Results 1 to 5 of 5

Math Help - Polynomials questions again... part 3

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    78

    Polynomials questions again... part 3

    5. Factorize p(x)= 8x^4-4x^3-2x^2+7x-3 completely. Show that the equation has only two real roots, and find these both roots.
    Deduce both the real roots of the equation x^4-x^3-x^2+7x-6=0

    6. Using the substitution u=x+\frac{1}{x}, find the values of a and b such that x^2-5x+6-\frac{5}{x}+\frac{1}{x^2}=u^2+au+b. Subsequently, determine all the roots of the equation x^4-5x^3+6x^2-5x+1=0

    Answer:
    5. p(x)=(2x-1)(x+1)( 4x^2-4x+3); x=-1, x= \frac{1}{2}

    6. a=-5, b=4; x=2 \pm \sqrt{3}, x=\frac{1}{2} \pm \frac{\sqrt{3}}{2}i
    Last edited by cloud5; June 27th 2009 at 08:46 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by cloud5 View Post

    6. Using the substitution u=x+\frac{1}{x}, find the values of a and b such that x^2-5x+6-\frac{5}{x}+\frac{1}{x^2}=u^2+au+b.

    x^2-5x+6-\frac{5}{x}+\frac{1}{x^2}
    =x^2+\frac{1}{x^2}-5x-\frac{5}{x}+6
    =(x+\frac{1}{x})^2-2-5(x+\frac{1}{x})+6
    =u^2-2-5u+6
    =u^2-5u+4

    hence, a=-5, b=4

    Subsequently, determine all the roots of the equation x^4-5x^3+6x^2-5x+1=0
    x^4-5x^3+6x^2-5x+1=0

    Let x is not equal to zero.
    x^2(x^2-5x^1+6-\frac{5}{x}+\frac{1}{x^2})=0
    x^2(u^2-5u+4)=0

    Solving for u gives u=1,4
    so, x+\frac{1}{x}=1,4

    Solving by quadratic formula, you will get

    Answer:
    x=2 \pm \sqrt{3}, x=\frac{1}{2} \pm \frac{\sqrt{3}}{2}i
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by cloud5 View Post
    5. Factorize p(x)= 8x^4-4x^3+ax^2+bx-3 completely. Show that the equation has only two real roots, and find these both roots.
    [snip]
    I don't see how you can expect this equation to be factorised when the coefficients of x^2 and x are not known ....
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2009
    Posts
    78
    Quote Originally Posted by mr fantastic View Post
    I don't see how you can expect this equation to be factorised when the coefficients of x^2 and x are not known ....
    Opss... made a mistake there...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    May 2009
    Posts
    527
    Quote Originally Posted by cloud5 View Post
    5. Factorize p(x)= 8x^4-4x^3-2x^2+7x-3 completely. Show that the equation has only two real roots, and find these both roots.
    Now that's more like it...

    5) By the rational roots theorem, the possible rational roots are
    \frac{\text{factors of}\; 3}{\text{factors of}\; 8} = \frac{\pm 1, \pm 3}{\pm 1, \pm 2, \pm 4, \pm 8}
    = 1, 3, 1/2, 3/2, 1/4, 3/4, 1/8, 3/8.

    Start testing these factors by using the Factor Theorem or synthetic division. You'll find that x = -1 is a root:
    Code:
    -1| 8  -4  -2   7  -3
    ---    -8  12 -10   3
    ----------------------
        8 -12  10  -3   0
    x = 1/2 is also a root:
    Code:
    1/2| 8 -12  10  -3
    ----     4  -4   3
    -------------------
         8  -8   6   0
    So
    \begin{aligned}<br />
p(x) &= 8x^4-4x^3-2x^2+7x-3 \\<br />
&= (x - 1/2)(x + 1)(8x^2 - 8x + 6) \\<br />
&= 2(x - 1/2)(x + 1)(4x^2 - 4x + 3) \\<br />
&= (2x - 1)(x + 1)(4x^2 - 4x + 3)<br />
\end{aligned}

    The quadratic is irreducible over the reals -- the discriminant is negative:
    b^2 - 4ac = (-4)^2 - 4(4)(3) = 16 - 48 = -32
    We have two real roots, x = 1/2 and x = -1.

    Deduce both the real roots of the equation x^4-x^3-x^2+7x-6=0
    Do this the same way as I did above.


    01
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Polynomials questions again... part 6
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: July 15th 2009, 08:50 AM
  2. Polynomials questions again... part 5
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 12th 2009, 04:37 AM
  3. Polynomials questions again... part 4
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: June 27th 2009, 05:09 AM
  4. Polynomials questions again... part 1
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 27th 2009, 04:48 AM
  5. Polynomials questions again... part 2
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 25th 2009, 07:44 AM

Search Tags


/mathhelpforum @mathhelpforum