A School Algebra

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A School Algebra
George Albert Wentworth
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Must still be contained in 2a; — 5, and therefore the process may be continued with 2 a; — 5 for a divisor.
2a;-5)4:r^- 8a:-5(2:r+l 4:x' — 10x 2x-b 2x-5 . -. TheH. G. F. = 2a:-5.
(2) Find the H. C. F. Of 21:i;^ - 4a;' - 15a; - 2 and 21 x' -32x''~ 54a; - 2lx^ - 4a;' - 15a; - 2) 21 x' - 32a;' - 54a; - 7(1 21a;^- 4a;' -15a; -2 -28a;'-39a;-5. J ^ ^^ The difficulty here cannot be obviated by removing a simple factor from the remainder, for — 28 a;'-^- 39a; — 5 has no simple factor. In this case, the
... expression 21 a;^ — 4 a;^— 15a; — 2 must be multiplied by the simple factor 4 to make its first term exactly divisible by — 28 3;^^.
The introduction of such a factor can in no way affect the H. C. F. Sought, for 4 is not a factor of the remainder.
The signs of all the terms of the remainder may be changed ; for if an expression A is divisible by —F, it is divisible by + F.
The process then is continued by changing the signs of the re- mainder and multiplying the divisor by 4.


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