Digitized by CjOOQ IC



COMllOlf FACTORS Al^D MULTIPLES. Hg

3. Find the H.C.F. of

21a;« - 4aj« - 15x - 2 and 21 «» - 32a:« - 54a; - 7.
21 OB* - 4i^ - 16* - 2)21x8 - 32aj2 - 64« - 7 (1
21x8- 4x2-16x-2
— 28x« — S9x — 5

The difficulty here cannot be obviated by removing a simple>factor
from the remainder, for — 28 x^ — 39 x — 5 has no simple factor. In
this case, to avoid the inconvenience of fractions we multiply the
expression 21 x8 — 4 x^ ~ 15x — 2 by the simple factor 4 to make its
first term exactly divisible by — 28 x*.
The introdiiction of such a factor can in no way affect the H.C.F.
sought, for 4 is not a factor of either of the given expressions ; and if
we multiply only one of the expressions by 4 we do not introduce a
common factor.
The 8ign8 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 all the terms
of the remainder and multiplying the divisor by 4.
28x2 + 39X + 6)84x8- 16x2- 60x- 8(8x
84x8 + 117x2+ 15x



Multiply by — 4,



Divide by — (



-133x2-


76 X


- 8


4, -4






632x2 + 300X


+ 32(1


632x2 + 741X


+ 96


-63)^


441x


-63




7x


+ 1


7x + 1)28x2 + 39x


+ 5(4x + 6


28x2+ 4aj






35 X


+ 6




86 X


+ 6





.theH.C.F. = 7aj + l.
Digitized by CjOOQ IC



116



COMMON FACTORS AND -MULTIPLES.
