Sec 3 A Math: Polynomials & Partial Fractions, practice questions & worked solutions

When $x=-2$, $2(-2)^3+a(-2)^2+b(-2)+10=0$: $$\begin{aligned} 4a-2b &= 6\\ 2a-b &= 3 \text{- eqn 1} \end{aligned}$$ When $x=3$, $2(3)^3+a(3)^2+b(3)+10=10$: $$\begin{aligned} 9a+3b &= -54\\ 3a+b &= -18 \text{- eqn 2} \end{aligned}$$ eqn 1 + eqn 2: $$\begin{aligned} 5a &= -15\\ a &= -3\\ \therefore b &= -9 \end{aligned}$$

(a)

$$\begin{aligned} (2x + 1)^3 - (2x - 1)^3 &= \left[(2x + 1) - (2x - 1)\right]\left[(2x + 1)^2 + (2x + 1)(2x - 1) + (2x - 1)^2\right]\\ &= (2)\left[4x^2 + 4x + 1 + 4x^2 - 1 + 4x^2 - 4x + 1\right]\\ &= 2(12x^2 + 1) \end{aligned}$$

(b)

$$2(2x + 1)(12x^2 + 1) = 0$$ $$12x^2 + 1 = 0 \qquad\text{or}\qquad 2x + 1 = 0$$ $$\begin{aligned} \text{Discriminant} &= 0^2 - 4(12)(1)\\ &= -48 < 0 \end{aligned}$$ No real roots $$x = -\frac{1}{2} \text{ is the only real root.}$$

By long division, $$\begin{aligned} \frac{2x^3 + 4}{x(x + 2)^2} &= \frac{2x^3 + 4}{x^3 + 4x^2 + 4x}\\ &= 2 + \frac{-8x^2 - 8x + 4}{x(x + 2)^2} \end{aligned}$$ $$\frac{-8x^2 - 8x + 4}{x(x + 2)^2} = \frac{A}{x} + \frac{B}{x + 2} + \frac{C}{(x + 2)^2}$$ $$-8x^2 - 8x + 4 = A(x + 2)^2 + Bx(x + 2) + Cx$$ Let $x = 0$: $$4 = 4A, A = 1$$ Let $x = -2$: $$-12 = -2C, C = 6$$ Let $x = 1$: $$-8 - 8 + 4 = 9 + 3B + 6, B = -9$$ $$\therefore \frac{2x^3 + 4}{x(x + 2)^2} = 2 + \frac{1}{x} - \frac{9}{x + 2} + \frac{6}{(x + 2)^2}$$

$$x^2+2x-3 = (x-1)(x+3)$$ $$f(x) = (x-1)(x+3)Q(x)+(ax+b)$$ $$\begin{aligned} f(1) &= 61\\ a+b &= 61\\ f(-3) &= -123\\ -3a+b &= -123\\ a &= 46\\ b &= 15 \end{aligned}$$ The remainder is $46x+15$.

$$\frac{3x^2+5x-2}{(x+1)(x^2+4)} = \frac{A}{x+1}+\frac{Bx+C}{x^2+4}$$ $$3x^2+5x-2 = A(x^2+4)+(Bx+C)(x+1)$$ Sub $x=-1$, $3-5-2=5A$, $$A=-\frac{4}{5}$$ Compare constant: $-2=4A+C$, $$C=-2-4\!\left(-\frac{4}{5}\right)=\frac{6}{5}$$ Compare coeff of $x$: $5=B+C$, $$B=5-\frac{6}{5}=\frac{19}{5}$$ $$\frac{3x^2+5x-2}{(x+1)(x^2+4)} = -\frac{4}{5(x+1)}+\frac{19x+6}{5(x^2+4)}$$

Long division: $$\therefore\ \frac{2x^3 - 2x^2 + 7x - 12}{x^3 + 4x} = 2 + \frac{-2x^2 - x - 12}{x^3 + 4x}$$ $$\frac{-2x^2 - x - 12}{x^3 + 4x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 4}$$ Multiply by $\left(x^3 + 4x\right)$ throughout, $$-2x^2 - x - 12 = A\left(x^2 + 4\right) + x(Bx + C)$$ Sub $x = 0$, $$-12 = A(4) + 0, A = -3$$ Sub $x = 1$, $$\begin{aligned} -2 - 1 - 12 &= -3\left(1^2 + 4\right) + (1)(B + C)\\ -15 &= -15 + B + C\\ B + C &= 0 \text{------- (1)} \end{aligned}$$ Sub $x = -1$, $$\begin{aligned} -2 + 1 - 12 &= -3\left(1^2 + 4\right) + (-1)(-B + C)\\ -13 &= -15 + B - C\\ B - C &= 2 \text{------- (2)} \end{aligned}$$ $$\begin{aligned} (1) + (2): 2B &= 2\\ B &= 1 \end{aligned}$$ Sub $B = 1$ into (1): $$\begin{aligned} 1 + C &= 0\\ C &= -1 \end{aligned}$$ $$\frac{2x^3 - 2x^2 + 7x - 12}{x^3 + 4x} = 2 + \frac{x - 1}{x^2 + 4} - \frac{3}{x}$$

(a)

$$\begin{aligned} \mathrm{f}(x) &= x^3 + px^2 + qx - 12\\ \mathrm{f}(-1) &= 0\\ -1 + p - q - 12 &= 0\\ p - q &= 13 \text{------ (1)} \end{aligned}$$ $$\begin{aligned} \mathrm{f}(3) &= 36\\ 27 + 9p + 3q - 12 &= 36\\ 9p + 3q &= 21\\ 3p + q &= 7 \text{------ (2)} \end{aligned}$$ $(1) + (2)$: $$\begin{aligned} 4p &= 20\\ p &= 5\\ q &= 5 - 13\\ q &= -8\\ p = 5 \text{and} q &= -8 \end{aligned}$$

(b)

When $p = 5$ and $q = -8$, $$\begin{aligned} \mathrm{f}(x) &= 0\\ x^3 + 5x^2 - 8x - 12 &= 0\\ (x + 1)\left(x^2 + 4x - 12\right) &= 0\\ (x + 1)(x + 6)(x - 2) &= 0\\ x = -1 \text{or} x = -6 & \text{or} x = 2 \end{aligned}$$

Let $\dfrac{15x^2-13x+18}{(x-2)(3x^2+1)} = \dfrac{A}{x-2}+\dfrac{Bx+C}{3x^2+1}$. $$15x^2-13x+18 = A(3x^2+1)+(Bx+C)(x-2)$$ $$\begin{aligned} \text{when } x=2, 13A &= 52, A=4\\ \text{when } x=0, -2C+4 &= 18, C=-7\\ \text{when } x=1, 16-B+7 &= 20, B=3 \end{aligned}$$ $$\frac{15x^2-13x+18}{(x-2)(3x^2+1)} = \frac{4}{x-2}+\frac{3x-7}{3x^2+1}$$

(a)

$$x^3 + 27k^3 = (x + 3k)(x^2 - 3kx + 9k^2)$$

(b)

$$\begin{aligned} k = 1 : \ x^3 + 27 &= (x + 3)(x^2 - 3x + 9)\\ (x + 3)(x^2 - 3x + 9) &= (x + 3)(x + 10)\\ x + 3 = 0 \ \text{or} \ x^2 - 3x + 9 &= x + 10\\ x = -3 \ \text{or} \ x^2 - 4x - 1 &= 0 \end{aligned}$$ $$\begin{aligned} x &= \frac{4 \pm \sqrt{16 - 4(-1)}}{2}\\ &= 2 \pm \sqrt{5} \end{aligned}$$

(c)

Let $\mathrm{f}(x) = x^3 + 27k^3$ $$\begin{aligned} \mathrm{f}(2) = 2^3 + 27k^3 &= 351\\ k &= \sqrt[3]{\frac{351 - 8}{27}} = \frac{7}{3} \end{aligned}$$

(a)

Given $\mathrm{f}(x) = 3x^3 + 2x^2 + 16$. $$\begin{aligned} \mathrm{f}(3a) &= \mathrm{f}\!\left(-\frac{2}{3}\right)\\ 3(3a)^3 + 2(3a)^2 + 16 &= 3\left(-\frac{2}{3}\right)^3 + 2\left(-\frac{2}{3}\right)^2 + 16\\ 3(27a^3) + 2(9a^2) &= 3\left(-\frac{8}{27}\right) + 2\left(\frac{4}{9}\right)\\ 81a^3 + 18a^2 &= -\frac{8}{9} + \frac{8}{9}\\ 9a^2(9a + 2) &= 0\\ a = 0 \ &\text{or}\ a = -\frac{2}{9} \end{aligned}$$

(b)

Subst $x = -2$ into $\mathrm{f}(x)$, $$\begin{aligned} \mathrm{f}(-2) &= 3(-2)^3 + 2(-2)^2 + 16\\ &= -24 + 8 + 16\\ &= 0 \end{aligned}$$ $\therefore x = -2$ is a solution of $\mathrm{f}(x)$.

(c)

From (b), $x + 2$ is a factor of $\mathrm{f}(x)$.

By inspection, $$\begin{aligned} \mathrm{f}(x) &= 3x^3 + 2x^2 + 16\\ &= (x + 2)(3x^2 + kx + 8) \end{aligned}$$ Equating coefficient of $$\begin{array}{c|l} x^2 : & k + 6 = 2\\ & k = -4\\ \hline & \text{or}\\ \hline x : & 2k + 8 = 0\\ & k = -4 \end{array}$$ $$\begin{aligned} \mathrm{f}(x) &= 3x^3 + 2x^2 + 16\\ &= (x + 2)(3x^2 - 4x + 8) \end{aligned}$$ For $\mathrm{f}(x) = 0$ $$\begin{aligned} (x + 2)(3x^2 - 4x + 8) &= 0\\ x + 2 &= 0\\ x &= -2 \end{aligned}$$ $$\begin{aligned} \text{or } 3x^2 - 4x + 8 &= 0\\ D &= (4)^2 - 4(3)(8)\\ D &= -80 \end{aligned}$$ Since $D < 0$, there is no real roots for the quadratic equation

$\therefore \mathrm{f}(x) = 0$ has only one real root. $x = -2$.

(d)

$$\begin{aligned} 16y^3 + 2y - 3 &= 0\\ 16 + \frac{2}{y^2} - \frac{3}{y^3} &= 0\\ 3\left(-\frac{1}{y}\right)^3 + 2\left(-\frac{1}{y}\right)^2 + 16 &= 0 \end{aligned}$$ Let $x = -\dfrac{1}{y}$ $$\begin{aligned} 3x^3 + 2x^2 + 16 &= 0\\ x &= -2\\ -\frac{1}{y} &= -2\\ y &= \frac{1}{2} \end{aligned}$$