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  1. #1
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    help solve 1

    The polynomial p(x)is given by p(x)=ax3 +16x2+cx 120, where a and c are constants. The three zeros of p(x)are 2, 3 and α. Find the value of α.

    with working out please
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  2. #2
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    Quote Originally Posted by andy69 View Post
    The polynomial p(x)is given by p(x)=ax3 +16x2+cx 120, where a and c are constants. The three zeros of p(x)are 2, 3 and α. Find the value of α.

    with working out please
    $\displaystyle P(x) = ax^3 + 16x^2 + cx - 120$.


    By the remainder and factor theorems, if $\displaystyle a$ is a root then $\displaystyle P(a) = 0$.

    So you should find:

    $\displaystyle P(-2) = a(-2)^3 + 16(-2)^2 - 2c - 120 = 0$

    $\displaystyle P(3) = a(3)^3 + 16(3)^2 + 3c - 120 = 0$.


    After you have simplified these, you have 2 equations in 2 unknowns. Solve them simultaneously for $\displaystyle a$ and $\displaystyle b$.


    Once you have $\displaystyle a$ and $\displaystyle b$, you have the polynomial.

    Since $\displaystyle -2$ is a root, $\displaystyle x + 2$ is a factor.

    Since $\displaystyle 3$ is a root, $\displaystyle x - 3$ is a factor.

    Long divide your polynomials by these factors to find the third factor. From there you can find the final root $\displaystyle \alpha$.


    If you have any more trouble, please show your working and exactly where you are stuck.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    $\displaystyle P(x) = ax^3 + 16x^2 + cx - 120$.


    By the remainder and factor theorems, if $\displaystyle a$ is a root then $\displaystyle P(a) = 0$.

    So you should find:

    $\displaystyle P(-2) = a(-2)^3 + 16(-2)^2 - 2c - 120 = 0$

    $\displaystyle P(3) = a(3)^3 + 16(3)^2 + 3c - 120 = 0$.


    After you have simplified these, you have 2 equations in 2 unknowns. Solve them simultaneously for $\displaystyle a$ and $\displaystyle b$.


    Once you have $\displaystyle a$ and $\displaystyle b$, you have the polynomial.

    Since $\displaystyle -2$ is a root, $\displaystyle x + 2$ is a factor.

    Since $\displaystyle 3$ is a root, $\displaystyle x - 3$ is a factor.

    Long divide your polynomials by these factors to find the third factor. From there you can find the final root $\displaystyle \alpha$.


    If you have any more trouble, please show your working and exactly where you are stuck.
    ok is this right um i subbed -2 and 3 into the equations
    and got -8a+64-2c-120=0(1) and 27a+144+3c-120=0(2)
    then i did -8a+64-2c-120(times by -3) and 27a+144+3c-120(times by 2)
    which is 24a+6c+168=0(3) and 54a+6c+44=0(4)
    then i did (4)-(3)
    which is 30a-124=0
    30a=124
    a=124 divide by 30
    =4.13
    sub a=4.13 into (4)
    can you tell me if this is right so far.
    Last edited by andy69; May 20th 2010 at 12:29 AM.
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  4. #4
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    Quote Originally Posted by andy69 View Post
    ok is this right um i subbed -2 and 3 into the equations
    and got -8a+64-2c-120=0(1) and 27a+144+3c-120=0(2)
    then i did -8a+64-2c-120(times by -3) and 27a+144+3c-120(times by 2)
    which is 24a+6c+168=0(3) and 54a+6c+44=0(4)
    then i did (4)-(3)
    which is 30a-124=0
    30a=124
    a=124 divide by 30
    =4.13
    sub a=4.13 into (4)
    can you tell me if this is right so far.
    Incorrect.

    You are right that you should transform equation 1 into

    $\displaystyle 24a + 6c + 168 = 0$

    but you are incorrect with equation 2.


    You should have gotten:

    $\displaystyle 27a + 144 + 3c - 120 = 0$

    $\displaystyle 27a + 3c + 24= 0$

    $\displaystyle 54a + 6c + 48 = 0$.


    So the two equations are:

    $\displaystyle 24a + 6c + 168 = 0$

    $\displaystyle 54a + 6c + 48 = 0$.


    Go from here.
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  5. #5
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    Quote Originally Posted by Prove It View Post
    Incorrect.

    You are right that you should transform equation 1 into

    $\displaystyle 24a + 6c + 168 = 0$

    but you are incorrect with equation 2.


    You should have gotten:

    $\displaystyle 27a + 144 + 3c - 120 = 0$

    $\displaystyle 27a + 3c + 24= 0$

    $\displaystyle 54a + 6c + 48 = 0$.


    So the two equations are:

    $\displaystyle 24a + 6c + 168 = 0$

    $\displaystyle 54a + 6c + 48 = 0$.


    Go from here.
    24a + 6c + 168 = 0(1)
    54a + 6c + 48 = 0(2)
    (2)-(1)
    30a-120=0
    30a=120
    a=120/30
    a=4
    sub a=4 into (1)
    96+6c+168=0
    6c=-168-96
    6c=-264
    c=-264/6
    c=-44

    is this working out right
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  6. #6
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    Quote Originally Posted by andy69 View Post
    24a + 6c + 168 = 0(1)
    54a + 6c + 48 = 0(2)
    (2)-(1)
    30a-120=0
    30a=120
    a=120/30
    a=4
    sub a=4 into (1)
    96+6c+168=0
    6c=-168-96
    6c=-264
    c=-264/6
    c=-44

    is this working out right
    Yes, well done.

    Now long divide by $\displaystyle x + 2$ and $\displaystyle x - 3$. From the third factor you'll be able to find $\displaystyle \alpha$.
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  7. #7
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    Quote Originally Posted by Prove It View Post
    Yes, well done.

    Now long divide by $\displaystyle x + 2$ and $\displaystyle x - 3$. From the third factor you'll be able to find $\displaystyle \alpha$.
    so do i divide these 2 factors with p(x)=ax3 +16x2+cx 120
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  8. #8
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    Quote Originally Posted by andy69 View Post
    so do i divide these 2 factors with p(x)=ax3 +16x2+cx 120
    You know what $\displaystyle a$ and $\displaystyle c$ are now... Substitute them into your polynomial first.
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  9. #9
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    Quote Originally Posted by Prove It View Post
    You know what $\displaystyle a$ and $\displaystyle c$ are now... Substitute them into your polynomial first.
    4x^3+16x^2+44x-120 what do i do now?
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