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Downloaded from https:// www.studiestoday.com, (iii) Since, product = — 6 and sum = 5., , The product of two numbers is negative and their sum is positive only when the larger, of the two numbers is positive and the smaller is negative., , By trial, we find that the required two numbers are 6 and — 1. (Ans.), , (iv) Since, product = — 6 and sum = — 5. The product of two numbers is negative and their, sum is also negative only when the /arger is negative and the smaller is positive., , :. By trial, we find that the required two numbers are — 6 and 1. (Ans.), , , , TEST YOURSELF, , , , , , Note : The standard forms of a trinomial are :, , (i) 6x? +11x+3 i.e. descending order of the powers of its literal coefficients., (ii) 9+ 11x+ 6x? i.e. ascending order of the powers of its literal coefficients., , To factorise a given trinomial, the following steps should be adopted :, , 1. Find the product of the first and the last terms of the trinomial with their signs. In case, of trinomial 6x2 + 11x + 3, the product of its first and last terms = 6x2 x 3 = 18x?., , 2. Split the middle term of the given trinomial (i.e. + 11x) such that the sum of these two, terms is equal to the middle term and their product is equal to the product obtained in, step 1 (i.e. 18x?), , By trial, we find that the two such terms are + 9x and + 2x., 3. Now by forming the suitable groups, factorise the given trinomial., je. 6x24+11x+3 = 6x?+9x+2x+3, = 3x(2x + 3) + 1(2x + 3), , = (2x + 3)(3x + 1) (Ans.), Example 8 :, Factorise :, (i) x? - 9x + 20 (ii) y? + 5y — 24 (iii) 1 — 3a — 28a?, » Solution :, , (i) Given trinomial = x? — 9x + 20, The product of its first and the last terms = x2 x 20 = 20x?, , Splitting the middle term (i.e. —9x) into two terms so that their product is 20x? and sum, is -9x; we get : -5x and —4x., , x? —9x +20 = x?-5x— 4x + 20, = x(x — 5) -4(x — 5), = (x — 5)(x - 4) (Ans.), , cases tom niet”,, loaded from http:, , s:// www.studiestoday.com :