7.4 Partial fractions  (Page 5/7)

Can any quotient of polynomials be decomposed into at least two partial fractions? If so, explain why, and if not, give an example of such a fraction

Can you explain why a partial fraction decomposition is unique? (Hint: Think about it as a system of equations.)

Can you explain how to verify a partial fraction decomposition graphically?

You are unsure if you correctly decomposed the partial fraction correctly. Explain how you could double-check your answer.

Once you have a system of equations generated by the partial fraction decomposition, can you explain another method to solve it? For example if you had $\frac{7x+13}{3x^2+8x+15}=\frac{A}{x+1}+\frac{B}{3x+5},$ we eventually simplify to $7x+13=A(3x+5)+B(x+1).$ Explain how you could intelligently choose an $x$ -value that will eliminate either $A$ or $B$ and solve for $A$ and $B.$

$\frac{5x+16}{x^2+10x+24}$

$\frac{3x\mathrm{-79}}{x^2\mathrm{-5}x\mathrm{-24}}$

$\frac{-x\mathrm{-24}}{x^2\mathrm{-2}x\mathrm{-24}}$

$\frac{10x+47}{x^2+7x+10}$

$\frac{x}{6x^2+25x+25}$

$\frac{32x\mathrm{-11}}{20x^2\mathrm{-13}x+2}$

$\frac{x+1}{x^2+7x+10}$

$\frac{5x}{x^2\mathrm{-9}}$

$\frac{10x}{x^2\mathrm{-25}}$

$\frac{6x}{x^2\mathrm{-4}}$

$\frac{2x\mathrm{-3}}{x^2\mathrm{-6}x+5}$

$\frac{4x\mathrm{-1}}{x^2-x\mathrm{-6}}$

$\frac{4x+3}{x^2+8x+15}$

$\frac{3x\mathrm{-1}}{x^2\mathrm{-5}x+6}$

$\frac{\mathrm{-5}x\mathrm{-19}}{{\left(x+4\right)}^2}$

$\frac{x}{{\left(x\mathrm{-2}\right)}^2}$

$\frac{7x+14}{{\left(x+3\right)}^2}$

$\frac{\mathrm{-24}x\mathrm{-27}}{{\left(4x+5\right)}^2}$

$\frac{\mathrm{-24}x\mathrm{-27}}{{\left(6x\mathrm{-7}\right)}^2}$

$\frac{5-x}{{\left(x\mathrm{-7}\right)}^2}$

$\frac{5x+14}{2x^2+12x+18}$

$\frac{5x^2+20x+8}{2x{\left(x+1\right)}^2}$

$\frac{4x^2+55x+25}{5x{\left(3x+5\right)}^2}$

$\frac{54x^3+127x^2+80x+16}{2x^2{\left(3x+2\right)}^2}$

$\frac{x^3\mathrm{-5}{x}^2+12x+144}{x^2\left(x^2+12x+36\right)}$

$\frac{4x^2+6x+11}{\left(x+2\right)\left(x^2+x+3\right)}$

$\frac{4x^2+9x+23}{\left(x\mathrm{-1}\right)\left(x^2+6x+11\right)}$

$\frac{\mathrm{-2}{x}^2+10x+4}{\left(x\mathrm{-1}\right)\left(x^2+3x+8\right)}$

$\frac{x^2+3x+1}{\left(x+1\right)\left(x^2+5x\mathrm{-2}\right)}$

$\frac{4x^2+17x\mathrm{-1}}{\left(x+3\right)\left(x^2+6x+1\right)}$

$\frac{4x^2}{\left(x+5\right)\left(x^2+7x\mathrm{-5}\right)}$

$\frac{4x^2+5x+3}{x^3\mathrm{-1}}$

$\frac{\mathrm{-5}{x}^2+18x\mathrm{-4}}{x^3+8}$

$\frac{3x^2\mathrm{-7}x+33}{x^3+27}$

$\frac{x^2+2x+40}{x^3\mathrm{-125}}$

$\frac{4x^2+4x+12}{8x^3\mathrm{-27}}$

$\frac{\mathrm{-50}{x}^2+5x\mathrm{-3}}{125x^3\mathrm{-1}}$

$\frac{\mathrm{-2}{x}^3\mathrm{-30}{x}^2+36x+216}{x^4+216x}$

$\frac{3x^3+2x^2+14x+15}{{\left(x^2+4\right)}^2}$

$\frac{x^3+6x^2+5x+9}{{\left(x^2+1\right)}^2}$

$\frac{x^3-x^2+x\mathrm{-1}}{{\left(x^2\mathrm{-3}\right)}^2}$

$\frac{x^2+5x+5}{{\left(x+2\right)}^2}$

$\frac{x^3+2x^2+4x}{{\left(x^2+2x+9\right)}^2}$

$\frac{x^2+25}{{\left(x^2+3x+25\right)}^2}$

$\frac{2x^3+11x+7x+70}{{\left(2x^2+x+14\right)}^2}$

$\frac{5x+2}{x{\left(x^2+4\right)}^2}$

$\frac{x^4+x^3+8x^2+6x+36}{x{\left(x^2+6\right)}^2}$

$\frac{2x\mathrm{-9}}{{\left(x^2-x\right)}^2}$

$\frac{5x^3\mathrm{-2}x+1}{{\left(x^2+2x\right)}^2}$

$\frac{x^2+4}{{\left(x+1\right)}^3}$

$\frac{x^3\mathrm{-4}{x}^2+5x+4}{{\left(x\mathrm{-2}\right)}^3}$

$\frac{7}{x+8}+\frac{5}{x\mathrm{-2}}-\frac{x\mathrm{-1}}{x^2\mathrm{-6}x\mathrm{-16}}$

$\frac{1}{x\mathrm{-4}}-\frac{3}{x+6}-\frac{2x+7}{x^2+2x\mathrm{-24}}$

$\frac{2x}{x^2\mathrm{-16}}-\frac{1\mathrm{-2}x}{x^2+6x+8}-\frac{x\mathrm{-5}}{x^2\mathrm{-4}x}$