Results 1 to 6 of 6

Math Help - 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
    527
    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
    527
    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
    11,659
    Thanks
    600
    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
    15,306
    Thanks
    1283
    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