Theory of Intelligence
11TH•Mathematics•Partial Fractions•5.3 Irreducible Quadratic Factors

Case III: Non-Repeated Irreducible Quadratic Factors

Partial fractions with non-repeated irreducible quadratic factors.

Case III: Non-Repeated Irreducible Quadratic Factors

A quadratic factor is irreducible if it cannot be written as a product of linear factors with real coefficients (e.g. x2+x+1x^2 + x + 1, x2+3x^2 + 3).

For each non-repeated irreducible factor ax2+bx+cax^2 + bx + c, include a term

Ax+Bax2+bx+c\dfrac{Ax + B}{ax^2 + bx + c}

Example 6

Resolve 3x−11(x2+1)(x+3)\dfrac{3x-11}{(x^2+1)(x+3)} into partial fractions.

Solution

3x−11(x2+1)(x+3)=Ax+Bx2+1+Cx+3\dfrac{3x-11}{(x^2+1)(x+3)} = \dfrac{Ax+B}{x^2+1} + \dfrac{C}{x+3} 3x−11=(Ax+B)(x+3)+C(x2+1)3x - 11 = (Ax+B)(x+3) + C(x^2+1)

x=−3x = -3 ⇒\Rightarrow C=−2C = -2. Equating coefficients: A=2A = 2, B=−3B = -3.

Hence 2x−3x2+1−2x+3\dfrac{2x-3}{x^2+1} - \dfrac{2}{x+3}.

Example 7

Resolve 4x2+8xx4+2x2+9\dfrac{4x^2+8x}{x^4+2x^2+9} into partial fractions.

Solution

x4+2x2+9=(x2+2x+3)(x2−2x+3)x^4 + 2x^2 + 9 = (x^2+2x+3)(x^2-2x+3), so

4x2+8x(x2+2x+3)(x2−2x+3)=Ax+Bx2+2x+3+Cx+Dx2−2x+3\dfrac{4x^2+8x}{(x^2+2x+3)(x^2-2x+3)} = \dfrac{Ax+B}{x^2+2x+3} + \dfrac{Cx+D}{x^2-2x+3}

Equating coefficients yields A=1A = 1, B=2B = 2, C=−1C = -1, D=−2D = -2.

Hence

x+2x2+2x+3+−x−2x2−2x+3\dfrac{x+2}{x^2+2x+3} + \dfrac{-x-2}{x^2-2x+3}

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