1969-01-28 — Page 25

FREE

頁一第張六第日一十月二十年申戊歷 WAH KIU YAT PO

WAH KIU YAT PO 報日橋

二期星

日八廿月一年九六九一曆公年八十五國民華中

$8#94

*#$%$#3 #8#%$%$#3 #73 #ÄLISES AIZS #D

[ $#%$#$%$#$%# $%$#T°#£#£#%##$%$#$%$#$#%#£#*%$%#$@$#$%$#$%8715-

1969

[C數中學會考試題預習

數學科

MATHMATICS (13)

(十三) ̇歐陽鐇文

LESSON. 13: FACTORISATION.

REMAINDER THEOREM

EXAMPLE 1: Divide 4x+2x-5x-9 by x-1. What is the

remainder? What is the value of the

dividend when 1 ?

SOLUZION we find the quotient by using the method of

detached coefficients

(4x2+2x2-5x-9)=(x-1)

4x+6x+1 rem.

1-114±2-5-9|4+6+1

(ii) Let F(x)=4x+2x =

when x1

)=4(1)+3(1)=5(1)-9

*+6-5-9

1-9

1-1

+8

(II) FACTORISATION

If twa or more integral expressions give a product) which equals to another given expression, then these integral expressions are called factors of the given expression. And the process of finding factors is called resolution inte factor, or factorisation

., Resolution inte factor is an inverse operation and is differ from the direct operation of multipli cation. Any two or more expression can be multiplied together and their product can be easily found. But in general, an algebraical expression written down at random, has no simple factor. E.g. 3x4 cannot be expressed as the product of two simple expressions (other than 1 and itself).

Again, there is no general method of factorisation: In multiplication, we have a definite process which, always gives us the product. But in order to factorise. a given expression, we have to learn a nimber of special devices. It should be noted that when one factor of an expression is known, the other factor(s) can also be obtained by division.

Expression in which 'eich term has a monorial factor: EXAMPLE 4. Resolve 2a3 −3ab2 +62° into factors SOLUTION

P=28b-Jab tha

=ab(2a-3b+6n, bởi

EXAMPLE 5:Evaluate by using facters

of 1970-3 of 370 SOLUTION: EXP -3 of 1976-3-df 370

of (1670-370) of 1500-1000 ||

EXAMPLE 12: Factorise X-2x=6x-9' SOLUTION: Let F(x) = x*-2x2-6x-9

As shown, F(-1)=8(3)=0

x-1x-3, and x+3

are factorstof[F(x);

And, the product of the

degree. 4

*.

( x ) = k(x+1)(x−3 ) { x+3)men,

140 +3.

By comparing one of the terms (says, the teria) _____? = k(x+1)(x-3)(x+3)

onstant

An expression e.s.ath-c7th-le-altera-b), is unaltered by changing a →by hccat is called a cyclic expression. w.r.b, a,b, Such an interchange of letters is called a cyclic interchange. This kind of expression. should always be factorised by actor theorem. (2)It is sometimes a good plan to group together,

terms of the same degree..

e.g. x*+2x+4x+2y*+3xy=(x*+3xy+2y2)+(2x+4y).

The following hints may help you to factorise any given algebraic expression:

1. Write down first the common factor in any

2. Make use of the difference(or sun) of two squares

(or cubes), if possible.

3. It may be possible to factorise by grouping,

expression contains four or more terms

thej

If the expression contains brackets, and no common. factor (or standard form) is seen, the brackets

should usually be removed and the expression there-fore simplified.

Make use of the relation, ir possible, that

? 23+b3+c3=3abo if a+b+c=00 se.g. • (x-y)2+(y+z)2 +(z-x)2 =3(x+y) (y-z]{z−x). 6. Use the factor theorem and all the related pro

whenever required.

From which we find that the remainder F(x)=(x-1) is the SAME as F(1) In general F(x) (x=K}}; where F(x) Then

a-ah

ax + bx tax

+ (bran)

(b+ah)+c

(b+2h)-h(bah)

(c+bh+ah?)+d

cbh+ah-hc+bhtah

d+ch+bb

In which, the quotient=ax*(bran)x+(c+bh+abTM)

the remainder= dichibh” kaha

•F(x)= ax + bx2+cx+da

We find F(h) ah+bh2+ch+a=the remainder

Hence, if a polynomial F(x) is divided by (x-h), and Q(x); represent the quotient and reminder rest.

We have F(x)+(x-h)=Q(x) ren. R

or. "F(x)= (x-h) •Q(x) + R; P ispindependent

of x.(a constant

In

RF(h). This is

which is an identity (1.e. true for all x) particular, put xh, then find called the remainder theorem

EXAMPLE 2: Without actual division, rind the remainder on dividing 3x1-5x2-6x+1 by (a) x+3; (b)3x-1

SOLUTION:

Let F(x)=3x2 -5x2-6x+1.

(a) Divide F(X) by x+3

By Remainder theorem, then

Remandera F(-3).

)3(-3)-5(-3))-6(-3)41

-81 −45 +18, +1 -107·

ByExpressions in which the terns can be

groups, which have a common factor,

EXAMPLE 6: Factorise (a) x(y-3)+6(3-y)

(b) ab÷ac-bd4cd

SOLUTION: (a) EXP. • x(y-3)+6(3-3)

=x-3)=6(y-3) =(x-2)(x-6)-

(b) EXP-a(b+c)−d(b−c)

Impassible to resolve this expression into factors

Note: (1) In (b), b-c¥ -(b+c).

(ii) As a general rule, it is better to select,.

whenever possible, a letter which occurs only to the first degree.

(Expressions in quadratic form:ax+bx+c

Consider the indentity (hp)(k+q) = ax + bx+c Then, from which, we find ashk

b=hotko

Hence, in case of factorising ax +bx+o, it is necessary

to replace by by two equivalent terms hax and kox such

that (1) hqtkpb

(11) (hk) (on)=ac.

It is advisible that you should systematically and completely do your working especially in this tonic

EXAMPLE 7: FactoriseTM 63+2y-y*,

·SOLUTION :

-(y-2-63) -(x2-95 +7 +63) = -Ly(y-9 )+7(y-9)}

-(y+7)(y-9)

Note: Some quadratic expressions can often be factorises.

by inspection. Yet, the answers should always be

checked mentally by multiplication, whichever whata method is used. The above example will be simple by using the cross product, method.e.g.

HINTS "&" "ANS. TO

216 of lead 20

"EMECKTIM}

and we're cu ft

st. the volume of the lead hipelis "20

cu.ft.

rne@cross-¿sectional

the length of the lead

(f)

Let time taken

mauber the distance

A to B is a hr

{total time? taken for double journeysis (a

Aviaspeed =

(a)=201 (b) When = 30-

When A = 401

2870)

55=12645

ANS:

IF

(b) Divide F(x) by 3x-1

By Remainder theorem ther

3(+3)

the divisor is ax'+b, then R=F(~~ (** x-h wax+b=a £x-(- 5-1)] ·. h=-}

It is perferable to find the remainder by the following division methad

e.g. the remainder on dividing 3x −5x −6x

by x-3.

355-16+1

-9+42-108 13-14+36-107-

This method is called Sythetic-division; "In which the third raw was obtained by adding the first twe rews together [e.g. (-5)+(-9)-14] Which is also method of division.

EXAMPLE&Factoris

SOLUTION: EXP.

(7x2+16xy-15y? y(7x-5y)(x+3y)*

rence or two squares? A.

EXAMPLE 9: Factorise 4(3x-2y)-(3x+2y). SOLUTION: EXP. =[2(3x-2y)} -(3x+2y)2;

=(2(3x-2y)+(3x+2y)]{2(3x-2y)=(3x+2y),

!= (9x+2y)(3x-by);" =3(9x-2y)(x+2y)

E)In the form of Perfect Square: Aa±2AB+B2=(A+B)►

EXAMPLE 10: Factorise 4x2+9y*~252*-12x3 SOLUTION: EXP = {4x2+9y*-12xy)-252

=(2x-3y)*+(5z)". =(2x-3x+52)(2x-33-5%);

⇒the sun or difference of two cubes

A**B3 = (A±8) ( K2 = AB¶Ð ́)

EXAMPLE 11: Factorise

'SOLUTION: EXP. =(x2}~~(y2)3

=(x* -y* ) ( x + x =(x+5)(x−5 ) {{(x2+y 2)* -(xy)"]

This expression can also be factoris-d by:,

EXP. = (x3)* - ( y3 y*.

b*(x− a)}=(p=q}(x+a}}

[b7=10p+8}}]}^}=\a (p = q ) + ab".

When p=13x10

8-109

x = a[b2 + (p-8)

EXXERCISE 13

3) If}F(k) = (3k+4}{2k-5),5find[F(-3); [F(X-1) (F(2t=3)..

2) If G(x,y)=x-ying &(2,-2);}G{X+1,742}

G(-28,-38).

3) Find the remainder or the first degree in x obtained]

on dividing 3x++7x+15x+2x-15.

Determine a and b in order that.

-3x2+x2+ax” 45ixth

3x

may be exactly divisible both by x-1 and x-2)

5) Factorise completely the following, if possible

-2xy - 41'y" - 6xy

- 24 (512 - 37)

veterains a and b In order that

ax-2x+bx-6x-9

Bay be exactly divisible by x+1 and x-3]

SOLUTION: Let F(xax-2x2+bx2-6x-9

(1) Since F(x) is divisible by xil

RF(-1)=a(−1)"−2(-1)*® +b(−1)-6(−1)-9

a+b-10%

• +0-1=04-

(11) Since F(x) is divisible

R=F(3)= a(3)′′-213)*+b(3)-6(3)-9

781a+9b-81=0

19a+b-9=0K

Eq. (1) Eq. (2): -8a +8=0

bst.jaelfinte Eq(1);;1+b=1=0

eFor the expression

in general,

(a) if any positive integer then x-ye (x−3 ) {x""'+x^ (b) if n=even positive integer

then, "x-y^= (x+y) (X^^

(c) if neodd positive integer

then,¡x"43" = {x+y)(x**

KEY the factor theorem.

We have proved in "Remainder Theorem" that if the rational integer function f(x) is divided by x-n, then] the remainder RF(h). In particular, f = F(h)=0"

then F(X) is exactly divisible brix-h Iniother words x-h is a factor of F(x)'.--

In more generallet F(x) be any rational integer, function, and, if}}(-2)=0. Then kx-h is a factor 01]F(x);\thialis therefore called the "Factor Theorem"

72+ 27%

·a* - 144* - 8a7;

-'}} 2* --- 372.70)

15(c+d) ~19 (d=c)*

Fina the two values of p for which the quadratic expression 10x-21x-10 and 100% +10x+p_have common factor.

7} LIX-Y-2=1+++ =1 then at lesatione of]

x,y,z.is.unity.

Page 25Page 26

真二第張六第

日一十月二十年申戊展翼