Sec 4 A Math: Polynomials & Partial Fractions, practice questions & worked solutions

Method 1

$$\frac{2x^3-5}{x^2(x-1)} = 2+\frac{2x^2-5}{x^2(x-1)}$$ $$\frac{2x^2-5}{x^2(x-1)} = \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}$$ $$2x^2-5 = Ax(x-1)+B(x-1)+Cx^2$$ Let $x=1$, $-3=C$.

Let $x=0$, $-5=B(-1)$, $B=5$.

Let $x=-1$, $-3=-A(-2)+5(-2)-3(1)$, $A=5$. $$\frac{2x^3-5}{x^2(x-1)} = 2+\frac{5}{x}+\frac{5}{x^2}-\frac{3}{x-1}$$

Method 2

$$\frac{2x^3-5}{x^2(x-1)} = 2+\frac{2x^2-5}{x^2(x-1)}$$ $$\frac{2x^2-5}{x^2(x-1)} = \frac{Ax+B}{x^2}+\frac{C}{x-1}$$ $$2x^2-5 = (Ax+B)(x-1)+Cx^2$$ Let $x=1$, $-3=C$.

Let $x=0$, $-5=B(-1)$, $B=5$.

Let $x=-1$, $-3=-A(-2)+5(-2)-3(1)$, $A=5$. $$\begin{aligned} \frac{2x^3-5}{x^2(x-1)} &= 2+\frac{5x+5}{x^2}-\frac{3}{x-1}\\ &= 2+\frac{5}{x}+\frac{5}{x^2}-\frac{3}{x-1} \end{aligned}$$

(a)

Note $x(x + 1)^2 = x^3 + 2x^2 + x$. By long division: $$\begin{array}{r} 2\phantom{x^3 + 12x^2 + 14x + 5} \\[-2pt] x^3 + 2x^2 + x\ \overline{\smash{)}\ 2x^3 + 12x^2 + 14x + 5}\phantom{} \\[-1pt] \underline{-(2x^3 + 4x^2 + 2x)}\phantom{+ 14x + 5} \\[-1pt] 8x^2 + 12x + 5\phantom{} \end{array}$$ $$\frac{2x^3 + 12x^2 + 14x + 5}{x(x + 1)^2} = 2 + \frac{8x^2 + 12x + 5}{x(x + 1)^2}$$

(b)

$$\frac{8x^2 + 12x + 5}{x(x + 1)^2} = \frac{A}{x} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}$$ $$8x^2 + 12x + 5 = A(x + 1)^2 + Bx(x + 1) + Cx$$ When $x = 0$, $A = 5$.

When $x = -1$, $C = -1$.

When $x = 1$, $2B = 6$ $$B = 3$$ $$\frac{2x^3 + 12x^2 + 14x + 5}{x(x + 1)^2} = 2 + \frac{5}{x} + \frac{3}{x + 1} - \frac{1}{(x + 1)^2}$$

(a)

$$2x^3 + ax^2 + bx - 10 = (x - 1)^2(2x + c)$$ Comparing constant, $$c = -10$$ Comparing $x^2$, $$a = -10 - 4 = -14$$ Comparing $x$, $$b = 20 + 2 = 22$$

(b)

$$\begin{aligned} 2\left(\tfrac{1}{2}\right)^3 - 14\left(\tfrac{1}{2}\right)^2 + 22\left(\tfrac{1}{2}\right) - 10 + k &= 0\\ k &= \frac{9}{4} \end{aligned}$$

(a)

$$\begin{aligned} \mathrm{f}(-1) &= 3(-1)^3 + a(-1)^2 - 7(-1) + 3\\ 0 &= -3 + a + 7 + 3\\ a &= -7 \end{aligned}$$ By long division, $$\begin{aligned} \mathrm{f}(x) &= 3x^3 - 7x^2 - 7x + 3\\ &= (x + 1)(3x^2 - 10x + 3)\\ &= (x + 1)(3x - 1)(x - 3) \end{aligned}$$

(b)

$$\begin{aligned} \cot^2\theta = -1\ (\text{rej}) \text{or} \cot^2\theta = \frac{1}{3} &\text{or} \cot^2\theta = 3\\ \cot\theta = \pm\frac{1}{\sqrt{3}} &\phantom{\text{or} } \cot\theta = \pm\sqrt{3}\\ \theta = \pm 60^\circ &\phantom{\text{or} } \theta = \pm 30^\circ \end{aligned}$$

(a)

Let $f(x)=3x^3+8x^2-5x-6$. $$\begin{aligned} f(1) &= 3+8-5-6 = 0 \end{aligned}$$ $\therefore$ By factor theorem, $x-1$ is a factor of $f(x)$.

(b)

$$\begin{aligned} f(x)=3x^3+8x^2-5x-6 &= (x-1)(Ax^2+Bx+C) \end{aligned}$$ Comparing coefficients of $x^3$: $A=3$.

Comparing coefficients of $x^2$: $8=B-3$, $B=11$.

Compare constant: $C=6$. $$\begin{aligned} f(x)=(x-1)(3x^2+11x+6) &= 0\\ (x-1)(3x+2)(x+3) &= 0\\ x=1,\ x=-\frac{2}{3},\ &x=-3 \end{aligned}$$

(c)

$$3(y-2)^3+8(y-2)^2-5(y-2)-6 = 0$$ $$\begin{aligned} y-2=1, y-2=-\frac{2}{3}, &y-2=-3\\ y=3, y=\frac{4}{3}, &y=-1 \end{aligned}$$

(a)

(i) $$\begin{aligned} f(x) &= (2x+3)h(x)-1\\ f(x)-1 &= (2x+3)h(x)-1-1\\ &= (2x+3)h(x)-2 \end{aligned}$$ Remainder $=-2$.

(ii) Polynomial $=f(x)+1$.

(b)

$$\begin{aligned} g(x) &= 2(x+1)(x-m)(x-2m)\\ g(1) &= 2(2)(1-m)(1-2m)=12\\ 4(1-3m+2m^2) &= 12\\ 2m^2-3m+1 &= 3\\ 2m^2-3m-2 &= 0\\ (2m+1)(m-2) &= 0\\ m=2 \;&\text{or}\; -\frac{1}{2}\ (reject)\\ g(x) &= 2(x+1)(x-2)(x-4)\\ &= 2(x^2-x-2)(x-4)\\ &= 2(x^3-4x^2-x^2+4x-2x+8)\\ &= 2(x^3-5x^2+2x+8)\\ &= 2x^3-10x^2+4x+16 \end{aligned}$$

(a)

$$\begin{array}{r} x - 2 \phantom{{}+ 000000} \\[2pt] 2x^2 + 1 \, \overline{)\, 2x^3 - 4x^2 + x - 2\phantom{0}} \\[2pt] \underline{-(2x^3 \phantom{{}- 4x^2} + x)}\phantom{{}- 2} \\[2pt] -4x^2 \phantom{{}+ x} - 2 \\[2pt] \underline{-(-4x^2 \phantom{{}+ x} - 2)} \\[2pt] 0 \end{array}$$ $\therefore 2x^2 + 1$ is a factor since remainder $= 0$

(b)

$$\begin{aligned} \frac{11x - 5x^2 - 11}{(x - 2)(2x^2 + 1)} &= \frac{A}{x - 2} + \frac{Bx + C}{2x^2 + 1}\\ 11x - 5x^2 - 11 &= A(2x^2 + 1) + (Bx + C)(x - 2) \end{aligned}$$ Sub $x = 2$, $$-9 = 9A \implies A = -1$$ Compare $x^2$, $$-5 = 2A + B \implies B = -3$$ Compare constants, $$-11 = A - 2C \implies C = 5$$ $$\therefore \frac{11x - 5x^2 - 11}{(x - 2)(2x^2 + 1)} = -\frac{1}{x - 2} + \frac{5 - 3x}{2x^2 + 1}$$

$$\begin{aligned} (x - 3)(x + 1)^2 &= (x - 3)\left(x^2 + 2x + 1\right)\\ &= x^3 - x^2 - 5x - 3 \end{aligned}$$ Long division: $$\frac{x^3 + x^2 - 10x - 22}{x^3 - x^2 - 5x - 3} = 1 + \frac{2x^2 - 5x - 19}{x^3 - x^2 - 5x - 3}$$ $$\frac{x^3 + x^2 - 10x - 22}{(x - 3)(x + 1)^2} = 1 + \frac{2x^2 - 5x - 19}{(x - 3)(x + 1)^2}$$ Let $$\begin{aligned} \frac{2x^2 - 5x - 19}{(x - 3)(x + 1)^2} &= \frac{A}{x - 3} + \frac{B}{x + 1} + \frac{C}{(x + 1)^2}\\ &= \frac{A(x + 1)^2 + B(x - 3)(x + 1) + C(x - 3)}{(x - 3)(x + 1)^2} \end{aligned}$$ $$2x^2 - 5x - 19 = A(x + 1)^2 + B(x - 3)(x + 1) + C(x - 3)$$ Sub $x = -1$ $$\begin{aligned} 2(-1)^2 - 5(-1) - 19 &= C(-1 - 3)\\ -12 &= -4C\\ C &= 3 \end{aligned}$$ Sub $x = 3$ $$\begin{aligned} 2(3)^2 - 5(3) - 19 &= A(3 + 1)^2\\ -16 &= 16A\\ A &= -1 \end{aligned}$$ Sub $x = 0$ $$\begin{aligned} -19 &= A + B(-3)(1) + C(-3)\\ -19 &= -1 - 3B + 3(-3)\\ 3B &= 9\\ B &= 3 \end{aligned}$$ Hence $$\frac{x^3 + x^2 - 10x - 22}{(x - 3)(x + 1)^2} = 1 - \frac{1}{x - 3} + \frac{3}{x + 1} + \frac{3}{(x + 1)^2}$$

(a)

$$\begin{aligned} \mathrm{f}(x) &= 2\left(x - \tfrac{1}{2}\right)(x + 3)(x^2 + 1)\\ &= 2x^4 + 5x^3 - x^2 + 5x - 3 \end{aligned}$$

(b)

By factor theorem, $$\begin{aligned} \mathrm{f}(3) &= 0\\ (3)^2 + 3(k - 1) + k^2 - 16 &= 0\\ 9 + 3k - 3 + k^2 - 16 &= 0\\ k^2 + 3k - 10 &= 0\\ (k + 5)(k - 2) &= 0\\ k = -5 \ &\text{or}\ k = 2 \end{aligned}$$ When $k = -5$, When $k = 2$, $$\mathrm{f}(-4) = 49 \qquad \mathrm{f}(-4) = 0$$ $$\therefore k = -5$$

(a)

$$\begin{array}{r} x^2-\frac{3}{2}x-2\phantom{xxxxxxxx}\\[2pt] 2x-4\,\overline{\smash{)}\,2x^3-7x^2+2x+3\phantom{x}}\\[2pt] \underline{-(2x^3-4x^2)}\phantom{xxxxxxxxxx}\\[2pt] -3x^2+2x\phantom{xxxx}\\[2pt] \underline{-(-3x^2+6x)}\phantom{xxx}\\[2pt] 3-4x\phantom{xx}\\[2pt] \underline{-(8-4x)}\phantom{x}\\[2pt] -5\phantom{xx} \end{array}$$

(b)

$$\begin{aligned} 2x^3-7x^2+2x+3 &= 0 \end{aligned}$$ Let $f(x)=2x^3-7x^2+2x+3$.

When $x=3$, $f(3)=2(3)^3-7(3)^2+2(3)+3=0$.

By factor theorem, $(x-3)$ is a factor of $f(x)$. $$\begin{aligned} (x-3)(2x^2-x-1) &= 0\\ (x-3)(2x+1)(x-1) &= 0\\ x=-\frac{1}{2},\ 1 \;&\text{or}\; 3 \end{aligned}$$

(c)

$$\begin{aligned} 3y^6+2y^4-7y^2+2 &= 0\\ 3+\frac{2}{y^2}-\frac{7}{y^4}+\frac{2}{y^6} &= 0\\ 3+2\!\left(\frac{1}{y^2}\right)-7\!\left(\frac{1}{y^4}\right)+2\!\left(\frac{1}{y^6}\right) &= 0\\ 3+2\!\left(\frac{1}{y^2}\right)-7\!\left(\frac{1}{y^2}\right)^2+2\!\left(\frac{1}{y^2}\right)^3 &= 0 \end{aligned}$$ $\therefore x=\dfrac{1}{y^2}$, Kun Ye is correct.