082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

28 Chapter 1 Review of algebraic techniques
Solution We factorize the numerator and denominator of the second fraction:
x 2 +4x+3 (x +1)(x +3)
=
x 2 +10x+21 (x +7)(x +3)
Dividingbothnumeratoranddenominatorby (x +3)resultsin x +1
x +7 .Sothetwogiven
fractions are equivalent.
Dividingbothnumeratoranddenominatorbyx+3isoftenreferredtoas‘cancelling
x+3’.
1.5.3 Expressingafractioninitssimplestform
Consider the numerical fraction 6 . To simplify this we factorize both numerator and
10
denominator and then cancel any common factors.Thus
6
10 = 2 ×3
2 ×5 = 3 5
Thefractions 6
10 and 3 5 haveidenticalvaluesbut 3 5 isinasimplerformthan 6
10 .Itisimportanttostressthatonlyfactorswhicharecommontobothnumeratoranddenominator
can becancelled.
Example1.27 Simplify
(a)
(b)
6x
18x 2
12x 3 y 2
4x 2 yz
Solution (a) Notethat18canbefactorizedto6×3andso6isafactorcommontobothnumerator
and denominator. Alsox 2 isx×x and soxis also a common factor. Cancelling the
common factors,6andx, produces
6x
18x 2 =
6x
(6)(3)(x)(x) = 1 3x
(b) The common factors are4,x 2 andy. Cancelling these factors gives
12x 3 y 2
4x 2 yz = 3xy
z
Example1.28 Simplify (a)
4
6x+4
(b) 6t3 +3t 2 +6t
3t 2 +3t
Solution (a) Factorizingbothnumeratoranddenominatorandcancellingcommonfactorsyields
4
6x+4 = (2)(2)
2(3x +2) = 2
3x+2