Results 1 to 3 of 3

Math Help - Polynomials questions again... part 4

  1. #1
    Member
    Joined
    Jun 2009
    Posts
    78

    Polynomials questions again... part 4

    7. a) Show that the polynomial x^4+6x^3+13x^2+12x+4 can be expressed in the form (x^2+bx+c)^2, where b and c are constants which must be evaluated. Subsequently, find all the roots of the equation x^4+6x^3+13x^2+12x+4=0

    b) By substitution y=x+\frac{1}{x}, transform the equation 6x^4+5x^3-38x^2+5x+6=0 into quadratic equation in y and both values of y that satisfy the quadratic equation. Hence solve the equation 6x^4+5x^3-38x^2+5x+6=0.

    Answer:
    7a) b=3, c=2; x=-2, -2, -1, -1
    b) 6y^2+5y-50=0
    y=-\frac{10}{3}, \frac{5}{2}; x=-3, -\frac{1}{3}, \frac{1}{2}, 2
    Last edited by mr fantastic; June 27th 2009 at 04:38 AM. Reason: Fixed obvious typos
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    May 2009
    Posts
    527
    Quote Originally Posted by cloud5 View Post
    7. a) Show that the polynomial x^4+6x^3+13x^2+12x+4 can be expressed in the form (x^2+bx+c)^2, where b and c are constants which must be evaluated. Subsequently, find all the roots of the equation x^4+6x^3+13x^2+12x+4=0
    (x^2 + bx + c)^2
    \begin{aligned}<br />
&= x^4 + bx^3 + cx^2 + bx^3 + b^2x^2 + bcx + cx^2 + bcx + c^2 \\<br />
&= x^4 + 2bx^3 + (b^2 + 2c)x^2 + 2bcx + c^2 \\<br />
&= x^4+6x^3+13x^2+12x+4<br />
\end{aligned}
    By comparing coefficients, you can see that b = 3 and c = 2.

    So,
    x^4+6x^3+13x^2+12x+4
    \begin{aligned}<br />
&= (x^2 + 3x + 2)(x^2 + 3x + 2) \\<br />
&= (x + 2)(x + 1)(x + 2)(x + 1)<br />
\end{aligned}

    So x = -2, -2, -1, -1.


    01
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    May 2009
    Posts
    527
    Quote Originally Posted by cloud5 View Post
    b) By substitution y=x+\frac{1}{x}, transform the equation 6x^4+5x^3-38x^2+5x+6=0 into quadratic equation in y and both values of y that satisfy the quadratic equation. Hence solve the equation 6x^4+5x^3-38x^2+5x+6=0.
    6x^4+5x^3-38x^2+5x+6=0
    Divide both sides by x^2:
    6x^2+5x-38+\frac{5}{x}+\frac{6}{x^2}=0

    So,
    6x^2+5x-38+\frac{5}{x}+\frac{6}{x^2}
    \begin{aligned}<br />
&= 6x^2 + \frac{6}{x^2} + 5x + \frac{5}{x} - 38 \\<br />
&= 6\left(x^2 + \frac{1}{x^2}\right) + 5\left(x + \frac{1}{x}\right) - 38 \\<br />
&= 6\left(x^2 + 2 + \frac{1}{x^2}\right) + 5\left(x + \frac{1}{x}\right) - 38 - 12 \\<br />
&= 6\left(x + \frac{1}{x}\right)^2 + 5\left(x + \frac{1}{x}\right) - 50 \\<br />
&= 6y^2 + 5y - 50<br />
\end{aligned}

    Now solve the quadratic:
    \begin{aligned}<br />
6y^2 + 5y - 50 &= 0 \\<br />
(3y + 10)(2y - 5) &= 0 \\<br />
3y + 10 &= 0 \\<br />
y &= -\frac{10}{3} \\<br />
2y - 5 &= 0 \\<br />
y &= \frac{5}{2}<br />
\end{aligned}

    Replace y with x + \frac{1}{x}:
    \begin{aligned}<br />
x + \frac{1}{x} &= -\frac{10}{3} \\<br />
3x^2 + 3 &= -10x \\<br />
3x^2 + 10x + 3 &= 0 \\<br />
(3x + 1)(x + 3) &= 0 \\<br />
3x + 1 &= 0 \\<br />
x &= -\frac{1}{3} \\<br />
x + 3 &= 0 \\<br />
x &= -3<br />
\end{aligned}

    \begin{aligned}<br />
x + \frac{1}{x} &= \frac{5}{2} \\<br />
2x^2 + 2 &= 5x \\<br />
2x^2 - 5x + 2 &= 0 \\<br />
(2x - 1)(x - 2) &= 0 \\<br />
2x - 1 &= 0 \\<br />
x &= \frac{1}{2} \\<br />
x - 2 &= 0 \\<br />
x &= 2<br />
\end{aligned}

    So your answers are x = -3,\; -\frac{1}{3},\; \frac{1}{2},\; 2.


    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 3
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: June 27th 2009, 03:46 PM
  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