# Thread: Polynomials questions again... part 4

1. ## 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 $\displaystyle 6x^4+5x^3-38x^2+5x+6=0$.

7a) b=3, c=2; x=-2, -2, -1, -1
b) $\displaystyle 6y^2+5y-50=0$
$\displaystyle y=-\frac{10}{3}, \frac{5}{2}; x=-3, -\frac{1}{3}, \frac{1}{2}, 2$

2. Originally Posted by cloud5
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.

01

3. Originally Posted by cloud5
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}

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