Fac­tor­iza­tion

Jamaica Gleaner - - YL: FEATURE - CLE­MENT RAD­CLIFFE Con­trib­u­tor

THIS WEEK we will com­plete the re­view of fac­tor­iza­tion by go­ing through the home­work from last week’s les­son. Please be re­minded that:

Fac­tor­iza­tion con­verts an al­ge­braic ex­pres­sion to the prod­uct of the fac­tors.

It is by study­ing the var­i­ous meth­ods that you will be able to select the ap­pro­pri­ate one.

The CXC math­e­mat­ics re­ports in­di­cate that most dif­fi­culty is ex­pe­ri­enced in the group­ing and qua­dratic fac­tors meth­ods.

More time should be spent pre­par­ing these.

Be guided by this and be sure to ex­press the fac­tors in the sim­plest form, for ex­am­ple, 2(x + 3) in­stead of (2x + 6).

The fol­low­ing are the so­lu­tions: (a) Fac­tor­ize: x2 + 9x + 14 In this case, we find two num­bers, a and b, such that a+ b = 9 and ax b = 14. The num­bers are 7 and 2. x2 + 5x +6 = (x + 7) (x + 2).

(b) Fac­tor­ize: 3x2 - 7x -6 Given 3x2 - 7x -6 and us­ing the method sim­i­lar to that used above, 3x2 - 7x -6 = (3x +2) (x - 3).

Please clear the brack­ets to ver­ify your an­swer as fol­lows, (3x +2) (x - 3) = 3x2 + 2x - 9x - 6 = 3x2 - 7x -6.

(c) Fac­tor­ize: 3x -8y - 4xy + 6 Us­ing the group­ing method: The terms are re-ar­ranged such that both terms have the same com­mon fac­tor. 3x + 6 -8y - 4xy = (3x + 6) - (8y + 4xy) 3(x + 2) - 4y(2 + x) N.B. ( x + 2 ) is com­mon to both. =(x + 2) (3 - 4y) You may ex­pand the an­swer to en­sure that the an­swer is the same as the given ex­pres­sion. In­deed, this is re­quired if time al­lows.

(d) Fac­tor­ize: 25a2 - 9 b2 The dif­fer­ence of two squares method is used. As a re­minder, we must find the square root of 25a2 and 9b2. Since the square roots are 5a and 3b, re­spec­tively, then: 25a2 - 9b2 = (5a - 3b)(5a + 3b)

I am sure that you re­al­ize that (5a + 3b)(5a - 3b) is also cor­rect.

Find­ing square root pro­vides a chal­lenge for some, please prac­tice and check your re­sults.

(e) Fac­tor­ize: 3x2 - 3y2 - 4x + 4y Us­ing the group­ing method: 3x2 - 3y2 - 4x + 4y = (3x2 - 3y2) - (4x - 4y) Fac­tor­iz­ing: = 3(x2 - y2) - 4(x - y) = 3(x - y)(x + y) - 4(x - y) = 3(x +y) (x - y) - 4(x - y) =( x - y) ( 3x + 3y - 4) An­swer : (x - y) (3x + 3y - 4)

(f) Fac­tor­ize: 1 - 64x2 By us­ing the dif­fer­ence of two squares method, the square roots of 1 and 36x2 are 1 and 6x, re­spec­tively, 1 - 36x2 = ( 1 - 6x)( 1 + 6x) = ( 1 - 6x)( 1 + 6x)

On your own, please fac­tor­ize: 1. x2 + 4x - 32 2. 2x2 + 5x - 12

You must en­sure that you are fa­mil­iar with the four meth­ods, com­mon fac­tor, group­ing, qua­dratic fac­tors and dif­fer­ence of two squares demon­strated above, and know when to use each. You can­not af­ford to not earn the marks avail­able from these ques­tions.

Now let us re­view an­other topic, si­mul­ta­ne­ous lin­ear equa­tions. The spe­cific ob­jec­tive was pre­sented pre­vi­ously.

SI­MUL­TA­NE­OUS LIN­EAR EQUA­TIONS

The solution of the si­mul­ta­ne­ous equa­tions is the pair of x and y val­ues which sat­isfy both equa­tions.

If both equa­tions are plot­ted on a graph, it is the point of in­ter­sec­tion of both lines.

You may use the elim­i­na­tion or sub­sti­tu­tion method. Both meth­ods should be stud­ied. You should also know the ap­pro­pri­ate cir­cum­stances to use each.

It would be ap­pro­pri­ate to re­view di­rected num­bers and the sim­pli­fi­ca­tion of al­ge­braic ex­pres­sions at the out­set.

The elim­i­na­tion method is il­lus­trated as fol­lows:

EX­AM­PLE 1

Solve the si­mul­ta­ne­ous equa­tions: 3x + 2y = 13 ....... (1) x - 2y =-1 ........ (2) Add equa­tions (2) and (1) This is done as the co­ef­fi­cients of y are plus and mi­nus 2.

4x = 12 y is elim­i­nated. x = 12 = 3

4 Sub­sti­tut­ing x = 3 into (1) 3x 3 + 2y = 13 9 + 2y = 13 2 y = 13 - 9 = 4 y= 4

2 y = 2 An­swer is x= 3, y = 2

You may check your an­swer by sub­sti­tut­ing the val­ues x = 3 and y = 2 into both equa­tions to

show that they sat­isfy the equa­tions.

Do you re­alise that in ex­am­ple 1, since the co­ef­fi­cients of y are - 2 and 2, re­spec­tively, in both equa­tions you elim­i­nate y by adding? If the co­ef­fi­cients are the same, that is, if the co­ef­fi­cients of y are both 2, then you elim­i­nate by sub­tract­ing.

EX­AM­PLE 2

Solve the si­mul­ta­ne­ous equa­tions: 5x + 3y = 31 ........ (1) 2x +y = 12 ......... (2)

Mul­ti­ply equa­tion (2) by 3 to form equa­tion (3) and then sub­tract equa­tion (1) from equa­tion (3). 6x + 3y = 36 ......... (3) 5x + 3y = 31 ......... (1) Both equa­tions have the same co­ef­fi­cient of y x = 5 Sub­sti­tut­ing x = 5 in (2) 10 +y = 12 ........ (2) y = 12 - 10 = 2. An­swer: x = 5 and y = 2.

The fol­low­ing is an ex­am­ple of the sub­sti­tu­tion method:

EX­AM­PLE 3

Solve the si­mul­ta­ne­ous equa­tions: 5x + 3y = 31 2x +y = 12

5x + 3y = 31 ........... (1) 2x +y = 12 ........... (2) Us­ing the sub­sti­tu­tion method, From equa­tion (2),y = 12 - 2x Sub­sti­tut­ing into (1) 5x + 3(12 - 2x) =31 Clear­ing the brack­ets, en­sure that this is done cor­rectly 5x + 36 - 6x = 31 - x = 31 - 36 -x= -5 or x = 5

Sub­sti­tut­ing x = 5 into equa­tion (2) 10 +y = 12 y = 12 - 10 = 2 An­swer is: x = 5 and y = 2

Let us try an­other ex­am­ple. 3x - 2y =7 ....... (1) -x + 3y =-7 ....... (2) The elim­i­na­tion method is the ap­pro­pri­ate one to be used here.

Mul­ti­ply equa­tion (2) x 3 to form equa­tion (3) -3x + 9y = -21 ...... (3) Add equa­tions (1) and (3) 7y = -14 y = -14 = - 2 7 Sub­sti­tute y= -2 in equa­tion (1) 3x +-2 x -2 = 7 3x + 4 = 7 3x =7 - 4 = 3 x = 1 An­swer x= 1, y = - 2.

AC­TIV­ITY

Please at­tempt to solve the fol­low­ing si­mul­ta­ne­ous equa­tions: (a) 2n +m = 3

5n - 2m = 12

(b) 2x + 3y = 9 3x - y= 8

(c) 3x - 4y = 32 5x + 2y = 10

(d) 3x +y = 2 4x + 3y = 3

(d) 2x + 3y = 3 5x - 2y = 17

(e) 2x = 11 + 3y x + 2y + 12 = 0

(f) 3x + 2y = 1 4x - y = 16

(h) 2x + 3y = 9 3x - y = 8

Con­tinue to prac­tise these and keep your an­swers for future ref­er­ence. Have a good week.

Cle­ment Rad­cliffe is an in­de­pen­dent con­trib­u­tor. Send ques­tions and com­ments to kerry-ann.hep­burn@glean­erjm.com

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