Factorization - Exercise 7.8 | Mathematics (Maths) Class 8 PDF Download

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Q u e s t i o n : 1 6 0
Resolve each of the following quadratic trinomial into factor:
2x
2
 + 5x + 3
S o l u t i o n :
The given expression is 2x2+5x+3.                        (Coefficient of x2=2, coefficient of x=5 and constant term=3)We will split the coefficient of x into two parts such that their sum is 5 and their
Now, 2+3=5 and 2×3=6Replacing the middle term 5x by 2x+3x, we have:2x2+5x+3=2x2+2x+3x+3                    =(2x2+2x)+(3x+3)                    =2x(x+1)+3(x+1)                    =(x+1)(2x+3)
Q u e s t i o n : 1 6 1
Resolve each of the following quadratic trinomial into factor:
2x
2
 - 3x - 2
S o l u t i o n :
The given expression is 2x2-3x-2.                             (Coefficient of x2=2, coefficient of x=-3 and constant term=-2)We will split the coefficient of x into two parts such that their sum is -3 and
Now, (-4)+1=-3 and (-4)×1=-4Replacing the middle term 3x by -4x+x, we have:2x2-3x-2=2x2-4x+x-2                    =(2x2-4x)+(x-2)                    =2x(x-2)+(x-2)                    =(2x+1)(x-2)
Q u e s t i o n : 1 6 2
Resolve each of the following quadratic trinomial into factor:
3x
2
 + 10x + 3
S o l u t i o n :
The given expression is 3x2+10x+3.          (Coefficient of x2=3, coefficient of x=10 and constant term=3)We will split the coefficient of x into two parts such that their sum is 10 and their product
Now, 9+1=10 and 9×1=9Replacing the middle term 10x by 9x+x, we have:3x2+10x+3=3x2+9x+x+3                      =(3x2+9x)+(x+3)                     =3x(x+3)+(x+3)                     =(3x+1)(x+3)
Q u e s t i o n : 1 6 3
Resolve each of the following quadratic trinomial into factor:
7x - 6 - 2x
2
S o l u t i o n :
The given expression is 7x-6-2x2.                                        (Coefficient of x2=-2, coefficient of x=7 and constant term=-6)We will split the coefficient of x into two parts such that their sum
Now, 4+3=7 and 4×3=12Replacing the middle term 7x by 4x+3x, we have:7x-6-2x2=-2x2+4x+3x-6                    =(-2x2+4x)+(3x-6)                    =2x(2-x)-3(2-x)                    =(2x-3)(2-x)
Q u e s t i o n : 1 6 4
Resolve each of the following quadratic trinomial into factor:
7x
2
 - 19x - 6
S o l u t i o n :
The given expression is 7x2-19x-6.              (Coefficient of x2=7, coefficient of x=-19 and constant term=-6)We will split the coefficient of x into two parts such that their sum is -19 and their
Now, (-21)+2=-19 and (-21)×2=-42Replacing the middle term -19x by -21x+2x, we have:7x2-19x-6=7x2-21x+2x-6                      =(7x2-21x)+(2x-6)                      =7x(x-3)+2(x-3)                      =
Q u e s t i o n : 1 6 5
Resolve each of the following quadratic trinomial into factor:
28 - 31x - 5x
2
S o l u t i o n :
The given expression is 28-31x-5x2.                (Coefficient of x2=-5, coefficient of x=-31 and constant term=28)We will split the coefficient of x into two parts such that their sum is -31 and
Now,(-35)+4=-31 and (-35)×4=-140Replacing the middle term -31x by -35x+4x, we have:-5x2-31x+28=-5x2-35x+4x+28                           =(-5x2-35x)+(4x+28)                           =-5x(x+7)+4(x+7
                           =(4-5x)(x+7)
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Q u e s t i o n : 1 6 0
Resolve each of the following quadratic trinomial into factor:
2x
2
 + 5x + 3
S o l u t i o n :
The given expression is 2x2+5x+3.                        (Coefficient of x2=2, coefficient of x=5 and constant term=3)We will split the coefficient of x into two parts such that their sum is 5 and their
Now, 2+3=5 and 2×3=6Replacing the middle term 5x by 2x+3x, we have:2x2+5x+3=2x2+2x+3x+3                    =(2x2+2x)+(3x+3)                    =2x(x+1)+3(x+1)                    =(x+1)(2x+3)
Q u e s t i o n : 1 6 1
Resolve each of the following quadratic trinomial into factor:
2x
2
 - 3x - 2
S o l u t i o n :
The given expression is 2x2-3x-2.                             (Coefficient of x2=2, coefficient of x=-3 and constant term=-2)We will split the coefficient of x into two parts such that their sum is -3 and
Now, (-4)+1=-3 and (-4)×1=-4Replacing the middle term 3x by -4x+x, we have:2x2-3x-2=2x2-4x+x-2                    =(2x2-4x)+(x-2)                    =2x(x-2)+(x-2)                    =(2x+1)(x-2)
Q u e s t i o n : 1 6 2
Resolve each of the following quadratic trinomial into factor:
3x
2
 + 10x + 3
S o l u t i o n :
The given expression is 3x2+10x+3.          (Coefficient of x2=3, coefficient of x=10 and constant term=3)We will split the coefficient of x into two parts such that their sum is 10 and their product
Now, 9+1=10 and 9×1=9Replacing the middle term 10x by 9x+x, we have:3x2+10x+3=3x2+9x+x+3                      =(3x2+9x)+(x+3)                     =3x(x+3)+(x+3)                     =(3x+1)(x+3)
Q u e s t i o n : 1 6 3
Resolve each of the following quadratic trinomial into factor:
7x - 6 - 2x
2
S o l u t i o n :
The given expression is 7x-6-2x2.                                        (Coefficient of x2=-2, coefficient of x=7 and constant term=-6)We will split the coefficient of x into two parts such that their sum
Now, 4+3=7 and 4×3=12Replacing the middle term 7x by 4x+3x, we have:7x-6-2x2=-2x2+4x+3x-6                    =(-2x2+4x)+(3x-6)                    =2x(2-x)-3(2-x)                    =(2x-3)(2-x)
Q u e s t i o n : 1 6 4
Resolve each of the following quadratic trinomial into factor:
7x
2
 - 19x - 6
S o l u t i o n :
The given expression is 7x2-19x-6.              (Coefficient of x2=7, coefficient of x=-19 and constant term=-6)We will split the coefficient of x into two parts such that their sum is -19 and their
Now, (-21)+2=-19 and (-21)×2=-42Replacing the middle term -19x by -21x+2x, we have:7x2-19x-6=7x2-21x+2x-6                      =(7x2-21x)+(2x-6)                      =7x(x-3)+2(x-3)                      =
Q u e s t i o n : 1 6 5
Resolve each of the following quadratic trinomial into factor:
28 - 31x - 5x
2
S o l u t i o n :
The given expression is 28-31x-5x2.                (Coefficient of x2=-5, coefficient of x=-31 and constant term=28)We will split the coefficient of x into two parts such that their sum is -31 and
Now,(-35)+4=-31 and (-35)×4=-140Replacing the middle term -31x by -35x+4x, we have:-5x2-31x+28=-5x2-35x+4x+28                           =(-5x2-35x)+(4x+28)                           =-5x(x+7)+4(x+7
                           =(4-5x)(x+7)
Q u e s t i o n : 1 6 6
Resolve each of the following quadratic trinomial into factor:
3 + 23y - 8y
2
S o l u t i o n :
The given expression is 3+23y-8y2.              (Coefficient of y2=-8, coefficient of y=23 and constant term=3)We will split the coefficient of y into two parts such that their sum is 23 and their
Now, (-1)+24=23 and (-1)×24=-24Replacing the middle term 23y by -y+24y, we have:3+23y-8y2=-8y2+23y+3=-8y2-y+24y+3=(-8y2-y)+(24y+3)=-y(8y+1)+3(8y+1)=(3-y)(8y+1)
Q u e s t i o n : 1 6 7
Resolve each of the following quadratic trinomial into factor:
11x
2
 - 54x + 63
S o l u t i o n :
The given expression is 11x2-54x+63.                  (Coefficient of x2=11, coefficient of x=-54 and constant term=63)We will split the coefficient of x into two parts such that their sum is -54
Now, (-33)+(-21)=-54 and (-33)×(-21)=693Replacing the middle term-54x by -33x-21x, we have:11x2-54x+63= 11x2-33x-21x+63                          =(11x2-33x)+(-21x+63)                          =11x(
                          =(11x-21)(x-3)
Q u e s t i o n : 1 6 8
Resolve each of the following quadratic trinomial into factor:
7x - 6x
2
 + 20
S o l u t i o n :
The given expression is 7x-6x2+20.                     (Coefficient of x2=-6, coefficient of x=7 and constant term=20)We will split the coefficient of x into two parts such that their sum is 7 and 
Now,15+(-8)=7 and 15×(-8)=-120Replacing the middle term 7x by 15x-8x, we get:7x-6x2+20=-6x2+7x+20=-6x2+15x-8x+20=(-6x2+15x)+(-8x+20)=3x(-2x+5)+4(-2x+5)=(3x+4)(-2x+5)
Q u e s t i o n : 1 6 9
Resolve each of the following quadratic trinomial into factor:
3x
2
 + 22x + 35
S o l u t i o n :
The given expression is 3x2+22x+35.                                         (Coefficient of x2=3, coefficient of x=22 and constant term=35)We will split the coefficient of x into two parts such that their
Now, 15+7=22 and 15×7=105Replacing the middle term 22x by 15x+7x, we get:3x2+22x+35=3x2+15x+7x+35                         =(3x2+15x)+(7x+35)                         =3x(x+5)+7(x+5)               
Q u e s t i o n : 1 7 0
Resolve each of the following quadratic trinomial into factor:
12x
2
 - 17xy + 6y
2
S o l u t i o n :
The given expression is 12x2-17xy+6y2.                          (Coefficient of x2=12, coefficient of x=-17y and constant term=6y2)We willsplit the coefficient of x into two parts such that their sum
Now, (-9y)+(-8y)=-17y and (-9y)×(-8y)=72y2Replacing the middle term -17xy by -9xy-8xy, we get:12x2-17xy+6y2=12x2-9xy-8xy+6y2                              =(12x2-9xy)-(8xy-6y2)                       
                              =(3x-2y)(4x-3y)
Q u e s t i o n : 1 7 1
Resolve each of the following quadratic trinomial into factor:
6x
2
 - 5xy - 6y
2
S o l u t i o n :
The given expression is 6x2-5xy-6y2.                           (Coefficient of x2=6, coefficient of x=-5y and constant term=-6y2)We will split the coefficient of x into two parts such that their sum 
Now, (-9y)+4y=-5y and(-9y)×4y=-36y2 Replacing the middle term -5xy by -9xy+4xy, we get:6x2-5xy-6y2= 6x2-9xy+4xy-6y2                          =(6x2-9xy)+(4xy-6y2)                          =3x(2x-3y
                          =(3x+2y)(2x-3y)
Q u e s t i o n : 1 7 2
Resolve each of the following quadratic trinomial into factor:
6x
2
 - 13xy + 2y
2
S o l u t i o n :
The given expression is 6x2-13xy+2y2.                        (Coefficient of x2=6, coefficient of x=-13y and constant term=2y2)We will split the coefficient of x into two parts such that their sum
Now,(-12y)+(-y)=-13y and(-12y)×(-y)=12y2Replacing the middle term -13xy by -12xy-xy, we get: 6x2-13xy+2y2= 6x2-12xy-xy+2y2                            =(6x2-12xy)-(xy-2y2)                            =6
                            =(6x-y)(x-2y)
Q u e s t i o n : 1 7 3
Resolve each of the following quadratic trinomial into factor:
14x
2
 + 11xy - 15y
2
S o l u t i o n :
The given expression is 14x2+11xy-15y2.        (Coefficient of x2=14, coefficient of x=11y and constant term=-15y2)Now, we will split the coefficient of x into two parts such that their sum
Now,21y+(-10y)=11y and21y×(-10y)=-210y2Replacing the middle term -11xy by -10xy+21xy, we get:14x2+11xy-15y2= 14x2-10xy+21xy-15y2                               =(14x2-10xy)+(21xy-15y2
                               =2x(7x-5y)+3y(7x-5y)                              =(2x+3y)(7x-5y)
Q u e s t i o n : 1 7 4
Resolve each of the following quadratic trinomial into factor:
6a
2
 + 17ab - 3b
2
S o l u t i o n :
The given expression is 6a2+17ab-3b2.             (Coefficient  of a2=6, coefficient of a=17b and constant term=-3b2)Now, we will split the coefficient of a into two parts such that their sum is
Now,18b+(-b)=17b and18b×(-b)=-18b2Replacing the middle term 17ab by -ab+18ab, we get:16a2+17ab-3b2=6a2+-ab+18ab-3b2                               =(6a2-ab)+(18ab-3b2)                              =
                             =(a+3b)(6a-b)
Q u e s t i o n : 1 7 5
Resolve each of the following quadratic trinomial into factor:
36a
2
 + 12abc - 15b
2
c
2
S o l u t i o n :
The given expression is 36a2+12abc-15b2c2.          (Coefficient of a2=36, coefficient of a=12bc and constant term=-15b2c2)Now, we will split the coefficient of a into two parts such that their
Now,(-18bc)+30bc=12bc and(-18bc)×30bc=-540b2c2Replacing the middle term 12abc by -18abc+30abc, we get:36a2+12abc-15b2c2=36a2-18abc+30abc-15b2c2                                     =(
                                     =18a(2a-bc)+15bc(2a-bc)                                     =(18a+15bc)(2a-bc)                                    =3(6a+5bc)(2a-bc)
Q u e s t i o n : 1 7 6
Resolve each of the following quadratic trinomial into factor:
15x
2
 - 16xyz - 15y
2
z
2
S o l u t i o n :
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Q u e s t i o n : 1 6 0
Resolve each of the following quadratic trinomial into factor:
2x
2
 + 5x + 3
S o l u t i o n :
The given expression is 2x2+5x+3.                        (Coefficient of x2=2, coefficient of x=5 and constant term=3)We will split the coefficient of x into two parts such that their sum is 5 and their
Now, 2+3=5 and 2×3=6Replacing the middle term 5x by 2x+3x, we have:2x2+5x+3=2x2+2x+3x+3                    =(2x2+2x)+(3x+3)                    =2x(x+1)+3(x+1)                    =(x+1)(2x+3)
Q u e s t i o n : 1 6 1
Resolve each of the following quadratic trinomial into factor:
2x
2
 - 3x - 2
S o l u t i o n :
The given expression is 2x2-3x-2.                             (Coefficient of x2=2, coefficient of x=-3 and constant term=-2)We will split the coefficient of x into two parts such that their sum is -3 and
Now, (-4)+1=-3 and (-4)×1=-4Replacing the middle term 3x by -4x+x, we have:2x2-3x-2=2x2-4x+x-2                    =(2x2-4x)+(x-2)                    =2x(x-2)+(x-2)                    =(2x+1)(x-2)
Q u e s t i o n : 1 6 2
Resolve each of the following quadratic trinomial into factor:
3x
2
 + 10x + 3
S o l u t i o n :
The given expression is 3x2+10x+3.          (Coefficient of x2=3, coefficient of x=10 and constant term=3)We will split the coefficient of x into two parts such that their sum is 10 and their product
Now, 9+1=10 and 9×1=9Replacing the middle term 10x by 9x+x, we have:3x2+10x+3=3x2+9x+x+3                      =(3x2+9x)+(x+3)                     =3x(x+3)+(x+3)                     =(3x+1)(x+3)
Q u e s t i o n : 1 6 3
Resolve each of the following quadratic trinomial into factor:
7x - 6 - 2x
2
S o l u t i o n :
The given expression is 7x-6-2x2.                                        (Coefficient of x2=-2, coefficient of x=7 and constant term=-6)We will split the coefficient of x into two parts such that their sum
Now, 4+3=7 and 4×3=12Replacing the middle term 7x by 4x+3x, we have:7x-6-2x2=-2x2+4x+3x-6                    =(-2x2+4x)+(3x-6)                    =2x(2-x)-3(2-x)                    =(2x-3)(2-x)
Q u e s t i o n : 1 6 4
Resolve each of the following quadratic trinomial into factor:
7x
2
 - 19x - 6
S o l u t i o n :
The given expression is 7x2-19x-6.              (Coefficient of x2=7, coefficient of x=-19 and constant term=-6)We will split the coefficient of x into two parts such that their sum is -19 and their
Now, (-21)+2=-19 and (-21)×2=-42Replacing the middle term -19x by -21x+2x, we have:7x2-19x-6=7x2-21x+2x-6                      =(7x2-21x)+(2x-6)                      =7x(x-3)+2(x-3)                      =
Q u e s t i o n : 1 6 5
Resolve each of the following quadratic trinomial into factor:
28 - 31x - 5x
2
S o l u t i o n :
The given expression is 28-31x-5x2.                (Coefficient of x2=-5, coefficient of x=-31 and constant term=28)We will split the coefficient of x into two parts such that their sum is -31 and
Now,(-35)+4=-31 and (-35)×4=-140Replacing the middle term -31x by -35x+4x, we have:-5x2-31x+28=-5x2-35x+4x+28                           =(-5x2-35x)+(4x+28)                           =-5x(x+7)+4(x+7
                           =(4-5x)(x+7)
Q u e s t i o n : 1 6 6
Resolve each of the following quadratic trinomial into factor:
3 + 23y - 8y
2
S o l u t i o n :
The given expression is 3+23y-8y2.              (Coefficient of y2=-8, coefficient of y=23 and constant term=3)We will split the coefficient of y into two parts such that their sum is 23 and their
Now, (-1)+24=23 and (-1)×24=-24Replacing the middle term 23y by -y+24y, we have:3+23y-8y2=-8y2+23y+3=-8y2-y+24y+3=(-8y2-y)+(24y+3)=-y(8y+1)+3(8y+1)=(3-y)(8y+1)
Q u e s t i o n : 1 6 7
Resolve each of the following quadratic trinomial into factor:
11x
2
 - 54x + 63
S o l u t i o n :
The given expression is 11x2-54x+63.                  (Coefficient of x2=11, coefficient of x=-54 and constant term=63)We will split the coefficient of x into two parts such that their sum is -54
Now, (-33)+(-21)=-54 and (-33)×(-21)=693Replacing the middle term-54x by -33x-21x, we have:11x2-54x+63= 11x2-33x-21x+63                          =(11x2-33x)+(-21x+63)                          =11x(
                          =(11x-21)(x-3)
Q u e s t i o n : 1 6 8
Resolve each of the following quadratic trinomial into factor:
7x - 6x
2
 + 20
S o l u t i o n :
The given expression is 7x-6x2+20.                     (Coefficient of x2=-6, coefficient of x=7 and constant term=20)We will split the coefficient of x into two parts such that their sum is 7 and 
Now,15+(-8)=7 and 15×(-8)=-120Replacing the middle term 7x by 15x-8x, we get:7x-6x2+20=-6x2+7x+20=-6x2+15x-8x+20=(-6x2+15x)+(-8x+20)=3x(-2x+5)+4(-2x+5)=(3x+4)(-2x+5)
Q u e s t i o n : 1 6 9
Resolve each of the following quadratic trinomial into factor:
3x
2
 + 22x + 35
S o l u t i o n :
The given expression is 3x2+22x+35.                                         (Coefficient of x2=3, coefficient of x=22 and constant term=35)We will split the coefficient of x into two parts such that their
Now, 15+7=22 and 15×7=105Replacing the middle term 22x by 15x+7x, we get:3x2+22x+35=3x2+15x+7x+35                         =(3x2+15x)+(7x+35)                         =3x(x+5)+7(x+5)               
Q u e s t i o n : 1 7 0
Resolve each of the following quadratic trinomial into factor:
12x
2
 - 17xy + 6y
2
S o l u t i o n :
The given expression is 12x2-17xy+6y2.                          (Coefficient of x2=12, coefficient of x=-17y and constant term=6y2)We willsplit the coefficient of x into two parts such that their sum
Now, (-9y)+(-8y)=-17y and (-9y)×(-8y)=72y2Replacing the middle term -17xy by -9xy-8xy, we get:12x2-17xy+6y2=12x2-9xy-8xy+6y2                              =(12x2-9xy)-(8xy-6y2)                       
                              =(3x-2y)(4x-3y)
Q u e s t i o n : 1 7 1
Resolve each of the following quadratic trinomial into factor:
6x
2
 - 5xy - 6y
2
S o l u t i o n :
The given expression is 6x2-5xy-6y2.                           (Coefficient of x2=6, coefficient of x=-5y and constant term=-6y2)We will split the coefficient of x into two parts such that their sum 
Now, (-9y)+4y=-5y and(-9y)×4y=-36y2 Replacing the middle term -5xy by -9xy+4xy, we get:6x2-5xy-6y2= 6x2-9xy+4xy-6y2                          =(6x2-9xy)+(4xy-6y2)                          =3x(2x-3y
                          =(3x+2y)(2x-3y)
Q u e s t i o n : 1 7 2
Resolve each of the following quadratic trinomial into factor:
6x
2
 - 13xy + 2y
2
S o l u t i o n :
The given expression is 6x2-13xy+2y2.                        (Coefficient of x2=6, coefficient of x=-13y and constant term=2y2)We will split the coefficient of x into two parts such that their sum
Now,(-12y)+(-y)=-13y and(-12y)×(-y)=12y2Replacing the middle term -13xy by -12xy-xy, we get: 6x2-13xy+2y2= 6x2-12xy-xy+2y2                            =(6x2-12xy)-(xy-2y2)                            =6
                            =(6x-y)(x-2y)
Q u e s t i o n : 1 7 3
Resolve each of the following quadratic trinomial into factor:
14x
2
 + 11xy - 15y
2
S o l u t i o n :
The given expression is 14x2+11xy-15y2.        (Coefficient of x2=14, coefficient of x=11y and constant term=-15y2)Now, we will split the coefficient of x into two parts such that their sum
Now,21y+(-10y)=11y and21y×(-10y)=-210y2Replacing the middle term -11xy by -10xy+21xy, we get:14x2+11xy-15y2= 14x2-10xy+21xy-15y2                               =(14x2-10xy)+(21xy-15y2
                               =2x(7x-5y)+3y(7x-5y)                              =(2x+3y)(7x-5y)
Q u e s t i o n : 1 7 4
Resolve each of the following quadratic trinomial into factor:
6a
2
 + 17ab - 3b
2
S o l u t i o n :
The given expression is 6a2+17ab-3b2.             (Coefficient  of a2=6, coefficient of a=17b and constant term=-3b2)Now, we will split the coefficient of a into two parts such that their sum is
Now,18b+(-b)=17b and18b×(-b)=-18b2Replacing the middle term 17ab by -ab+18ab, we get:16a2+17ab-3b2=6a2+-ab+18ab-3b2                               =(6a2-ab)+(18ab-3b2)                              =
                             =(a+3b)(6a-b)
Q u e s t i o n : 1 7 5
Resolve each of the following quadratic trinomial into factor:
36a
2
 + 12abc - 15b
2
c
2
S o l u t i o n :
The given expression is 36a2+12abc-15b2c2.          (Coefficient of a2=36, coefficient of a=12bc and constant term=-15b2c2)Now, we will split the coefficient of a into two parts such that their
Now,(-18bc)+30bc=12bc and(-18bc)×30bc=-540b2c2Replacing the middle term 12abc by -18abc+30abc, we get:36a2+12abc-15b2c2=36a2-18abc+30abc-15b2c2                                     =(
                                     =18a(2a-bc)+15bc(2a-bc)                                     =(18a+15bc)(2a-bc)                                    =3(6a+5bc)(2a-bc)
Q u e s t i o n : 1 7 6
Resolve each of the following quadratic trinomial into factor:
15x
2
 - 16xyz - 15y
2
z
2
S o l u t i o n :
The given expression is 15x2-16xyz-15y2z2.(Coefficient of x2=15, coefficient of x=-16yz and constant term=-15y2z2)Now, we will split the coefficient of x into two parts such that their su
Now,(-25yz)+9yz=-16yx and (-25yz)×9yz=-225y2z2Replacing the middle term -16xyz by -25xyz+9xyz, we have:15x2-16xyz-15y2z2=15x2-25xyz+9xyz-15y2z2                                        =(
                                        =5x(3x-5yz)+3yz(3x-5yz)                                        =(5x+3yz)(3x-5yz)
Q u e s t i o n : 1 7 7
Resolve each of the following quadratic trinomial into factor:
(x - 2y)
2
 - 5(x - 2y) + 6
S o l u t i o n :
The given expression is a2-5a+6.Assuming a=x-2y, we have: (x-2y)2-5(x-2y)+6=a2-5a+6       (Coefficient of a2=1, coefficient of a=-5 and constant term=6)Now, we will split the coefficient 
Clearly, (-2)+(-3)=-5 and(-2)×(-3)=6Replacing the middle term -5a by -2a-3a, we have:a2-5a+6=a2-2a-3a+6                  =(a2-2a)-(3a-6)                  =a(a-2)-3(a-2)                 =(a-3)(a-2)Replacing a
(a-3)(a-2)=(x-2y-3)(x-2y-2)
Q u e s t i o n : 1 7 8
Resolve each of the following quadratic trinomial into factor:
(2a - b)
2
 + 2(2a - b) - 8
S o l u t i o n :
Assuming x=2a-b, we have: (2a-b)2+2(2a-b)-8=x2+2x-8The given expression becomes x2+2x-8.    (Coefficient of x2=1 and that of x=2 ; constant term=-8)Now, we will split the coefficient o
Clearly,(-2)+4=2 and(-2)×4=-8Replacing the middle term 2x by -2x+4x, we get:x2+2x-8=x2-2x+4x-8                  =(x2-2x)+(4x-8)                  =x(x-2)+4(x-2)                  =(x+4)(x-2)Relacing x by 2
(x+4)(x-2)=(2a-b+4)(2a-b-2)