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Thread: Polynomials questions again... part 4

  1. #1
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    Polynomials questions again... part 4

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

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

    Answer:
    7a) b=3, c=2; x=-2, -2, -1, -1
    b) $\displaystyle 6y^2+5y-50=0$
    $\displaystyle y=-\frac{10}{3}, \frac{5}{2}; x=-3, -\frac{1}{3}, \frac{1}{2}, 2$
    Last edited by mr fantastic; Jun 27th 2009 at 03:38 AM. Reason: Fixed obvious typos
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  2. #2
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    Quote Originally Posted by cloud5 View Post
    7. a) Show that the polynomial $\displaystyle x^4+6x^3+13x^2+12x+4$ can be expressed in the form $\displaystyle (x^2+bx+c)^2$, where b and c are constants which must be evaluated. Subsequently, find all the roots of the equation $\displaystyle x^4+6x^3+13x^2+12x+4=0$
    $\displaystyle (x^2 + bx + c)^2$
    $\displaystyle \begin{aligned}
    &= x^4 + bx^3 + cx^2 + bx^3 + b^2x^2 + bcx + cx^2 + bcx + c^2 \\
    &= x^4 + 2bx^3 + (b^2 + 2c)x^2 + 2bcx + c^2 \\
    &= x^4+6x^3+13x^2+12x+4
    \end{aligned}$
    By comparing coefficients, you can see that b = 3 and c = 2.

    So,
    $\displaystyle x^4+6x^3+13x^2+12x+4$
    $\displaystyle \begin{aligned}
    &= (x^2 + 3x + 2)(x^2 + 3x + 2) \\
    &= (x + 2)(x + 1)(x + 2)(x + 1)
    \end{aligned}$

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


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

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

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

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

    $\displaystyle \begin{aligned}
    x + \frac{1}{x} &= \frac{5}{2} \\
    2x^2 + 2 &= 5x \\
    2x^2 - 5x + 2 &= 0 \\
    (2x - 1)(x - 2) &= 0 \\
    2x - 1 &= 0 \\
    x &= \frac{1}{2} \\
    x - 2 &= 0 \\
    x &= 2
    \end{aligned}$

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


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