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Heaviside’s Cover-up Method Example 1 (Only for linear terms in the denominator) 8x 2 3 x 21 8 x 2 3 x 21 x3 7 x 6 x 2 x 3 x 1 Now, we set this factored function equal to a sum of smaller fractions, each of which has one of the factored terms for a denominator. 8x 2 3x 21 A B C 3 x 7x 6 x 2 x 3 x 1 To find A (Think: Multiply both sides of the equation by (x+2) and then let x = -2, all of the terms on the right-hand side will go to zero except for A and the (x+2) will cancel on the left8x 2 3x 21 A B C side. x 2 x 2 x 2 x 2 x 3 x 1 x2 x 2 x 3 x 1 Do: 8 x 2 3x 21 x 2 x 3 x 1 x2 Calculate: B 8 x 2 3x 21 x 2 x 3 x 1 x3 Calculate C 8 x 2 3x 21 x 2 x 3 x 1 x1 8 2 3 2 21 2 2 3 2 1 1