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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
    P(x) = ax^3 + 16x^2 + cx - 120.


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

    So you should find:

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

    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 a and b.


    Once you have a and b, you have the polynomial.

    Since -2 is a root, x + 2 is a factor.

    Since 3 is a root, 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 \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
    P(x) = ax^3 + 16x^2 + cx - 120.


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

    So you should find:

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

    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 a and b.


    Once you have a and b, you have the polynomial.

    Since -2 is a root, x + 2 is a factor.

    Since 3 is a root, 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 \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 01:29 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

    24a + 6c + 168 = 0

    but you are incorrect with equation 2.


    You should have gotten:

    27a + 144 + 3c - 120 = 0

    27a + 3c + 24= 0

    54a + 6c + 48 = 0.


    So the two equations are:

    24a + 6c + 168 = 0

    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

    24a + 6c + 168 = 0

    but you are incorrect with equation 2.


    You should have gotten:

    27a + 144 + 3c - 120 = 0

    27a + 3c + 24= 0

    54a + 6c + 48 = 0.


    So the two equations are:

    24a + 6c + 168 = 0

    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 x + 2 and x - 3. From the third factor you'll be able to find \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 x + 2 and x - 3. From the third factor you'll be able to find \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 a and 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 a and c are now... Substitute them into your polynomial first.
    4x^3+16x^2+44x-120 what do i do now?
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