Results 1 to 6 of 6

Thread: Polynomials questions again... part 2

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    78

    Polynomials questions again... part 2

    3. Given that f(x)=x^n-nx+n-1 for integer n>1. By considering f(x) and f '(x), show that f(x)=(x-1)^2 g(x) is true for all polynomials g(x) with integer coefficients. Hence or otherwise
    a) Show that 3^{2n}-8n-1 is divisible by 64 for all integers n>1
    b) Show that the equation x^4-4x+3=0 does not have any real root other than 1

    4. If 3+\sqrt{3} is a root of the equation 2x^2+3bx+3c=0 where b and c are constants. Find the value of bc.

    Answer:
    4. bc=-16
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jan 2009
    Posts
    290
    I can't help you with 1 and 2, I hate those kind of questions.

    For 3, take the derivative of that function, find the critical points, and then you can show that 1 is the only time the function touch the x axis.

    For 4, you know one root is 3+ \sqrt 3, the other solution is 3- \sqrt 3, substitute these values to x, you get 2 equations, solve the simultaneous equations, and you'll get your b and c value.

    Another way of doing it is with the root coefficient relationship. In the quadratic equation ax^2+bx+c, suppose the roots are \alpha and \beta, \alpha + \beta = -b, \alpha \beta = c
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    May 2009
    Posts
    581
    Thanks
    294
    Quote Originally Posted by cloud5 View Post
    4. If 3+\sqrt{3} is a root of the equation 2x^2+3bx+3c=0 where b and c are constants. Find the value of bc.
    If 3+\sqrt{3} is a root, then so is 3-\sqrt{3}.
    For a quadratic in the form f(x) = ax^2 + bx + c, the sum of its roots would be -\frac{b}{a}, and the product of its roots would be \frac{c}{a}. Plug in (be careful that the b and c are not the same as the b and c in your problem, ie. a = 2, "b" = 3b, and "c" = 3c):
    \begin{aligned}<br />
(3 + \sqrt{3}) + (3 - \sqrt{3}) &= -\frac{3b}{2} \\<br />
6 &= -\frac{3b}{2} \\<br />
12 &= -3b \\<br />
b &= -4<br />
\end{aligned}

    \begin{aligned}<br />
(3 + \sqrt{3})(3 - \sqrt{3}) &= \frac{3c}{2} \\<br />
9 - 3 &= \frac{3c}{2} \\<br />
6 &= \frac{3c}{2} \\<br />
 12 &= 3c \\<br />
c &= 4<br />
\end{aligned}

    So bc = (-4)(4) = -16.


    01
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    May 2009
    Posts
    581
    Thanks
    294
    Quote Originally Posted by chengbin View Post
    Another way of doing it is with the root coefficient relationship. In the quadratic equation ax^2+bx+c, suppose the roots are \alpha and \beta, \alpha + \beta = -b, \alpha \beta = c
    This is only true if the leading coefficient is one, or {\color{red}x^2} + bx + c = 0.


    01
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    846
    Hello, cloud5!

    4. 3+\sqrt{3} is a root of the equation 2x^2+3bx+3c\:=\:0 where b and c are constants.
    Find the value of bc.

    Answer: bc=-16
    We expected to be familiar with this fact . . .


    Given: . x^2 + px + q \:=\:0, a quadratic equation with leading coefficient 1.

    . . If the roots are \alpha and \beta, then: . \begin{Bmatrix}\alpha + \beta &=& \text{-}p \\ \alpha\beta &=& q \end{Bmatrix}


    In English: . \begin{array}{ccc}\text{Sum of the roots} &=& \text{negative of the }x\text{-coefficient.} \\ \text{Product of the roots} &=& \text{the constant term.} \end{array}


    We have: . x^2 + \frac{3b}{2}x + \frac{3c}{2} \;=\;0

    If one root is (3 + \sqrt{3}), the other root is (3-\sqrt{3})
    . . The roots appear in "conjugate pairs."


    The sum of the roots is -\frac{3b}{2}\!:\quad (3+\sqrt{3}) + (3 - \sqrt{3}) \:=\:-\frac{3b}{2}
    . . Hence: . -\frac{3b}{2} \:=\:6 \quad\Rightarrow\quad\boxed{ b \:=\:-4}


    The product of the roots is \frac{3c}{2}\!:\quad (3+\sqrt{3})(3-\sqrt{3}) \:=\:\frac{3c}{2}
    . . Hence: . \frac{3c}{2} \:=\:6 \quad\Rightarrow\quad\boxed{ c \:=\:4}


    Therefore: . bc \;=\;(-4)(4) \;=\;{\color{blue}-16}

    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,076
    Thanks
    2390
    Quote Originally Posted by cloud5 View Post
    3. Given that f(x)=x^n-nx+n-1 for integer n>1. By considering f(x) and f '(x), show that f(x)=(x-1)^2 g(x) is true for all polynomials g(x) with integer coefficients.
    First, the way this is written, it is NOT true! That should be "for some polynomial g(x)". Obviously (x-1)^2g(x) cannnot always be equal to f(x) no matter what g is! What you are really asked to prove is that f(x) has 1 as a double zero. Don't look at f(x) and f'(x), look at f(1) and f'(1).

    Hence or otherwise
    a) Show that 3^{2n}-8n-1 is divisible by 64 for all integers n>1
    If x= 9= 3^2, f(9)= 9^n- 9n+ n- 1= 3^{2n}- n- 1 so f(x)= (x-1)g(x). Now, what does that look like when x= 3?

    b) Show that the equation x^4-4x+3=0 does not have any real root other than 1
    This is f(x)= x^4- 4x+ 4- 1 and so can be written as [tex](x-1)^2g(x)[tex] for some polynomial g. You can use polynomial division by (x-1)^2= x^2- 2x+ 1 (or synthetic division twice) to find g, a quadratic equation. What are the roots of g?

    4. If 3+\sqrt{3} is a root of the equation 2x^2+3bx+3c=0 where b and c are constants. Find the value of bc.

    Answer:
    4. bc=-16
    Last edited by mr fantastic; June 25th 2009 at 06:27 PM. Reason: Fixed a latex tag
    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, 07:50 AM
  2. Polynomials questions again... part 5
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: July 12th 2009, 03:37 AM
  3. Polynomials questions again... part 3
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 27th 2009, 02:46 PM
  4. Polynomials questions again... part 4
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: June 27th 2009, 04:09 AM
  5. Polynomials questions again... part 1
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 27th 2009, 03:48 AM

Search Tags


/mathhelpforum @mathhelpforum