8 Math Factor Is At Ion

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., rearrangement) 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 numerical 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 division either by dividing each term of the polynomial by the monomial or by the common factor method. 10. In the case of division of a polynomial by a polynomial, we cannot proceed by dividing 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 have Dividend = Divisor × Quotient. In general, however, the relation is Dividend = Divisor × Quotient + Remainder

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Finish Line & Beyond EXERCISE 1 1. Find the common factors of the given terms. (i) 12x, 36

12 x = 2 × 2 × 3 × x 36 = 2 × 2 × 3 × 3

Answer:

Here, common factor is 2 × 3 = 6 (ii) 2y, 22xy

2y = 2× y 22 y = 2 × 11 × y

Answer:

Common factor = 2 (iii) 14 pq, 28pq

14 pq = 2 × 7 × p × q 28 p ²q² = 2 × 2 × 7 × p × p × q × q Common Factor = 2 × 7 × p × q = 14 pq Answer:

(iv) 2x, 3x², 4 Answer: 2 x = 2 × x

3x² = 3 × x × x 4 = 2× 2

There is no common factor other than unity, so common factor= 1 (v) 6 abc, 24ab², 12 a²b Answer: 6abc = 2 × 3 × a × b × c

24ab ² = 2 × 2 × 2 × 3 × a × b × b 12a ²b = 2 × 2 × 3 × a × a × b Common factor = 2 × 3 × a × b = 6ab (vi) 16 x³, – 4x², 32x

16 x ³ = 2 × 2 × 2 × 2 × x × x × x − 4 x ² = -2 × 2 × x × x 32 x = 2 × 2 × 2 × 2 × 2 × x

Answer:

Common factor = 2x

(vii) 10 pq, 20qr, 30rp

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Finish Line & Beyond 10 pq = 2 × 5 × p × q 20qr = 2 × 2 × 5 × q × r 30rp = 2 × 3 × 5 × p × r Common factor = 2 × 5 = 10 Answer:

(viii) 3x² y³, 10x³ y²,6 x² y²z

3x² y ³ = 3 × x × x × y × y × y 10 x ³ y ³ = 2 × 5 × x × x × x × y × y × y 6 x² y ² z = 2 × 3 × x × x × y × y × z Common factor= x ² y ² Answer:

2. Factorise the following expressions. (i) 7x – 42

7 x − 42 = 7× x − 7× 6 = 7( x − 6)

Answer:

(ii) 6p – 12q

6 p − 12q = 6 × p − 6 × 2q = 6( p − 2q )

Answer:

(iii) 7a² + 14a

7 a ² + 14a = 7× a× a + 7× 2× a = 7a (a + 2)

Answer:

(iv) – 16 z + 20 z³ Answer: − 16 z + 20 z ³

= − 4× 4× z + 4× 5× z × z × z = 4 z (− 4 + 5 z × z ) = 4 z (5 z ² − 4)

(v) 20 l² m + 30 a l m

20l ² m + 30alm = 2× 2× 5× l × l × m + 2× 3× 5× a × l × m = 2lm(10l + 15a )

Answer:

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Finish Line & Beyond (vi) 5 x² y – 15 xy² Answer: 5 x ² y − 15 xy ²

= 5× x × x × y − 5× 3× x × y × y = 5 x ( xy − 3 y ² ) (vii) 10 a² – 15 b² + 20 c²

10a ² − 15b ² + 20c ² = 2 × 5× a × a − 3× 5× b × b + 4 × 5× c × c = 5(2a ² − 3b ² + 4c ²)

Answer:

(viii) – 4 a² + 4 ab – 4 ca Answer: − 4a ² + 4ab − 4ca

= − 4× a× a + 4× a× b − 4× a× c = − 4a ( a − b + c )

(ix) x² y z + x y²z + x y z²

x ² yz + xy ² z + xyz ² = x× x× y× z + x× y× y× z + x× y× z× z = xyz ( x + y + z )

Answer:

(x) a x² y + b x y² + c x y z²

ax ² yz + bxy ² z + cxyz ² = a× x× x× y× z + b× x× y× y× z + c× x× y× z× z = xyz (ax + by + cz )

Answer:

3. Factorise. (i) x² + x y + 8x + 8y

x ² + xy + 8 x + 8 y = x × x + x × y + 8× x + 8× y = x( x + y ) + 8( x + y ) = ( x + 8)( x + y )

Answer:

(ii) 15 xy – 6x + 5y – 2 Answer: 15 xy −

6x + 5 y − 2 = 3× 5× x × y − 2 × 3× x + 5× y − 2 = 3 x(5 y − 2) + 1(5 y − 2)

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Finish Line & Beyond = (3 x + 1)(5 y − 2) (iii) ax + bx – ay – by

ax + bx − ay − by = x ( a + b) − y ( a + b) = ( x − y )(a + b)

Answer:

(iv) 15 pq + 15 + 9q + 25p

15 pq + 15 + 9q + 25 p 5 × 3 × pq + 5 × 3 + 3 × 3q + 5 × 5 p 5 × 3 pq + 3 × 3q + 5 × 5 p + 5 × 3 3q (5 p + 3) + 5(5 p + 3) (3q + 5)(5 p + 3)

Answer:

= = = =

(v) z – 7 + 7 x y – x y z

z − 7 + 7 xy − xyz = z − xyz − 7 + 7 xy = z (1 − xy ) − 7(1 − xy ) = ( z − 7)(1 − xy )

Answer:

EXERCISE 2 1. Factorise the following expressions. (i) a² + 8a + 16 Answer: a ² + 8a + 16

= = = = =

a× a + 2× 4× a + 4× 4 a × a + 4a + 4a + 4 × 4 a(a + 4) + 4(a + 4) (a + 4)(a + 4) (a + 4)²

(ii) p² – 10 p + 25

p ² − 10 p + 25 p × p − 5p − 5p + 5× 5 p( p − 5) − 5( p − 5) ( p − 5)( p − 5) ( p − 5) ²

Answer:

= = = =

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Finish Line & Beyond (iii) 25m² + 30m + 9

25m² + 30m + 9 5× 5× m × m + 5× 3× m + 5× 3× m + 3× 3 5m(5m + 3) + 3(5m + 3) (5m + 3)(5m + 3) (5m + 3)²

Answer:

= = = =

(iv) 49y² + 84yz + 36z²

49 y ² + 84 yz + 36 z ² 7 × 7 × y × y + 42 yz + 42 yz + 6 × 6 × z × z 7 y (7 y + 6 z ) + 6 z (7 y + 6 z ) (7 y + 6 z )(7 y + 6 z ) (7 y + 6 z ) ²

Answer:

= = = =

(v) 4x² – 8x + 4 Answer: 4 x ² −

8x + 4 = 4 x² − 4 x − 4 x + 4 = 4 x( x − 1) − 4( x − 1) = (4 x − 4)( x − 1) (vi) 121b² – 88bc + 16c² Answer: 121b² − 88bc + 16c ²

= = = =

11 × 11 × b × b − 44bc − 44bc + 4 × 4 × c × c 11b(11b − 4c) − 4c(11b − 4c) (11b − 4c)(11b − 4c) (11b − 4c)²

(vii) (l + m) ² – 4lm Answer: (l +

m) ² − 4lm = l ² + m² + 2lm − 4lm = l ² + m² − 2lm

this can be facotrised using (a-b) ² Hence,

= l ² + m² − 2lm = (l − m)²

(viii) a4 + 2a²b² + b4

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Finish Line & Beyond Answer: This can be solved using (a+b) ²=a²+2ab+b² Hence, a4 + 2a²b² + b4 =(a²+b²)² 2. Factorise. (i) 4p² – 9q²

a ² - b² = (a + b)(a - b) Hence, 4 p ² − 9q ² = (2 p + 3q )(2 p − 3q) Answer: As you know:

(ii) 63a² – 112b²

63a ² − 112b ² = 7(9a ² − 16b ²) Using, a ² - b² = (a + b)(a - b) we get 9a ² − 16b² = (3a + 4b)(3a − 4b) Hence, 7(9a ² − 16b ² ) = 7(3a + 4b)(3a − 4b) Answer:

(iii) 49x² – 36 Answer:

49 x ² − 36 = (7 x + 6)(7 x − 6)

(iv) 16x – 144x³ Answer:16x5-144x3 = x³(16x²-144) = x³(4x+12)(4x-12) (v) (l + m) ² – (l – m) ²

(l + m)² − (l − m)² = (l + m + l − m)(l + m − l + m) = 2l × 2m = 4lm

Answer:

(vi) 9x² y² – 16

9 x ² y ² − 16 = (3 xy + 4)(3xy − 4)

Answer:

(vii) (x² – 2xy + y²) – z² Answer:The part in the bracket can be factorized using following identity:

(a − b)² = a ² − 2ab + b² Hence, x ² − 2 xy + y ² = ( x − y ) ² Now, ( x − y )² − z ² can be solved using a ² − b ² = (a + b)(a − b) Hence, ( x − y )² − z ² = ( x − y + z )( x − y − z )

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Finish Line & Beyond (viii) 25a² – 4b² + 28bc – 49c² Answer: (5a) ² −

(2b)² + 2 × 2b × 7c − (7c)² = (5a )² − {( 2b)² − 2 × 2b × 7c + (7c) ²} = (5a )² − (2b − 7c)² Using, a ² − b ² = (a + b)(a − b) the equation can be written as follows: (5a + 2b − 7c)(5a − 2b + 7c) 3. Factorise the expressions. (i) ax² + bx

ax ² + bx = x( ax + b)

Answer:

(ii) 7p² + 21q²

7 p ² + 21q ² = 7( p ² + 3q ² )

Answer:

(iii) 2x³ + 2xy² + 2xz² Answer: 2x³ + 2xy² + 2xz² =2x(x²+y²+z²) (iv) am² + bm² + bn² + an² Answer: am ² + an ² + bm ² + bn²

= a ( m ² + n ² ) + b( m ² + n ² ) = (a + b)(m² + n ² )

(v) (lm + l) + m + 1

(lm + l ) + m + 1 = l (m + 1) + m + 1 = l (m + 1) + 1( m + 1) = (l + 1)(m + 1)

Answer:

(vi) y (y + z) + 9 (y + z)

y ( y + z ) + 9( y + z ) = ( y + 9)( y + z )

Answer:

(vii) 5y² – 20y – 8z + 2yz

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Finish Line & Beyond Answer: 5 y ( y −

4) − 2 z (4 − y ) = 5 y ( y − 4) + 2 z ( y − 4) = (5 y + 2 z )( y − 4)

(viii) 10ab + 4a + 5b + 2

10ab + 4a + 5b + 2 = 2a (5b + 2) + 1(5b + 2) = (2a + 1)(5b + 2)

Answer:

(ix) 6xy – 4y + 6 – 9x Answer: 6 xy − 4 y + 6 −

9x

= 2 y (3 x − 2) + 3(2 − 3 x) = 2 y (3 x − 2) − 3(3 x − 2) = (2 y − 3)(3 x − 2) 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 (a − b)² = a ² − 2ab + b² Hence, the given equation can be written as follows: (a²-b²)² 5. Factorise the following expressions.

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Finish Line & Beyond (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 =

1 x 2

(ii) –36y³ ÷ 9y² Answer: -4y (iii) 66pq²r ÷ 11qr² Answer: 6pqr (iv) 34xyz ÷ 51xy²z Answer:

2 x²y 3

(v) 12a8b8 ÷ (– 6a6b4) Answer: -2a²b4 2. Divide the given polynomial by the given monomial. (i) (5x² – 6x) ÷ 3x Answer:

5 x-6 3

(ii) (3y8 – 4y6 + 5y4) ÷ y4

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Finish Line & Beyond Answer: 3y4-4y²+5 (iii) 8(xy²z² + x²yz² + x²y²z) ÷ 4x²y²z² Answer: 2(x+y+z) (iv) (x + 2x² + 3x) ÷ 2x Answer:

1 3 x²+2x+ 2 2

(v) (pq6 – p6q) ÷ pq 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:

1 xy (3) = xy 3

(v) 96abc(3a – 12) (5b – 30) ÷ 144(a – 4) (b – 6) Answer:

2 abc(3)(5) = 10abc 3

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