For a proper rational expression with two distinct linear factors, write one constant numerator over each factor. The numerator degree must be less than the denominator degree. px+q(ax+b)(cx+d)= Aax+b+ Bcx+d, where the two factors have different roots. Multiply by the full denominator to obtain an identity. Substitute the roots of the factors, or compare coefficients, to determine A and B. Restrictions on the original denominator remain. Example 5x+1(x-1)(x+2)= Ax-1+ Bx+2 5x+1=A(x+2)+B(x-1). At x=1, 6=3A, so A=2. At x=-2, -9=-3B, so B=3. 5x+1(x-1)(x+2)= 2x-1+ 3x+2, x 1,-2. Check by recombining: 2(x+2)+3(x-1)=5x+1. This form is useful before integrating a rational expression.

For a proper rational expression with two distinct linear factors, write one constant numerator over each factor. The numerator degree must be less than the denominator degree.

\(\dfrac{px+q}{(ax+b)(cx+d)}=\dfrac A{ax+b}+\dfrac B{cx+d}\), where the two factors have different roots.

Multiply by the full denominator to obtain an identity. Substitute the roots of the factors, or compare coefficients, to determine \(A\) and \(B\). Restrictions on the original denominator remain.

\(\begin{aligned}\dfrac{5x+1}{(x-1)(x+2)}&=\dfrac A{x-1}+\dfrac B{x+2}\\5x+1&=A(x+2)+B(x-1).\end{aligned}\)

At \(x=1\), \(6=3A\), so \(A=2\). At \(x=-2\), \(-9=-3B\), so \(B=3\).

\(\dfrac{5x+1}{(x-1)(x+2)}=\dfrac2{x-1}+\dfrac3{x+2},\qquad x\ne1,-2\).

Check by recombining: \(2(x+2)+3(x-1)=5x+1\). This form is useful before integrating a rational expression.

Polynomials, Cubic Equations and Partial Fractions

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Partial fractions

Partial fractions