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Thread: Quartic Polynomial Factorisation: 5-i is root of 2z^3+az^2+62z+(a-5)

  1. #1
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    Quartic Polynomial Factorisation: 5-i is root of 2z^3+az^2+62z+(a-5)

    Hi,

    I was presented with the following question (question 4).

    help4.jpg

    I was able to complete it using the below method.... Notice how in the bottom line, in the second lot of bracket, the z term has a coefficient of 2.

    HELP.png

    I was then presented with the following similar question (question 7).

    help3.jpg

    I was able to begin factorizing it using the below method. Notice how there is no coefficient of x in the second lot of brackets like last time. Now it is out the front of the entire equation.
    HELP2.png

    I understand the process of equating coefficents after this point to find the values of the unknowns. What I don't understand is in the first example why the coefficient of x can come into the second lot of brackets, while in the second example, the coefficient of x (2) has to stay out the front. When I attempted to use the alternate method in either example, it completely ruins it and gives a wrong answer. Can anyone please provide some explanation as to how/why/when you use either method or what I'm missing/forgetting/doing wrong?

    Thanks

  2. #2
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    It doesn't matter where the coefficient is as long as you solve for x correctly.
    (2x-2)(blah)=0, 2x-2=0, x=1.
    2(x-2)(blah)=0, x-2=0, x=2.

  3. #3
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    Quote Originally Posted by lev888 View Post
    It doesn't matter where the coefficient is as long as you solve for x correctly.
    (2x-2)(blah)=0, 2x-2=0, x=1.
    2(x-2)(blah)=0, x-2=0, x=2.
    Doesn't by having the coefficent of 2 out the front of the whole polynomial though, then effect the x values from the other factor?

  4. #4
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    Quote Originally Posted by Onigma View Post
    Doesn't by having the coefficent of 2 out the front of the whole polynomial though, then effect the x values from the other factor?
    Sure, different polynomials have different factors and zeros.
    2x-2 and 2(x-2) are different polynomials.

    2x-2 and 2(x-1) are the same.

    Please let me know if I misunderstood your question.

  5. #5
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    Quote Originally Posted by Onigma View Post
    Hi,

    I was presented with the following question (question 4).

    help4.jpg

    I was able to complete it using the below method.... Notice how in the bottom line, in the second lot of bracket, the z term has a coefficient of 2.

    HELP.png

    I was then presented with the following similar question (question 7).

    help3.jpg

    I was able to begin factorizing it using the below method. Notice how there is no coefficient of x in the second lot of brackets like last time. Now it is out the front of the entire equation.
    HELP2.png

    I understand the process of equating coefficents after this point to find the values of the unknowns. What I don't understand is in the first example why the coefficient of x can come into the second lot of brackets, while in the second example, the coefficient of x (2) has to stay out the front. When I attempted to use the alternate method in either example, it completely ruins it and gives a wrong answer. Can anyone please provide some explanation as to how/why/when you use either method or what I'm missing/forgetting/doing wrong?

    Thanks
    Why are you worrying about problem 7 when you have not completed problem 4.

    You were to find the both the other zeroes of P(z) and the value of a, given that:

    [tex]5 - i \text { is a zero of } P(z) = 2z^3 + az^2 + 62z + a - 5 \text { and } a \in \mathbb R.[/tex]

    You remembered that, because P(z) has only real coefficients, any complex roots come in conjugate pairs. So you correctly identified
    the other complex zero as 5 + i, which gave you one third of the answer. But because P(z) has only real coefficients and is of odd degree, it must have at least one real zero, making the third zero. You forgot that, and you forgot to find a.

    You started well by realizing that

    [tex]\{z - (5 - i)\}\{z - (5 + i)\} = z^2 - z(5 + i + 5 - i) + (5 - i)(5 + i) = z^2 - 10z + 26[/tex]

    factors P(z) so [tex]P(z) = (z^2 - 10z + 26)(2z + b).[/tex]

    But that does not finish up the two missing parts of the problem.

    [tex]P(z) = (z^2 - 10z + 26)(2z + b) = [/tex]

    [tex]2z^3 + (b - 20)z^2 + (52 - 10b)z + 26b \implies[/tex]

    [tex]62 = 52 - 10b \implies b = -\ 1 \text{, BUT}[/tex]

    [tex]26b = a - 5 \implies a = 26b + 5 = -\ 26 + 5 = -\ 21.[/tex]

    So [tex]a = - 21.[/tex] That is the second part of your answer.

    How about the real zero?

    [tex](2z + b) = 0 \implies 2z - 1 = 0 \implies z = \dfrac{1}{2}.[/tex]

    Thus the third zero = [tex]\dfrac{1}{2}[/tex].

    Let's check.

    [tex]2 * \left ( \dfrac{1}{2} \right )^3 - 21 * \left ( \dfrac{1}{2} \right )^2 + 62 * \left ( \dfrac{1}{2} \right ) - 21 - 5 =[/tex]

    [tex]2 * \dfrac{1}{8} - 21 * \dfrac{1}{4} + 62 * \dfrac{1}{2} - 26 =[/tex]

    [tex]\dfrac{1}{4} - \dfrac{21}{4} + 31 - 26 = -\ \dfrac{20}{4} + 5 = 0.[/tex]

    I am not at all sure what you are asking at the end. I think you are talking about treatment of the leading coefficient (the coefficient of the cubed term). If so, that is a good question. Please confirm
    Last edited by JeffM; 03-08-2018 at 05:29 PM. Reason: Fixed LaTeX error

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