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Thread: Plzz Help solving polynomial

  1. #1
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    Plzz Help solving polynomial

    I REALLY need help with these problems.. its factoring polynomial equations.. and i kinda understand how to do it BUT these problems are just way over my head would love some help plz here they are

    1. 8X^3 - 27 = 0

    2. X^3 - 3X^2 = 16x - 48

    3. X^4 - 2X - 8 = 0

    4. 16X^4 - 1 = 0

    5. -3X^3 - 18X^2 + 27X + 162 = 0

    These are the 5 problems that i cant seem to solve... plz help would appreciate it alot
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by inobd22 View Post
    I REALLY need help with these problems.. its factoring polynomial equations.. and i kinda understand how to do it BUT these problems are just way over my head would love some help plz here they are

    1. 8X^3 - 27 = 0

    2. X^3 - 3X^2 = 16x - 48

    3. X^4 - 2X - 8 = 0

    4. 16X^4 - 1 = 0

    5. -3X^3 - 18X^2 + 27X + 162 = 0

    These are the 5 problems that i cant seem to solve... plz help would appreciate it alot
    1. 1. 8X^3 - 27 = 0
    Notice that 8, x^3, and 27 are perfect cube terms. We can use the difference of cubes to solve this:
    (a^3 - b^3) = (a - b)(a^2 + ab + b^2)
    Notice that a is the cubed root of a^3 and b is the cubed root of b^3, so to solve this problem, we need the cubed roots of the two terms.
    8x^3 --> 2x
    27 --> 3
    Therefore,
    8x^3 - 27 = 0
    (2x - 3)[(2x)^2 + (2x)(3) + (3)^2] = 0
    (2x - 3)(4x^2 + 6x + 9) = 0

    2. X^3 - 3X^2 = 16x - 48
    First, we need to 'set this equal to 0' by moving everything to the left.
    x^3 - 3x^2 - 16x + 48 = 0
    Now, I'm going to try to factor this by grouping:
    (x^3 - 3x^2) + (-16x + 48) = 0
    x^2(x - 3) - 16(x - 3) = 0 ... Notice that (x - 3) is common in both terms, I can factor it to get:
    (x - 3)(x^2 - 16) = 0 ... Notice that x^2 - 16 is a difference of squares:
    (x - 3)(x + 4)(x - 4) = 0

    3. X^4 - 2X - 8 = 0
    Check to make sure you didn't write this one wrong. I doubt it should have an x^4 term.

    I have to go... hopefully someone else can help solve the rest.
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  3. #3
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    nope i wrote it right.. i no it confuses me too but thanks for the help that ya gave me hopefully ppl can help me with the rest..
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by inobd22 View Post
    3. X^4 - 2X - 8 = 0

    5. -3X^3 - 18X^2 + 27X + 162 = 0
    These are both going to involve the "Rational Root" theorem:
    Given a polynomial equation
    px^n + ax^{n-1} + ... + bx + q = 0

    any rational roots of this equation (if rational roots exist) will be of the form:
    x = (+/-){factor of q}/{factor of p}

    So for 3):
    x^4 - 2x - 8 = 0
    The possible rational roots will be:
    x = (+/-)1, 2, 4, 8

    Unfortunately none of these work. So a numerical solution is going to be your best bet. I get two real roots:
    x = -1.49595 and x = 1.84942

    Now you can do some polynomial division and use the quadratic formula to find the two remaining complex solutions:
    x = -0.1769734 (+/-)1.69126*I (where I = sqrt(-1), with due apologies to TPH!)

    For 5):
    -3x^3 - 18x^2 + 27x + 162 = 0
    The possible rational roots are:
    x = (+/-)1, 2, 3, 6, 27, 54, 81, 162

    We find that x = -6, (+/-)3. Since there are 3 solutions listed here, we are done.

    -Dan

    PS Heh. I always do things the hard way. You can divide the equation in 5 by a -3 to get:
    x^3 + 6x^2 - 9x - 54 = 0
    which is slightly easier to solve.

    -Dan
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by inobd22 View Post
    4. 16x^4 - 1 = 0
    This is the difference of two squares, and thus factors:
    16x^4 - 1 = 0

    (4x^2)^2 - 1^2 = 0

    (4x^2 - 1)(4x^2 + 1) = 0

    Set each factor to 0:
    a) 4x^2 - 1 = 0

    b) 4x^2 + 1 = 0

    Let's do a).
    4x^2 - 1 = 0 <-- Again the difference of two squares.

    (2x)^2 - 1^2 = 0

    (2x - 1)(2x + 1) = 0

    Set each factor to 0:
    2x - 1 = 0 ==> x = 1/2

    2x + 1 = 0 ==> x = -1/2

    Now do b).
    4x^2 + 1 = 0

    x^2 = -1/4

    x = (+/-)I/2 (where I = sqrt(-1).)

    So the 4 solutions are
    x = (+/-)1/2, (+/-)I/2

    -Dan
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  6. #6
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by topsquark View Post

    PS Heh. I always do things the hard way. You can divide the equation in 5 by a -3 to get:
    x^3 + 6x^2 - 9x - 54 = 0
    which is slightly easier to solve.

    -Dan
    Factor by grouping this one as well:

    (x^3 + 6x^2) + (-9x - 54) = 0
    x^2(x + 6) - 9(x + 6) = 0
    (x + 6)(x^2 - 9) = 0
    (x + 6)(x - 3)(x + 3) = 0

    By the way, thanks for helping in the last problems, topsquark.

    P.S. I really think problem 3 was a typo because I doubt that inobd22 knows about polynomial devision. Even then, the answer came out irrational - which makes it a 'slightly' harder problem than the other 4 .
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  7. #7
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    i double checked i wrote number 3 down right.. n polynomial division is easy and this stuff is easy to.. except these specific problems im having trouble with ..
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