# Thread: Polynomials questions again... part 4

1. ## 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) $6y^2+5y-50=0$
$y=-\frac{10}{3}, \frac{5}{2}; x=-3, -\frac{1}{3}, \frac{1}{2}, 2$

2. Originally Posted by cloud5
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}
&= 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,
$x^4+6x^3+13x^2+12x+4$
\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. Originally Posted by cloud5
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}
&= 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:
\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 $x + \frac{1}{x}$:
\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}

\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 $x = -3,\; -\frac{1}{3},\; \frac{1}{2},\; 2$.

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