Express (X²-16)/(X-1)(X+3) in partial fractions

(X2-16)/(X-1)(X+3) can be expressed as partial fractions as it is equivalent to A + B/(X-1) + C/(X+3) giving us : (X2-16)/(X-1)(X+3)≡ A + B/(X-1) +C/X+3). By multiplying both sides of this equation by (X-1) and (X+3) you get X2-16≡A(X-1)(X+3) + B(X+3) +C(X-1). This must be true for all values so to work out the variables A, B and C you start off by looking at the values of X which make the value of the bracket 0. These are X=1 and X=-3. When X=1: -15=4B, therefore B=-15/4. When X=-3: -7=-4C, therefore C=7/4. When the brackets are fully expanded the only X2 term is AX2 , therefore AX2=X2, therefore A=1.

SP

Related Further Mathematics A Level answers

All answers ▸

using an integrating factor, find the general solution of the differential equation dy/dx +y(tanx)=tan^3(x)sec(x)


Use de Moivre's theorem to calculate an expression for sin(5x) in terms of sin(x) only.


Can you express 3 + 4j in polar form?


Find the eigenvalues for the matrix (4/2/3,2/7/0,-2/1/8)