# Atge 8

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ALGEBRA ALGEBRA E - - *** be = x (x - 3) + 8 (x - 3) = (x-3) (x + 8) Example : Factorise : x - 4x-21 Sol. : We try to split -4 into two parts whose sum is -4 and product is -21. BEE Clearly, (-7) + 3 = -4 and (-7) x 3 =-21 :: x2 - 4x-21 = x – 7x+3x-21 = x (x - 7) + 3 (x-7) (x-7)(x+3) Example : Factorise : 6x2 + 7x-3 Sol. : Here, 6 X-3 =-18 So, we try to split 7 into two parts whose sum is 7 and product is -18. Clearly, 9 + (-2) = 7 and 9 X (-2) =-18 .:. 6x2 + 7x-3= 6x + 9x-2x-3 = 3x (2x + 3) - (2x + 3) = (3x-1) + (2x + 3) Example : Factorise : 2x2 - 7x - 39. Sol. : Here, 2 x (-39) = -78 So, we try to split -7 into two parts whose sum is -7 and product is -78. Clearly, (-13) + 6 =-7 and (-13) x6 =-78 .. 2x - 7x-39 = 2x2 – 13x + 6x-398 = x (2x - 13) + 3 (2x - 13) = (x + 3) + (2x-13) SW Example : Factorise : 9x2 – 22x+8. Method 4: Factorisation of forms X - Y and x + yº: Remember these formulae : x - y = (x - y) (x2 + xy + y4) x + y = (x + y) (x2 - xy + y^) a Example : Factorise : x - 27y3 Sol. : ** - 27y3 = (x)3 - (3y)3 = (x - 3y) {(x12 + xx 3y + (3y)2} = (x - 3y). (x2 + 3xy + 9y?) Example : 8x3+27 Sol. : 8x3 + 27 = (2x)3 + (3)3 = (2x + 3) {(2x)2 – 2xx 3 + (3)2} = (2x + 3) (4x2 - 6x + 9) Method 5 : Factorisation of x + y + 2-3xyz Theorem : prove that ** + y + 72 – 3 xyz = (x + y + 2) (x2 + y2 + z2 - xy - yz- zx) Proof : x + y + 2 - 3xyz = (x + y ) + 2-3xyz = [(x + 3xy (x + y)] + 2 – 3xyz - W - 3xYu + 7 - 3xyz, JEI (x + y) = u (u + Z) - 3xy (u + 2) - (u + z) (u? – uz + z2) - 3xy (u + 2) - (u + 2) (u? + z2 - uz- 3xy) = (x + y + 2) [(x + y)2 + 2 - (x +y) 2 - 3xy) = (x + y + 2) (x2 + y2 +- xy - y2 - 2x) Note : x2 + y2 + 22 - xy - yz - 2x + Z- = 1 ...
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