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Writing: 6 1 1 ------------- = ------- - ------- xUsually people use this method in conjunction with another method (e.g. substitution of values), however there is nothing wrong with using this method on its own if you prefer it.^{2}+ 2x - 8 x - 2 x + 4 6 means that you have expressed ----------- in partial fractions... x^{2}+ 2x - 8 There are three main "methods" that allow us to do this: (1) Substitution of strategic values (2) Solving with coefficients (3) Cover up method (can only be used on fractions without powers in the denominator) Method 1 Substitution of strategic values ======================== Problem: Express as a partial fraction x - 1 --------------- (3x - 5)(x - 3) ======================== Solution: First write the fraction as x - 1 A B --------------- = -------- + ------- (3x - 5)(x - 3) (3x - 5) (x - 3) Notice that I have taken the two terms that are in brackets and placed them on their own as seperate fractions using 'A' and 'B' as numerators. A and B are the mystery numbers we need to discover. Next, multiply denominator (the bottom bit of the big fraction) by both sides of the equation to cancel out terms So we now have x - 1 = A(x - 3) + B(3x - 5) In order to get rid of either the 'A' or 'B' term, we can substitute a 'strategic' value to make it equal to zero. For example, to get rid of A(x - 3), let's make x = 3, so (3 - 3) ends up as 0. Gone. Remember that you have to do this to all of the x's in the equation 3 - 1 = A(3 - 3) + B(3*3 - 5) you can see the 'A' component becomes zero. 3 - 1 = B(9 - 5) 2 = 4B so, 2/4 = B or, 1/2 = B We've discovered what B is, now we can repeat the process to discover A Let's make x = 5/3 to get rid of the 'B' part of the equation... so, B(5/3 * 3) = 5 and, B(5 - 5) = 0 The B(3x - 5) term is gone! So we can plug this value into the entire equation to find A 5/3 - 1 = A(5/3 - 3) 2/3 = A * -4/3 2/3 ------ = A -4/3 -1/2 = A We now have A and B, and the answer to our problem. x - 1 -1 1 --------------- = --------- + -------- (3x - 5)(x - 3) 2(3x - 5) 2(x - 3) ######################################################################################### Method 2 Solving by coefficients ========================

======================== Problem: Express as a partial fraction x - 1 --------------- (3x - 5)(x - 3) ======================== Solution: First write the fraction as x - 1 A B --------------- = -------- + ------- (3x - 5)(x - 3) (3x - 5) (x - 3) The first step (just as in the first method) is to multiply the denominator by both sides x - 1 = A(x - 3) + B(3x - 5) It may help if the A and B parts are expanded but this step can usually be missed. x - 1 = A*x - A*3 + B*3*x - 5B x - 1 = Ax - 3A + 3Bx - 5B now look at this and try to equate coefficients for 'x' Ok, we've now done the first step, let's do the same again but with something else (like xThis method is (in my opinion) the easiest of all the three methods, but it can be misleading if you follow it like a cookbook recipe without knowing how it really works. Be wary that it cannot be used with "non-linear" fractions (explained after this section).^{2}or constants (the constants are the actual number's, i.e not x's which can be anything). Coefficient for constants -1 = -3A - 5B we now have a simultanious equation... 1 = A + 3B --- equation (1) -1 = -3A + (-5B) --- equation (2) We can easily solve this First, multiply equation (1) by -3 -3 = -3A + (-9B) --- equation (1a) Next, subtract equation (1a) from equation (2) 2 = 4B 2/4 = B 0.5 = B Then we can plug this value into equation (1) to get A 1 = A + 3 * 0.5 1 - 3 * 0.5 = A -0.5 = A So we now know A and B x - 1 -0.5 0.5 --------------- = --------- + -------- (3x - 5)(x - 3) (3x - 5) (x - 3) Finally, the "0.5's" can be written with "1/2's" instead x - 1 -1 1 --------------- = --------- + -------- (3x - 5)(x - 3) 2(3x - 5) 2(x - 3) ######################################################################################### Method 3 Cover up method ========================

======================== Problem: Express as a partial fraction x - 1 --------------- (3x - 5)(x - 3) ======================== Solution: Delete first term in the denominator 3x - 5 3x - 5 = 0 3x = 5 x = 5/3 Plug 5/3 as 'x' into what is left of the fraction 5/3 - 1 2/3 -1 ------------- = ------------ = --- 5/3 - 3 -4/3 2 We now have the answer for the (3x - 5) expression in the partial fraction. Let's repeat for other expression First, delete second term x - 3 x - 3 = 0 x = 3 Plug in 3 as 'x' for what is left of the fraction 3 - 1 2 1 ----------- = --- = --- 3*3 - 5 4 2 This is the answer for the (x - 3) expression of the partial fraction so, x - 1 -1 1 --------------- = --------- + -------- (3x - 5)(x - 3) 2(3x - 5) 2(x - 3) Done. #########################################################################################Notice that becuase (1 + x)## Strange exception 1 (repeated-linear fractions)

Where a fraction has a repeated linear part, e.g. (1 + x)^{2}the format will look something like numerator A B C ------------------- = ------------ + --------- + --------- expression*(1 + x)^{2}expression (1 + x) (1 + x)^{2}

Example ======================== Problem: Express as a partial fraction 1 --------------- (x - 3)(x + 1)If you have got a quadratic in the denominator i.e. x^{2}======================== Solution: 1 A B C --------------- = ------- + ------- + -------- (x - 3)(x + 1)^{2}(x - 3) (x + 1) (x + 1)^{2}Multiply both sides by the denominator 1 = A(x + 1)^{2}+ B(x - 3)(x + 1) + C(x - 3) Expand out (in this case its only the 'A' term) 1 = A(x + 1)(x + 1) + B(x - 3)(x + 1) + C(x - 3) Find A by making x = 3 (this is an 'substitution of strategic values' part) 1 = A(3 + 1)(3 + 1) 1 = A*16 1/16 = A Lets also get C by making x = -1 1 = C(-1 - 3) 1 = C*-4 -1/4 = C We now know A and C, so can easily get B. Lets make x = 1, so nothing gets cancelled out. In full, 1 = A(1 + 1)(1 + 1) + B(1 - 3)(1 + 1) + C(1 - 3) 1 = A*4 + B*-4 + C*-2 1 = (1/16)*4 + B*-4 + (-1/4)*-2 1 = 1/4 + -4B + 2/4(-1/4)*-2 1 = 3/4 - 4B 1 - 3/4 = -4B 1/4 = -4B -1/16 = B Solved. 1 1 1 1 --------------- = ----------- - ----------- - ------------ (x - 3)(x + 1)^{2}16(x - 3) 16(x + 1) 4(x + 1)^{2}########################################################################################### Strange exception 2 (quadratics in the denominator)

A Bx + C fraction = --------- + ------------ factor quadractic Example ======================== Problem: Express as a partial fraction x - 1 -------------------- (x + 3)(xHope all of that helped!^{2}+ 3x + 2) ======================== Solution: x - 1 A Bx + C -------------------- = --------- + ------------- (x + 3)(x^{2}+ 3x + 2) x + 3 x^{2}+ 3x + 2 First multiply by denominator x - 1 = A(x^{2}+ 3x + 2) + (Bx + C)(x + 3) Using our substitution of strategic values method, we can make x = -3 (at this point be careful that your selected value doesn't also make the quadratic equal zero as well!) -3 - 1 = A((-3)^{2}+ 3*(-3) + 2) -4 = A(9 - 9 + 2) -4 = 2A -2 = A We can use the coefficient method, looking at the coefficients of x^{2}0 = A + B 0 = -2 + B 2 = B Similarly, looking at the coefficients of x 1 = 3A + 3B + C 1 = 3*(-2) + 3*(2) + C 1 = C Done. x - 1 -2 2x + 1 -------------------- = --------- + --------------- (x + 3)(x^{2}+ 3x + 2) (x + 3) x^{2}+ 3x + 2

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