How do you write the partial fraction decomposition of the rational expression 3/(x ^(2) + 16x +5)3x2+16x+5?

2 Answers
mason m
Dec 14, 2015

3/(x^2+16x+5)3x2+16x+5

Explanation:

Notice how the denominator cannot be factored (within the real numbers).

This fraction is already as simple as it can get.

George C.
Dec 14, 2015

3/(x^2+16x+5) = 3/(2 sqrt(59)(x+8-sqrt(59)))-3/(2 sqrt(59)(x+8+sqrt(59)))3x2+16x+5=32√59(x+8−√59)−32√59(x+8+√59)

Explanation:

x^2+16x+5 = x^2+16x+64-59x2+16x+5=x2+16x+64−59

=(x+8)^2-(sqrt(59))^2=(x+8)2−(√59)2

=(x+8-sqrt(59))(x+8+sqrt(59))=(x+8−√59)(x+8+√59)

So to find the partial fraction decomposition, solve:

3/((x+8-sqrt(59))(x+8+sqrt(59)))3(x+8−√59)(x+8+√59)

= A/(x+8-sqrt(59)) + B/(x+8+sqrt(59))=Ax+8−√59+Bx+8+√59

= (A(x+8+sqrt(59))+B(x+8-sqrt(59)))/((x+8-sqrt(59))(x+8+sqrt(59)))=A(x+8+√59)+B(x+8−√59)(x+8−√59)(x+8+√59)

=((A+B)x + 8(A+B) + (A-B)sqrt(59))/((x+8-sqrt(59))(x+8+sqrt(59)))=(A+B)x+8(A+B)+(A−B)√59(x+8−√59)(x+8+√59)

Equating the coefficient of xx, we find A+B = 0A+B=0. So B = -AB=−A.

Equating the remaining constant term we find 2 A sqrt(59) = 32A√59=3

Hence A=3/(2 sqrt(59))A=32√59 and:

3/(x^2+16x+5) = 3/(2 sqrt(59)(x+8-sqrt(59)))-3/(2 sqrt(59)(x+8+sqrt(59)))3x2+16x+5=32√59(x+8−√59)−32√59(x+8+√59)

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