Class 9 Advance Maths(Elective) Chapter 4 Exercise 4.1

4. SPECIAL PRODUCT AND FACTORIZTION

Exercise 4.1

1. Expand the following:

1. Expand the following- 
(i) (2x + 3y)² 
= (2x)² + 2.2x.3y + (3y)² 
= 4x² + 12xy + 9y² 

(ii) (x - 2y)³ 
= x³ - 3.x².2y + 3.x.(2y)² - (2y)³ 
= x³ - 6x²y + 3x.4y² - 8y³
=  x³ - 6x²y + 12xy² - 8y³

(iii) (x - y + 2z)²
= x² + (-y)² + (2z)² + 2.x.(-y) + 2.(-y).2z + 2.2z.x
= x² + y² + 4z² - 2xy - 4yz + 4zx

(iv) (a + 2b - 3c)²
= a² + (2b)² + (-3c)² + 2.a.2b + 2.2b.(-3c) + 2.(-3c).a
= a² + 4b² + 9c² + 4ab - 12bc - 6ca

(v) (3x - y)² (x + 3y)²
= { (3x - y) (x + 3y) }²
= { 3x (x + 3y) - y (x + 3y) }²    
= ( 3x² + 9xy - xy - 3y²)²
= ( 3x² + 8xy - 3y²)²
= (3x²)² + (8xy)² + (-3y²)² + 2.3x².8xy + 2.8xy.(-3y²) + 2.(-3y²).3x²
= 9x⁴ + 64x2y² + 9y⁴ + 48x³y - 48xy³ + 18x²y²

(vi) (x + y - a - b )²
= {(x + y) - (a + b)}²
= {(x + y)² - 2(x+y)(a+b) + (a+b)²}
= x² + 2xy + y² - 2{x(a+b) + y(a+b)} + a² + 2ab + b²
= x² + 2xy + y² - 2(ax + bx + ay + by) + a² + 2ab + b²
= x² + 2xy + y² - 2ax - 2bx - 2ay - 2by + a² + 2ab + b²

(vii) (- m - n)²
= [-(m + n)]²
= (m + n)²
= m² + 2mn + n²
    
(viii) (- m - n)³
= [-(m + n)]³
= -(m³ + 3m²n + 3mn² + n³)
= -m³ - 3m²n - 3mn² - n³

2. Simplify-

(i) (3a - 2b)² (3a + 2b)²
= {(3a - 2b)(3a + 2b)}²
= {(3a)² - (2b)²}²
= (9a² - 4b²)²
= (9a²)² - 2.9a².4b² + (4b)²
= 81a⁴ - 72a²b² + 16b⁴

(ii) (6x - y)² - (5x + 2y)²
= {(6x)² - 2.6x.y + y²} - {(5x)² + 2.5x.2y + (2y)²}
= 36x² - 12xy + y² - 25x² - 20xy - 4y²
= 11x² - 32 xy - 3y²

(iii)(6x - y)³ - (5x + 2y)³
= (6x)³ - 3.(6x)².y + 3.6x.y² - y³ - {(5x)³ + 3.(5x)².2y + 3.5x.(2y)² + (2y)³}
= 216x³ - 3.36x.y + 18xy² - y³ - 125x³ - 6y.25x² - 15x.4y² - 8y³
= 216x³ - 108xy + 18xy² - y³ - 125x³ - 150x²y - 60xy² - 8y³
= 91x³ - 9y³ - 258x²y - 42xy²

(iv) (x + y)³ + (x - y)³
= x³ + 3x²y + 3xy² + y³ + x³ - 3x²y + 3xy² - y³
= 2x³ + 6xy²

(v) (4 - b)³ + (4 + b)³ + 3(4 - b)² (4+b) + 3(4-b) (4+b)²
Solution:
Let 4 - b = x and 4 + b = y.
Therefore,   x³ + y³ + x²y + xy²
= (x + y)³
= (4 - b + 4 + b)³
= 8³
= 512


(vi) (x + y)(x - y) + (y + z)(y - z) + (z + x)(z - x)
= (x² - y²) + (y² - z²) + (z² - x²)
= x² - y² + y² - z² + z² - x²
= 0

(vii) (x² - 1 + 2x)(x² - 1 - 2x) 
= {(x² - 1) + 2x}{(x² - 1) - 2x}
= (x² - 1)² - (2x)²
= (x²)² - 2.x².1 + 1² - 4x²
= x⁴ - 2x² + 1 - 4x²
= x⁴ - 6x² + 1 


3. 

(i) If a+b+c=13 and ab+bc+ca=50 then find the value of a²+b²+c²=? 
Solution: We know that,
   (a+b+c)²=a²+b²+c²+2ab+2bc+2ca
=> (a+b+c)²=a²+b²+c²+2(ab+bc+ca)
=> 13=a²+b²+c²+2*50
=> 169=a²+b²+c²+100
=> 169-100=a²+b²+c²
=> a²+b²+c²=69

(ii) a+b+c =6, a²+b²+c² =14 then ab+bc+ca=?
Solution: We know that,
   (a+b+c)²=a²+b²+c²+2ab+2bc+2ca
=> (a+b+c)²=a²+b²+c²+2(ab+bc+ca)
=> 6²=14+2(ab+bc+ca)
=> 36-14=2(ab+bc+ca)
=> 22/2=ab+bc+ca
=> ab+bc+ca=11

(iii) a - b + c = 7 and a²+b²+c² =19, then what is the value of ca-ab-bc=?
Solution: We know that,
   {a+(-b)+c}²=a²+(-b)²+c²+2a(-b)+2(-b)c+2ca
=> (a-b+c)²=a²+b²+c²-2ab-2bc+2ca
=> 7²=19+2(ca-ab-bc)
=> 49=19+2(ca-ab-bc)
=> 49-19=2(ca-ab-bc)
=> 30/2=(ca-ab-bc)
=> ca-ab-bc=15

4. Resolve into factors
(i) (x+1)⁴+(x+1)²+1/4
= [4(x+1)⁴+4(x+1)²+1]/4
= 1/4[4(x+1)⁴+4(x+1)²+1]
= 1/4[{2(x+1)²}² + 2.2(x+1)².1 + 1²]
= 1/4{2(x+1)²+1}²
= 1/4{2(x²+2x+1²)+1}²
= 1/4{2x²+4x+2+1}²
= 1/4(2x²+4x+3)²
= 1/4(2x²+4x+3)(2x²+4x+3)

(ii) (3a-2b)²-4(3a-2b)-21 
= (3a-2b)²-4(3a-2b)+4-4-21
= (3a-2b)²-4(3a-2b)+2²-25
= (3a-2b-2)²-5²
= (3a-2b-2-5)(3a-2b-2+5)
= (3a-2b-7)(3a-2b+3) 

(iii) m⁴-9m²+20 
= m⁴-(4+5)m²+20 
= m⁴-4m²-5m²+20
= m²(m²-4)-5(m²+4)
= (m²-4)(m²-5)

(iv) 36-60y+25y²
= 6²-2.6.5y+(5y)²
= (6-5y)²
= (6-5y)(6-5y)

(v) x³-3x²y+2xy² 
= x³-(1+2)x²y+2xy² 
= x³-x²y-2x²y+2xy²
= x²(x-y)-2xy(x-y)
= (x-y)(x²-2xy)
= x(x-y)(x-2y)

(vi) 2ax²-8a-3x²+12
= 2ax²-3x²-8a+12
= x²(2a-3)-4(2a-3)
= (2a-3)(x²-4)

(vii)x²-y²+2x+2y 
= (x+y)(x-y)+2(x+y)
= (x+y){(x-y)+2}
= (x+y)(x-y+2)

(vii) 64x²-9y²+12yz-4z²
= (8x)²-(3y)²-(2z)²+12yz
= (8x)²-{(3y)²-2.3y.2z+(2z)²}
= (8x)²-(3y-2z)²
= (8x-3y+2z)(8x+3y-2z)

(ix) x²+12x+35 
= x²+(5+7)x+35
= x²+5x+7x+35
= x(x+5)+7(x+5)
= (x+5)(x+7)
 
(x) x²y+2x+xy+2x²+y+2 
= x²y+2x²+xy+2x+y+2
= x²(y+2)+x(y+2)+1(y+2)
= (y+2)(x²+x+1)

5. Dividing a⁷+b⁷ by a+b, prove the formula 12 given in the introduction of this chapter 6. 

Solution:




















6. Show that-

(i) (x+a)(x+b)=x²+(a+b)x+ab 

RHS = x²+(a+b)x+ab
    = x²+ax+bx+ab
    = x(x+a)+b(x+a)
    = (x+a)(x+b)
    = LHS

(ii) (x+a)(x+b)(x+c)=x³+(a+b+c)x²+(ab+bc+ca)x+abc 

RHS = x³+(a+b+c)x²+(ab+bc+ca)x+abc 
    = x³+x²a+x²b+x²c+xab+xbc+xca+abc
    = (x³+x²a)+(x²b+xab)+(x²c+xca)+(xbc+abc)
    = x²(x+a)+xb(x+a)+xc(x+a)+bc(x+a)
    = (x+a)(x²+xb+xc+bc)
    = (x+a){x(x+b)+c(x+b)}
    = (x+a)(x+b)(x+c)
    = LHS

7. Express as perfect square 

(i) a²+4b²+9c²-4ab+12bc-6ac

= (-a)²+ (2b)²+ (3c)²+ 2.(-a).2b + 2.2b.3c + 2.3c.(-a)
= (-a+2b+3c)²

 
(ii) (x⁴+2x³+3x²+2x+1) 

= (x²)²+x²+1+2x³+2x²+2x
= (x²)²+ x²+ 1²+ 2.x².x + 2.x².1 + 2.1.x
= (x²+x+1)


8. If (A)²=x⁴y⁸-2x²y⁴z⁵+z¹⁰ . Then find A.

Solution:
We know,(A)²=x⁴y⁸-2x²y⁴z⁵+z¹⁰
A²=(x²y⁴)²-2.x²y⁴.z⁵+(z⁵)²
A²=(x²y⁴-z⁵)²
A=x²y⁴-z


9.Examine whether (x-2) is a factor of x³+5x²-2x-24.

Solution:
Hence, x-2 is a factor of x³+5x²-2x-24 .

10.Expand 

(i) (2a)⁵+(3b)⁵
= (2a+3b)[(2a)⁴ - (2a)³.3b + (2a)².(3b)² - (2a).(3b)³ + (3b)⁴]
= (2a+3b)[16a⁴ - 8a³.3b + 4a².9b² - 2a.27b³ + 81b⁴]
= (2a+3b)[16a⁴ - 24a³b + 36a²b² - 54ab³ + 81b⁴]

(ii) (x+y)⁶ 
= {(x+y)³}²
= (x³+3x²y+3xy²+y³)²
= (x³+3x²y+3xy²+y³)(x³+3x²y+3xy²+y³)
= x³(x³+3x²y+3xy²+y³) + 3x²y(x³+3x²y+3xy²+y³) + 3xy²(x³+3x²y+3xy²+y³) + y³(x³+3x²y+3xy²+y³)
= x⁶ + 3x⁵y + 3x⁴y² + x³y³ + 3x⁵y + 9x⁴y² + 9x³y³ + 3x²y⁴ + 3x⁴y² + 9x³y³ + 9x²y⁴ + 3xy⁵ + x³y³ + 3x²y⁴ + 3xy⁵ + y⁶
= x⁶+y⁶+6x⁵y+15x⁴y²+20x³y³+15x²y⁴+6xy⁵ 

(iii) (2x)⁷-(3a)⁷ 
= (2x-3a){(2x)⁶+(2x)⁵.(3a)+(2x)⁴.(3a)²+(2x)³.(3a)³+(2x)².(3a)⁴+(2x).(3a)⁵+(3a)⁶}
= (2x-3a)(64x⁶ + 32x⁵.3a + 16x⁴.9a² + 8x³.27a³ + 4x².81a⁴ + 2x.243a⁵ + 729a⁶)
= (2x-3a)(64x⁶ + 96x⁵a + 144x⁴a² + 216x³a³ + 324x²a⁴ + 486xa⁵ + 729a⁶)

(iv) x⁸-y⁸ 
= (x-y)(x⁷+x⁶y+x⁵y²+x⁴y³+x³y⁴+x²y⁵+xy⁶+y⁷)


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