6. Factors of expression of the form x2 + px + q or x2 - px + q and x2 + px - q or x2 - px - q. If p and q stand for given positive numbers, 1) To factorise x2 + px + q or x2 - px + q, where the last term q is positive, we search for two numbers whose product is q and whose sum is p. These two numbers, with the same sign as that of the middle term of the expression ( i.e. trinomial), are the second terms of the two factors. 2) To factorise x2 + px - q or x2 - px - q, where the last term q of the trinomials is negative, we hunt for two numbers whose product is q and whose difference is p. These two numbers, with opposite sign are second terms of the two factors, the larger of the two numbers having the sign of the middle term of the trinomial. Example Resolve into factors x2 + 17x + 70 Solution: Since the last term is positive
and the middle term is also positive, Now we hunt for two numbers whose product is 70 and sum is 17; the numbers are 10 and 7 \ x2 + 17x + 70 = ( x + 10 ) ( x + 7 ) Example Factorise m2 - 6 mn + 8 n2 Solution: Since the last term is positive and middle terms is negative, m2 - 6 mn + 8 n2 = (m - ... n ) ( m - .. n ) Now we have to find two numbers such that their product is 8 and sum is 6; the two ;numbers are 4 and 2. \ m2 - 6 mn + 8 n2 = ( m - 4n ) ( m - 2n ) Example Factorise x2 + 9x - 36 Solution: Since the last term is negative and middle terms is positive; x2 + 9x - 36 = ( x + ...) ( x - ....) where the blanks are filled by two numbers such that their product is 36 and difference is 9; the numbers are 12 and 3. \ x2 + 9x - 36 = ( x + 12 ) ( x - 3 ) Example Resolve into factors : x2 - 8x - 105 Solution: Since the last term and the middle term both are negative, x2 - 8x - 105 = ( x - ...) ( x + ... ) ; where the blanks are to be filled by two numbers such that their product is 105 and difference is 8. The two numbers are 15 and 7. \ x2 - 8x - 105 = ( x - 15 ) ( x + 7 ) 
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