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Finish Line & Beyond
FACTORISATION
1. When we factorise an expression, we write it as a product of factors. These factors may be numbers, algebraic variables or algebraic expressions.2. An irreducible factor is a factor which cannot be expressed further as a product of factors.3. A systematic way of factorising an expression is the common factor method. It consists of three steps: (i) Write each term of the expression as a product of irreducible factors (ii) Look for and separate the common factors and (iii) Combine the remaining factors in each term in accordance with the distributive law.4. Sometimes, all the terms in a given expression do not have a common factor; but the terms can be grouped in such a way that all the terms in each group have a common factor. When we do this, there emerges a common factor across all the groups leading to the required factorisation of the expression. This is the method of regrouping.5. In factorisation by regrouping, we should remember that any regrouping (i.e., re-arrangement) of the terms in the given expression may not lead to factorisation. We must observe the expression and come out with the desired regrouping by trial and error.6. A number of expressions to be factorised are of the form or can be put into the form : a + 2ab + b,a – 2ab + b, a – b and x + (a + b) + ab. These expressions can be easily factorised using following identities:a + 2ab + b = (a + b) a – 2ab + b = (a – b) a – b = (a + b) (a – b)x + (a + b) x + ab = (x + a) (x + b)7. In expressions which have factors of the type (x + a) (x + b), remember the nu-merical term gives ab. Its factors, a and b, should be so chosen that their sum, with signs taken care of, is the coefficient of x.8. We know that in the case of numbers, division is the inverse of multiplication. This idea is applicable also to the division of algebraic expressions.9. In the case of division of a polynomial by a monomial, we may carry out the divi-sion either by dividing each term of the polynomial by the monomial or by the com-mon factor method.10. In the case of division of a polynomial by a polynomial, we cannot proceed by di-viding each term in the dividend polynomial by the divisor polynomial. Instead, we factorise both the polynomials and cancel their common factors.11. In the case of divisions of algebraic expressions that we studied in this chapter, we haveDividend = Divisor × Quotient.In general, however, the relation isDividend = Divisor × Quotient + Remainder
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EXERCISE 1
1. Find the common factors of the given terms.(i) 12x, 36
Answer: xx ×××= 32212332236 ×××=
Here, common factor is 632 =×
(ii) 2y, 22xy
Answer: yy ×= 22yy ××= 11222
Common factor = 2
(iii) 14 pq, 28pq
Answer: qppq ×××= 7214qp722²q²28 ××××××= qpp
Common Factor = pqqp 1472 =×××
(iv) 2x, 3x², 4
Answer: xx ×= 22xx3²3 ××=x
224 ×=
There is no common factor other than unity, so common factor= 1(v) 6 abc, 24ab², 12 a²b
Answer: cbaabc ××××= 326bba3222²24 ××××××=ab
baa322²b12 ×××××=aCommon factor = ba ××× 32 ab6=
(vi) 16 x³, – 4x², 32x
Answer: xxx2222³16 ××××××=xxx2-2²4 ×××=− x
xx ×××××= 2222232Common factor = 2x
(vii) 10 pq, 20qr, 30rp
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Answer: qppq ×××= 5210rqqr ××××= 52220rprp ××××= 53230
Common factor = 1052 =×
(viii) 3x² y³, 10x³ y²,6 x² y²z
Answer: yyyxxyx ×××××= 3³²3yyyxxxyx ×××××××= 52³³10
zyyxxzyx ××××××= 32²²6Common factor= ²² yx
2. Factorise the following expressions.(i) 7x – 42
Answer: 427 −x677 ×−×= x
)6(7 −= x
(ii) 6p – 12q
Answer: qp 126 −qp 266 ×−×=
)2(6 qp −=
(iii) 7a² + 14a
Answer: aa 14²7 +aaa ××+××= 277
)2(7 += aa
(iv) – 16 z + 20 z³
Answer: ³2016 zz +−zzzz ××××+××−= 5444
)54(4 zzz ×+−=)4²5(4 −= zz
(v) 20 l² m + 30 a l m
Answer: almml 30²20 +mlamll ×××××+×××××= 532522
)1510(2 allm +=
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(vi) 5 x² y – 15 xy² Answer: ²15²5 xyyx −
yyxyxx ××××−×××= 355)²3(5 yxyx −=
(vii) 10 a² – 15 b² + 20 c²
Answer: ²20²15²10 cba +−ccbbaa ×××+×××−×××= 545352
)²4²3²2(5 cba +−=
(viii) – 4 a² + 4 ab – 4 ca
Answer: caaba 44²4 −+−cabaaa ××−××+××−= 444
)(4 cbaa +−−=
(ix) x² y z + x y²z + x y z²
Answer: ²²² xyzzxyyzx ++zzyxzyyxzyxx ×××+×××+×××=
)( zyxxyz ++=
(x) a x² y + b x y² + c x y z²
Answer: ²²² cxyzzbxyyzax ++zzyxczyyxbzyxxa ××××+××××+××××=
)( czbyaxxyz ++=
3. Factorise.(i) x² + x y + 8x + 8y
Answer: yxxyx 88² +++yxyxxx ×+×+×+×= 88
)(8)( yxyxx +++=))(8( yxx ++=
(ii) 15 xy – 6x + 5y – 2
Answer: 25615 −+− yxxy253253 −×+××−×××= yxyx
)25(1)25(3 −+−= yyx
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)25)(13( −+= yx
(iii) ax + bx – ay – by
Answer: byaybxax −−+)()( baybax +−+=
))(( bayx +−=
(iv) 15 pq + 15 + 9q + 25p
Answer: pqpq 2591515 +++pqpq 55333535 ×+×+×+××=35553335 ×+×+×+×= pqpq
)35(5)35(3 +++= ppq)35)(53( ++= pq
(v) z – 7 + 7 x y – x y z
Answer: xyzxyz −+− 77xyxyzz 77 +−−=
)1(7)1( xyxyz −−−=)1)(7( xyz −−=
EXERCISE 2
1. Factorise the following expressions.(i) a² + 8a + 16
Answer: 168² ++ aa4442 ×+××+×= aaa4444 ×+++×= aaaa
)4(4)4( +++= aaa)4)(4( ++= aa
²)4( += a
(ii) p² – 10 p + 25
Answer: 2510² +− pp5555 ×+−−×= pppp
)5(5)5( −−−= ppp)5)(5( −−= pp
²)5( −= p
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(iii) 25m² + 30m + 9
Answer: 930²25 ++ mm33353555 ×+××+××+×××= mmmm
)35(3)35(5 +++= mmm)35)(35( ++= mm
²)35( += m
(iv) 49y² + 84yz + 36z²
Answer: ²3684²49 zyzy ++zzyzyzyy ×××+++×××= 66424277
)67(6)67(7 zyzzyy +++=)67)(67( zyzy ++=
²)67( zy +=
(v) 4x² – 8x + 4
Answer: 48²4 +− xx444²4 +−−= xxx)1(4)1(4 −−−= xxx
= )1)(44( −− xx
(vi) 121b² – 88bc + 16c²
Answer: ²1688²121 cbcb +−ccbcbcbb ×××+−−×××= 4444441111
)411(4)411(11 cbccbb −−−=)411)(411( cbcb −−=
²)411( cb −=
(vii) (l + m) ² – 4lm
Answer: lmml 4²)( −+lmlmml 42²² −++=
lmml 2²² −+=this can be facotrised using (a-b) ²Hence,
lmml 2²² −+=²)( ml −=
(viii) a4 + 2a²b² + b4
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Answer: This can be solved using (a+b) ²=a²+2ab+b²Hence, a4 + 2a²b² + b4
=(a²+b²)²
2. Factorise.(i) 4p² – 9q²
Answer: As you know: b)-b)(a(ab²-² +=aHence, )32)(32(²9²4 qpqpqp −+=−
(ii) 63a² – 112b²
Answer: )²16²9(7²112²63 baba −=−Using, b)-b)(a(ab²-² +=a we get
)43)(43(²16²9 bababa −+=−Hence, )43)(43(7)²16²9(7 bababa −+=−(iii) 49x² – 36
Answer: )67)(67(36²49 −+=− xxx
(iv) 16x – 144x³
Answer:16x5-144x3
= x³(16x²-144)= x³(4x+12)(4x-12)
(v) (l + m) ² – (l – m) ²
Answer: ²)(²)( mlml −−+))(( mlmlmlml +−+−++=
lmml 422 =×=
(vi) 9x² y² – 16
Answer: 16²²9 −yx)43)(43( −+= xyxy
(vii) (x² – 2xy + y²) – z²
Answer:The part in the bracket can be factorized using following identity:²2²²)( bababa +−=−
Hence, ²)(²2² yxyxyx −=+−Now, ²²)( zyx −− can be solved using ))((²² bababa −+=−Hence, ))((²²)( zyxzyxzyx −−+−=−−
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(viii) 25a² – 4b² + 28bc – 49c²
Answer: ²)7(722²)2(²)5( ccbba −××+−}²)7(722²)2{(²)5( ccbba +××−−=
²)72(²)5( cba −−=Using, ))((²² bababa −+=− the equation can be written as follows:
)725)(725( cbacba +−−+
3. Factorise the expressions.(i) ax² + bx
Answer: bxax +²)( baxx +=
(ii) 7p² + 21q²
Answer: ²21²7 qp +)²3²(7 qp +=
(iii) 2x³ + 2xy² + 2xz²
Answer: 2x³ + 2xy² + 2xz²=2x(x²+y²+z²)
(iv) am² + bm² + bn² + an²
Answer: ²²²² bnbmanam +++)²²()²²( nmbnma +++=
)²²)(( nmba ++=
(v) (lm + l) + m + 1
Answer: 1)( +++ mllm1)1( +++= mml)1(1)1( +++= mml
)1)(1( ++= ml
(vi) y (y + z) + 9 (y + z)
Answer: )(9)( zyzyy +++))(9( zyy ++=
(vii) 5y² – 20y – 8z + 2yz
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Answer: )4(2)4(5 yzyy −−−)4(2)4(5 −+−= yzyy
)4)(25( −+= yzy
(viii) 10ab + 4a + 5b + 2
Answer: 25410 +++ baab)25(1)25(2 +++= bba
)25)(12( ++= ba
(ix) 6xy – 4y + 6 – 9xAnswer: xyxy 9646 −+−
)32(3)23(2 xxy −+−=)23(3)23(2 −−−= xxy
)23)(32( −−= xy
4. Factorise.(i) a4 – b4
Answer: a4-b4 = (a²+b²)(a²-b²)
(ii) p4 – 81
Answer: p4 – 81 =(p²+9)(p²-9)
(iii) x4 – (y + z)4
Answer: x4 – (y + z)4
= (x²+(y+z) ²)(x²-(y+z) ²)= (x²+(y+z) ²)[(x+y+z)(x-y-z)]
(iv) x4 – (x – z)4
Answer: x4 – (x – z)4=(x²-(x-z) ²)(x²+(x-z) ²)=[(x+x-z)(x-x+z)](x²+(x-z) ²]
(v) a4 – 2a²b² + b4
Answer: a4 – 2a²b² + b4
This can be solved using ²2²²)( bababa +−=−Hence, the given equation can be written as follows:(a²-b²)²
5. Factorise the following expressions.
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(i) p² + 6p + 8
Asnwer: p²+6p+8=p(p+6)+8
(ii) q² – 10q + 21
Answer: q²-10q+21=q(q-10)+21
(iii) p² + 6p – 16
Answer: p²+6p-16=p(p+6)-16
EXERCISE 3
1. Carry out the following divisions.(i) 28x4 ÷ 56x
Answer: 28x ÷ 56x
= 21
x
(ii) –36y³ ÷ 9y²
Answer: -4y (iii) 66pq²r ÷ 11qr²
Answer: 6pqr
(iv) 34xyz ÷ 51xy²z
Answer: 32
x²y
(v) 12a8b8 ÷ (– 6a6b4)
Answer: -2a²b4
2. Divide the given polynomial by the given monomial.(i) (5x² – 6x) ÷ 3x
Answer: 35
x-6
(ii) (3y8 – 4y6 + 5y4) ÷ y4
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Answer: 3y4-4y²+5
(iii) 8(xy²z² + x²yz² + x²y²z) ÷ 4x²y²z²
Answer: 2(x+y+z) (iv) (x + 2x² + 3x) ÷ 2x
Answer: 21
x²+2x+23
(v) (pq6 – p6q) ÷ pq
Answer: q-p
3. Work out the following divisions.(i) (10x – 25) ÷ 5
Answer: 2x-5
(ii) (10x – 25) ÷ (2x – 5)
Answer: 5
(iii) 10y(6y + 21) ÷ 5(2y + 7)
Answer: 2y× 3 = 6y
(iv) 9x²y² (3z – 24) ÷ 27xy(z – 8)
Answer: xyxy =)3(31
(v) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4) (b – 6)
Answer: abcabc 10)5)(3(32 =
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