Partial Fractions
Partial Fractions
\(\frac{6x-2}{(x-3)(x+1)}=\frac{A}{x-3}+\frac{B}{x+1}\)
- \(A(x+1)+B(x-3)=6x-2\)
- \(Ax+A+Bx-3B=6x-2\)
- \(Ax+Bx+A-3B=6x-2\)
- \(A+B=6\)
- \(A-3B=-2\)
- \(4B=8\)
- \(B=2\)
- \(A=4\)
- Set \(x=3\) to cancel out one fraction.
- \(4A=16\)
- \(A=4\)
- Set \(x=-1\)
- \(-4B=-8\)
- \(B=2\)
\(\frac{6x^2+5x-2}{x(x-1)(2x+1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{2x+1}\)
- \(6x^2+5x-2=A(x-1)(2x+1)+Bx(2x+1)+Cx(x-1)\)
- Set \(x=1\)
- \(9=3B\)
- \(B=3\)
- Set \(x=0\)
- \(-2=-A\)
- \(A=2\)
- Set \(x=-1\)
- \(6-5-2=4+3+2C\)
- \(-1=7+2C\)
- \(2C=-8\)
- \(C=-4\)
\(\frac{5x+3}{(2x-3)(x+2)}=\frac{A}{2x-3}+\frac{B}{x+2}\)
- \(5x+3=A(x+2)+B(2x-3)\)
- Set \(x=-2\)
- \(-7=-7B\)
- \(B=1\)
- Set \(x=0\)
- \(3=2A-3(1)\)
- \(2A=6\)
- \(A=3\)
Express \(\frac{9x^2+20x-10}{(x+2)(3x-1)}\) in partial fractions.
- \(9x^2+20x-10=A(3x-1)+B(x+2)\)
- Set \(x=-2\)
- Come back to this later
Express \(\frac{4-2x}{(2x+1)(x+1)(x+3)}\)
- \(4-2x=A(x+1)(x+3)+B(2x+1)(x+3)+C(2x+1)(x+1)\)
- Sub in \(x=-\frac{1}{2}\)
- \(4-2\left( -\frac{1}{2} \right)=A\left( -\frac{1}{2} \right)\left( \frac{5}{2} \right)\)
- \(5=\frac{5}{4}A\)
- \(A=4\)
Sub in \(x=-1\)
- \(6=-2B\)
- \(B=-3\)
- Sub in \(x=-3\)
- \(10=10C\)
- \(C=1\)
- Answer: \(\frac{4}{2x+1}-\frac{3}{x+1}+\frac{1}{x+3}\)
Repeated Linear Factors
Split \(\frac{9x^2}{(x-1)^2(2x+1)}\) into partial fractions
- \(9x^2=A(x-1)(2x+1)+B(2x+1)+C(x-1)^2\)
- To find C, we sub in \(x=-\frac{1}{2}\)
- \(\frac{9}{4}=\frac{9}{4}C\)
- \(C=1\)
- To find B, we sub in \(x=1\)
- \(9=B(3)\)
- \(B=3\)
- To find A, we equate coefficients for \(x^2\)
- \(2A+C=9\)
- \(2A+1=9\)
- \(2A=8\)
- \(A=4\)
\(\frac{27x^2+32x+16}{(3x+2)^2(1-x)}\)
- \(27x^2+32x+16=\frac{A}{(3x+2)}+\frac{B}{(3x+2)^2}+\frac{C}{(1-x)}\)
\(\frac{3x+4}{(1+x)(2+x)^2}\)
\(3x+4=\frac{A}{(1+x)}+\frac{B}{(2+x)}+\frac{C}{(2+x)^2}\)
\(3x+4=A(2+x)^2+B(1+x)(2+x)+C(1+x)\)
- To find C we sub in \(x=-2\)
- \(-2=-C\)
- \(C=2\)
- To find A we sub in \(x=-1\)
- \(1=A\)
- \(A=1\)
- To find B we equate coefficients
- Set \(x=0\)
- \(4=4A+2B+C\)
- \(2B=-2\)
- \(B=-1\)
\(3x+4=\frac{1}{(1+x)}-\frac{1}{(2+x)}+\frac{2}{(2+x)^2}\)