Page 4.21

### Exercise 4.3

**Type 4**

**Problem based on finding factors of trinomial polynomial of the form ax ^{2}+bx+c**

**Working Rule:**

**1. Take out any variable or constant which is common factor in the given trinomial and write the remaining expression in the bracket.**

**2. Find a×c and let this be k . Now , find two factors p and q of k such that their sum is b . Now, ax ^{2}+bx+c=ax^{2}+(p+q)x+c=(ax^{2}+px)+(qx+c) take out common factor in each bracket.**

**3. How to find p and q.**

**(i) If k is positive, factors p and q will have same Sign as that of b. That is, if b is positive. p and q should be taken as positive. If b is negative p and q should be taken as negative.**

**(ii) If k is negative, find two factors p and q of |k|. Bigger factor of |k| will have sign same as that of b and its smaller factor will have Sign opposite to that of b**

#### Question 1

**Factorize the following :**

**(i) x ^{2}+14x+45**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=14,c=45

where k=ac=1×45=45(positive number) and b is also positive

factors of 45=5×9 Also b=+5+9=14

⇒x^{2}+14x+45

⇒x^{2}+5x+9x+45

⇒x(x+5)+9(x+5)

⇒(x+9)(x+5)

**(ii) x ^{2}+11x+24**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=11,c=24

where k=ac=1×24=24(positive number) and b is also positive

factors of 24=3×8 Also b=3+8=11

⇒x^{2}+11x+24

⇒x^{2}+3x+8x+24

⇒x(x+3)+8(x+3)

⇒(x+3)(x+8)

**(iii) x ^{2}-5x+6**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=-5,c=6

where k=ac=1×6=6(positive number) and b is negative

factors of 6=2×3 Also b=-3-2=-5

⇒x^{2}-5x+6

⇒x^{2}-3x-2x+6

⇒x(x-3)-2(x-3)

⇒(x-2)(x-3)

**(iv) x ^{2}-22x+120**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=-22,c=120

where k=ac=1×120=120(positive number) and b is negative

factors of 120=12×10 Also b=-12-10=-22

⇒x^{2}-22x+120

⇒x^{2}-12x-10x+120

⇒x(x-12)-10(x+12)

⇒(x-10)(x-12)

**(v) x ^{2}+6x-40**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=+6,c=-40

where k=ac=1×-40=-40(negative) and b is positive

factors of 40=4×10 Also b=+10-4=6

⇒x^{2}+6x-40

⇒x^{2}+10x-4x-40

⇒x(x+10)-4(x+10)

⇒(x+10)(x-4)

**(vi) x ^{2}+8x-48**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=8,c=-48

where k=ac=1×-48=-48(negative) and b is positive

factors of 48=4×12 Also b=+12-4=8

⇒x^{2}+8x-48

⇒x^{2}+12x-4x-48

⇒x(x+12)-4(x+12)

⇒(x+12)(x-4)

**(vii) y ^{2}-4y-21**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=-4,c=-21

where k=ac=1×-21=-21(negative) and b is negative

factors of 21=3×7 Also b=-7+3=-4

⇒y^{2}-4y-21

⇒y^{2}-7y+3y-21

⇒y(y-7)+3(y-7)

⇒(y-7)(y+3)

**(viii) x ^{2}-5x-14**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=-5,c=-14

where k=ac=1×-14=-14(negative) and b is negative

factors of 14=2×7 Also b=-7+2=-5

⇒x^{2}-5x-14

⇒x^{2}-7x+2x-14

⇒x(x-7)+2(x-7)

⇒(x-7)(x+2)

#### Question 2

**Factorize the following :**

**(i) p ^{2}-8p-65**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=-8,c=-65

where k=ac=1×-65=-65(negative) and b is negative

factors of 65=5×13 also b=-13+5=-8

⇒p^{2}-8p-65

⇒p^{2}-13p+5p-65

⇒p(p-13)+5(p-13)

⇒(p-13)(p+5)

**(ii) x ^{2}-x-132**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=-1,c=-132

where k=ac=1×-132=-132(negative) and b is negative

factors of 132=11×12 also b=-12+11=-1

⇒x^{2}-x-132

⇒x^{2}-12x+11x-132

⇒x(x-12)+11(x-12)

⇒(x-12)(x+11)

**(iii) p ^{2}+3p-108**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=1,b=3,c=-108

where k=ac=1×-108=-108(negative) and b is positive

factors of 108=12×9 also b=+12-9=+3

⇒p^{2}+3p-108

⇒p^{2}+12p-9p-108

⇒p(p+12)-9(p-12)

⇒(p+12)(p-9)

**(iv) 14-3a-5a ^{2}**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=-5,b=-3,c=+14

where k=ac=-5×14=-70(negative) and b is negative

factors of 70=7×10 also b=-10+7=-3

⇒14-3a-5a^{2}

⇒-5a^{2}-3a+14

⇒-5a^{2}-10a+7a+14

⇒-5a(a+2)+7(a+2)

⇒(a+2)(7-5a)

**(v) 35-2b-b ^{2}**

Sol :

⇒-b^{2}-2b+35

On comparing with general formula

ax^{2}+bx+c

we get a=-1,b=-2,c=+35

where k=ac=-1×35=-35(negative) and b is negative

factors of 35=7×5 also b=-7+5=-2

⇒-b^{2}-2b+35

⇒-b^{2}-7b+5b+35

⇒-b(b+7)+5(b+7)

⇒(5-b)(7+b)

**(vi) 96-4b-b ^{2}**

Sol :

⇒-b^{2}-4b+96

On comparing with general formula

ax^{2}+bx+c

we get a=-1,b=-4,c=+96

where k=ac=-1×96=-96(negative) and b is negative

factors of 96=12×8 also b=-12+8=-4

⇒-b^{2}-4b+96

⇒-b^{2}-12b+8b+96

⇒-b(b+12)+8(b+12)

⇒(12+b)(8-b)

#### Question 3

**Factorize the following :**

**(i) 2x ^{2}+3x+1**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=2,b=3,c=1

where k=ac=2×1=2(positive) and b is positive

factors of 2=1×2 also b=+2+1=3

⇒2x^{2}+3x+1

⇒2x^{2}+2x+x+1

⇒2x(x+1)+1(x+1)

⇒(2x+1)(x+1)

**(ii) 3x ^{2}+4x+1**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=3,b=4,c=1

where k=ac=3×1=3(positive) and b is positive

factors of 3=1×3 also b=+3+1=4

⇒3x^{2}+4x+1

⇒3x^{2}+3x+x+1

⇒3x(x+1)+1(x+1)

⇒(3x+1)(x+1)

**(iii) 2x ^{2}+7x+3**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=2,b=7,c=3

where k=ac=2×3=6(positive) and b is positive

factors of 6=6×1 also b=+6+1=7

⇒2x^{2}+7x+3

⇒2x^{2}+6x+x+3

⇒2x(x+3)+1(x+3)

⇒(2x+1)(x+3)

**(iv) 6u ^{2}+17u+12**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=6,b=17,c=12

where k=ac=6×12=72(positive) and b is positive

factors of 72=9×8 also b=+9+8=17

⇒6u^{2}+17u+12

⇒6u^{2}+9u+8u+12

⇒3u(2u+3)+4(2u+3)

⇒(2u+3)(3u+4)

**(v) 3u ^{2}-10u+8**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=3,b=-10,c=8

where k=ac=3×8=24(positive) and b is negative

factors of 24=6×4 also b=-6-4=-10

⇒3u^{2}-10u+8

⇒3u^{2}-6u-4u+8

⇒3u(u-2)-4(u-2)

⇒(u-2)(3u-4)

**(vi) 12x ^{2}-25x+12**

Sol :

On comparing with general formula

ax^{2}+bx+c

we get a=12,b=-25,c=12

where k=ac=12×12=144(positive) and b is negative

factors of 144=16×9 also b=-16-9=-25

⇒12x^{2}-25x+12

⇒12x^{2}-16x-9x+12

⇒4x(3x-4)-3(3x-4)

⇒(3x-4)(4x-3)

Type 5

**Problems based on factorization by taking out common factor from each term of given expression.**

**WORKING RULE:**

**1. Observe the given expression attentively and find out, which number or algebraic term or expression is a common factor in each term of the given expression.**

**2. Take this common factor outside the bracket and write within bracket the expression (quotient, obtained after dividing each term by the above common factor).**

#### Question 4

**Factorize :**

**(i) (p ^{2}+2p^{2}q)**

Sol :

Common factor p^{2}

⇒p^{2}(1+2q)

**(ii) y ^{5}-5y^{2}**

Sol :

Common factor y^{2}

⇒y^{2}(y^{3}-5)

**(iii) a ^{3}b^{3}-a^{2}b^{2}+2ab**

Sol :

Common factor ab

⇒ab(a^{2}b^{2}-ab+2)

Type 6

#### Question 5

**Factorize :**

**(i) ax-bx-ay+by**

Sol :

Common factors x , y

⇒x(a-b)-y(a-b)

⇒(a-b)(x-y)

**(ii) pq+qr-pr-r ^{2}**

Sol :

Common factors q , r

⇒pq+qr-pr-r^{2}

⇒q(p+r)-r(p+r)

⇒(p+r)(q-r)

**(iii) (x-y) ^{2}+2x-2y**

Sol :

⇒(x-y)^{2}+2x-2y

⇒(x-y)^{2}+2(x-y)

Common factor (x-y)

⇒(x-y)(x-y+2)

**(iv) (a ^{2}-b^{2})c+(b^{2}-c^{2})a**

Sol :

⇒(a^{2}-b^{2})c+(b^{2}-c^{2})a

⇒ca^{2}-cb^{2}+ab^{2}-ac^{2}

⇒ca^{2}+b^{2}(-c+a)-ac^{2}

⇒ca^{2}-ac^{2}+b^{2}(-c+a)

⇒ac(a-c)+b^{2}(a-c)

⇒(a-c)(ac+b^{2})