Factorisation Class 9 RS Aggarwal Exe-4D Goyal Brothers ICSE Maths Solutions Ch-4. In this article you will learn Factorisation by splitting middle term easily. Visit official Website CISCE for detail information about ICSE Board Class-9.
Factorisation Class 9 RS Aggarwal Exe-4D Goyal Brothers ICSE Maths Solutions
| Board | ICSE |
| Subject | Maths |
| Class | 9th |
| Chapter-4 | Factorisation |
| Writer | RS Aggrawal |
| Topics | Factorisation by splitting middle term |
| Academic Session | 2024-2025 |
Factorisation by splitting middle term
(Class 9 RS Aggarwal Exe-4D Goyal Brothers ICSE Maths Solutions Ch-4)
Rule: to factorise trinomial 𝑎𝑥2+𝑏𝑥+𝑐 where 𝑎,𝑏 and 𝑐 are real numbers, is to split 𝑏 (the coefficient of 𝑥) into two real numbers such that the algebraic sum of these two numbers is 𝑏 and their product is 𝑐, then factorise by grouping method
Exercise- 4D
Factorise : Page- 65
Que-1: x²+9x+18
Sol: x²+9x+18
x²+3x+6x+18
x(x+3)+6(x+3)
(x+3)(x+6).
Que-2: x²-5x-6
Sol: x²-5x-6
x²+x-6x-6
x(x+1)-6(x+1)
(x+1)(x-6).
Que-3: x²+6x-7
Sol: x²+6x-7
x²+7x-x-7
x(x+7)-1(x+7)
(x-1)(x+7).
Que-4: x²+7x-18
Sol: x²+7x-18
x²+9x-2x-18
x(x+9)-2(x+9)
(x+9)(x-2).
Que-5: x²-3x-54
Sol: x²-3x-54
x²+6x-9x-54
x(x+6)-9(x+6)
(x+6)(x-9).
Que-6: x²-17x-84
Sol: x²-17x-84
x²+4x-21x-84
x(x+4)-21(x+4)
(x+4)(x-21).
Que-7: 6x²+11x-10
Sol: 6x²+11x-10
6x²-4x+15x-10
2x(3x-2)+5(3x-2)
(2x+5)(3x-2).
Que-8: 3x²-4x-7
Sol: 3x²-4x-7
3x²+3x-7x-7
3x(x+1)-7(x+1)
(3x-7)(x+1).
Que-9: 2x²-7x-39
Sol: 2x²-7x-39
2x²+6x-13x-39
2x(x+3)-13(x+3)
(2x-13)(x+3).
Que-10: 2x²+3x-90
Sol: 2x²+3x-90
2x²-12x+15x-90
2x(x-6)+15(x-6)
(2x+15)(x-6).
Que-11: 2x²+5x-3
Sol: 2x²+5x-3
2x²+6x-x-3
2x(x+3)-1(x+3)
(x+3)(2x-1).
Que-12: 10+3x-x²
Sol: 10+3x+x²
10-2x+5x-x²
2(5-x)+x(5-x)
(2+x)(5-x).
Que-13: 7-12x-4x²
Sol: 7-12x-4x²
7-14x+2x-4x²
7(1-2x)+2x(1-2x)
(7+2x)(1-2x).
Que-14: 5-4y-12y²
Sol: 5-4y-12y²
5-10y+6y-12y²
5(1-2y)+6y(1-2y)
(5+6y)(1-2y).
Que-15: 1-18z-63z²
Sol: 1-18z-63z²
1+3z-21z-63z²
1(1+3z)-21z(1+3z)
(1+3z)(1-21z).
Que-16: x²-3ax-88a²
Sol: x²-3ax-88a²
x²-11ax+8ax-88a²
x(x-11a)+8a(x-11a)
(x+8a)(x-11a).
Que-17: 3-x(4+7x)
Sol: 3-x(4+7x)
3-4x-7x²
3+3x-7x-7x²
3(1+x)-7x(1+x)
(1+x)(3-7x)
Que-18: 24x³+37x²-5x
Sol: 24x³+37x²-5x
x[24x²+37x-5]
x[24x²+40x-3x-5]
x[8x(3x+5)-1(3x+5)]
x(8x-1)(3x+5).
Que-19: x²y²-12xy-45
Sol: x²y²-12xy-45
x²y²-15xy+3xy-45
xy(xy-15)+3(xy-15)
(xy+3)(xy-15).
Que-20: x(2x-y)-y²
Sol: x(2x-y)-y²
2x²-xy-y²
2x²-2xy+xy-y²
2x(x-y)+y(x-y)
(2x+y)(x-y).
Que-21: 5x²+17xy-12y²
Sol: 5x²+17xy-12y²
5x²+20xy-3xy-12y²
5x(x+4y)-3y(x+4y)
(5x-3y)(x+4y).
Que-22: 5(3a+b)²+6(3a+b)-8
Sol: 5(3a+b)²+6(3a+b)-8
Let (3a+b) = x
5x²+6x-8
5x²+10x-4x-8
5x(x+2)-4(x+2)
(5x-4)(x+2)
Now, put the value of x
[5(3a+b)-4][(3a+b)+2]
(15a+5b-4)(3a+b+2).
Que-23: 3(2a-b)²-19(2a-b)+28
Sol: 3(2a-b)²-19(2a-b)+28
Let (2a-b) = x
3x²-19x+28
3x²-12x-7x+28
3x(x-4)-7(x-4)
(x-4)(3x-7)
Now, put the value of x
[(2a-b)-4][3(2a-b)-7]
(2a-b-4)(6a-3b-7).
Que-24: 1-2(a+2b)-3(a+2b)²
Sol: 1-2(a+2b)-3(a+2b)²
Let (a+2b) = x
1-2x-3x²
1+x-3x-3x²
1(1+x)-3x(1+x)
(1-3x)(1+x)
[1-3(a+2b)][1+(a+2b)]
(1-3a-6b)(1+a+2b).
Que-25: (a+4)²-5ab-20b-6b²
Sol: (a+4)²-5ab-20b-6b²
(a+4)²-5b(a+4)-6b²
(a+4)²-6b(a+4)+b(a+4)-6b²
(a+4)(a+4-6b)+b(a+4-6b)
(a+4-6b)(a+4+b).
Que-26: 8(a-2b)²-2a+4b-1
Sol: 8(a-2b)²-2a+4b-1
8(a-2b)²-2(a-2b)-1
8(a-2b)²-4(a-2b)+2(a-2b)-1
4(a-2b)[2(a-2b)-1]+1[2(a-2b)-1]
(4a-8b)(2a-4b-1)+(2a-4b-1)
(2a-4b-1)(4a-8b+1).
Que-27: 4(2a-3)²-3(2a-3)(a-1)-7(a-1)²
Sol: 4(2a-3)²-3(2a-3)(a-1)-7(a-1)²
Let (2a-3) = x and (a-1) = y
4x²-3xy-7y²
4x²-7xy+4xy-7y²
x(4x-7y)+y(4x-7y)
(x+y)(4x-7y)
Now, put the value of x and y
[(2a-3)+(a-1)][4(2a-3)-7(a-1)]
(2a-3+a-1)(8a-12-7a+7)
(3a-4)(a-5).
Que-28: (a²-3a)(a²-3a+7)+10
Sol: (a²-3a)(a²-3a+7)+10
(a²-3a)(a²-3a)+7(a²-3a)+10
(a²-3a)²+7(a²-3a)+10
(a²-3a) = x
x²+7x+10
x²+2x+5x+10
x(x+2)+5(x+2)
(x+5)(x+2)
Now, put the value of x
(a²-3a+5)(a²-3a+2)
(a²-3a+5)(a²-a-2a+2)
(a²-3a+5)[a(a-1)-2(a-1)]
(a²-3a+5)(a-1)(a-2)
Que-29: (a²-a)(4a²-4a-5)-6
Sol: (a²−a)(4a²−4a−5)−6
= (a²−a)[4(a²−a)−5]−6
Let (a²−a) = x
Thus,
x(4x−5)−6
= 4x²−5x−6
= 4x²−8x+3x−6
= 4x(x−2)+3(x−2)
= (x−2)(4x+3)
Putting the value of x
= (a²−a−2)[4(a²−a)+3]
= (a²−2a+a−2)(4a²−4a+3)
= (a(a−2)+1(a−2))(4a²−4a+3)
= (a−2)(a+1)(4a²−4a+3)
Que-30: a^4 – 11a²+10
Sol: a^4 – 11a²+10
Factorising we get,
a^4 – 11a² + 10
= a^ – 10a² – a² + 10.
= a²(a² – 10) – 1(a² – 10)
= (a² – 10)(a² – 1)
= (a² – 10)(a – 1)(a + 1)
Que-31: 5-(3x²-2x)(6-3x²+2x)
Sol: 5-(3x²-2x)(6-3x²+2x)
Let (3x²-2x) = y
5-(y)(6-y)
5-(6y+y²)
5-6y+y²
5-5y-y+y²
5(1-y)-y(a-y)
(5-y)(1-y)
Now, put the of y
(1-3x²+2x)(5-3x²+2x)
(1+2x-3x²)(5+2x-3x²)
(1+3x-x-3x²)(5+5x-3x-3x²)
[1(1+3x)-x(1+3x)][5(1+x)-3x(1+x)]
(1+3x)(1-x)(5-3x)(1+x).
– : End of Factorisation Class 9 RS Aggarwal Exe-4D Goyal Brothers ICSE Maths Solutions Ch-4 :–
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