (k-1) is a factor but i get a reaminder when i divide the equation by it,
    may i get some help with this?
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  2. #2
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    Quote Originally Posted by sigma1 View Post
    given that f(x)= x^3+kx^2-2x+1 and when f(x) is divided by (x-k) the remainder is k. find the possible values for k.

    i have k^3+k(k)^2-kx+1= k
    2k^3 - 2k+1 = k

    am stuck here. i know that (k-1) is a factor but i get a reaminder when i divide the equation by it,
    may i get some help with this?
    Try this:
    2k^3 - 2k+1 - k = 0
    2k(k^2-1)-(k-1)=0
    2k(k-1)(k+1) - (k-1)=0
    (k-1)(2k(k+1)-1)=0
    etc
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  3. #3
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    Quote Originally Posted by sigma1 View Post
    given that f(x)= x^3+kx^2-2x+1 and when f(x) is divided by (x-k) the remainder is k. find the possible values for k.

    i have k^3+k(k)^2-kx+1= k
    2k^3 - 2k+1 = k

    am stuck here. i know that (k-1) is a factor but i get a reaminder when i divide the equation by it,
    may i get some help with this?
    Perhaps you tried to divide 2k^3-2k+1 by k-1? - That doesn't work out as expected to be sure. But what about dividing 2k^3-3k+1 by k-1?
    In this case I get k-1)=2k^2+2k-1" alt="(2k^3-3k+1)k-1)=2k^2+2k-1" />.
    Moral: you have to move everything to one side of the equation before trying the division by k-1.
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  4. #4
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    Quote Originally Posted by Failure View Post
    Perhaps you tried to divide 2k^3-2k+1 by k-1? - That doesn't work out as expected to be sure. But what about dividing 2k^3-3k+1 by k-1?
    In this case I get k-1)=2k^2+2k-1" alt="(2k^3-3k+1)k-1)=2k^2+2k-1" />.
    Moral: you have to move everything to one side of the equation before trying the division by k-1.

    ok thanks for that but how do you factorise that to get the other values for k?
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  5. #5
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    Quote Originally Posted by Debsta View Post
    Try this:
    2k^3 - 2k+1 - k = 0
    2k(k^2-1)-(k-1)=0
    2k(k-1)(k+1) - (k-1)=0
    (k-1)(2k(k+1)-1)=0
    etc
    thanks . i follow your steps but how do i solve for the values of k
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  6. #6
    Super Member Failure's Avatar
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    Quote Originally Posted by sigma1 View Post
    ok thanks for that but how do you factorise that to get the other values for k?
    The other two values of k that are solutions to the original equation are irrational, so you can't find them by "educated guessing": you have to solve the quadratic equation 2k^2+2k-1=0 by whatever other means available (such as applying a memorized formula for that, or simply "completing the square").
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  7. #7
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    Quote Originally Posted by Failure View Post
    The other two values of k that are solutions to the original equation are irrational, so you can't find them by "educated guessing": you have to solve the quadratic equation 2k^2+2k-1=0 by whatever other means available (such as applying a memorized formula for that, or simply "completing the square").
    i have tried completing the square but was unable to over several attempts.
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  8. #8
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    Quote Originally Posted by sigma1 View Post
    i have tried completing the square but was unable to over several attempts.
    2k^2+2k-1=2(k^2+k)-1=2\left[\left(k+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\right]-1=2\left(k+\frac{1}{2}\right)^2-\frac{3}{2}.
    So to solve that quadratic equation, you solve

    2\left(k+\frac{1}{2}\right)^2-\frac{3}{2}=0 for k, which gives k_{2,3}=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}.
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  9. #9
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    Quote Originally Posted by sigma1 View Post
    thanks . i follow your steps but how do i solve for the values of k
    (k-1)(2k^2+2k-1)=0
    Use the null factor law to get:
    k-1=0 {or} 2k^2+2k-1=0
    The first will lead to the solution k=1
    The second can be solved using the quadratic formula, so you will end up with 3 solutions overall.
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  10. #10
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    Quote Originally Posted by sigma1 View Post
    given that f(x)= x^3+kx^2-2x+1 and when f(x) is divided by (x-k) the remainder is k. find the possible values for k.

    i have k^3+k(k)^2-kx+1= k
    2k^3 - 2k+1 = k
    2k^3- 3k+ 1= 0
    By the rational root theorem, the only possible rational roots are 1, -1, 1/2, and -1/2. Putting those into the equation, 2(1)^3- 3(1)+ 1= 0, 2(-1)^3- 3(-1)+ 1= 2, 2(1/2)^3- 3(1/2)+ 1= -1/4, and 2(-1/2)^3- 3(-1/2)+ 1= 2.

    Since 1 is a root, 2k^3- 3k+ 1 has k- 1 as a factor: 2k^3- 3k+ 1= (k- 1)(2k^2+ 2k+ 1). You can use the quadratic formula to find the other two roots.

    am stuck here. i know that (k-1) is a factor but i get a reaminder when i divide the equation by it,
    may i get some help with this?
    Then youare dividing wrong!
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