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KC Sinha Mathematics Solution Class 9 Chapter 4 Algebraic identities exercise 4.3

Page 4.21

Exercise 4.3


Type 4

Problem based on f‌inding factors of trinomial polynomial of the form ax2+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, ax2+bx+c=ax2+(p+q)x+c=(ax2+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) x2+14x+45

Sol :

On comparing with general formula

ax2+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

⇒x2+14x+45

⇒x2+5x+9x+45

⇒x(x+5)+9(x+5)

⇒(x+9)(x+5)

 


(ii) x2+11x+24

Sol :

On comparing with general formula

ax2+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

⇒x2+11x+24

⇒x2+3x+8x+24

⇒x(x+3)+8(x+3)

⇒(x+3)(x+8)


(iii) x2-5x+6

Sol :

On comparing with general formula

ax2+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

⇒x2-5x+6

⇒x2-3x-2x+6

⇒x(x-3)-2(x-3)

⇒(x-2)(x-3)


(iv) x2-22x+120

Sol :

On comparing with general formula

ax2+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

⇒x2-22x+120

⇒x2-12x-10x+120

⇒x(x-12)-10(x+12)

⇒(x-10)(x-12)


(v) x2+6x-40

Sol :

On comparing with general formula

ax2+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

⇒x2+6x-40

⇒x2+10x-4x-40

⇒x(x+10)-4(x+10)

⇒(x+10)(x-4)


(vi) x2+8x-48

Sol :

On comparing with general formula

ax2+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

⇒x2+8x-48

⇒x2+12x-4x-48

⇒x(x+12)-4(x+12)

⇒(x+12)(x-4)


(vii) y2-4y-21

Sol :

On comparing with general formula

ax2+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

⇒y2-4y-21

⇒y2-7y+3y-21

⇒y(y-7)+3(y-7)

⇒(y-7)(y+3)


(viii) x2-5x-14

Sol :

On comparing with general formula

ax2+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

⇒x2-5x-14

⇒x2-7x+2x-14

⇒x(x-7)+2(x-7)

⇒(x-7)(x+2)


Question 2

Factorize the following :

(i) p2-8p-65

Sol :

On comparing with general formula

ax2+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

⇒p2-8p-65

⇒p2-13p+5p-65

⇒p(p-13)+5(p-13)

⇒(p-13)(p+5)


(ii) x2-x-132

Sol :

On comparing with general formula

ax2+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

⇒x2-x-132

⇒x2-12x+11x-132

⇒x(x-12)+11(x-12)

⇒(x-12)(x+11)


(iii) p2+3p-108

Sol :

On comparing with general formula

ax2+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

⇒p2+3p-108

⇒p2+12p-9p-108

⇒p(p+12)-9(p-12)

⇒(p+12)(p-9)


(iv) 14-3a-5a2

Sol :

On comparing with general formula

ax2+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-5a2

⇒-5a2-3a+14

⇒-5a2-10a+7a+14

⇒-5a(a+2)+7(a+2)

⇒(a+2)(7-5a)


(v) 35-2b-b2

Sol :

⇒-b2-2b+35

On comparing with general formula

ax2+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

⇒-b2-2b+35

⇒-b2-7b+5b+35

⇒-b(b+7)+5(b+7)

⇒(5-b)(7+b)


(vi) 96-4b-b2

Sol :

⇒-b2-4b+96

On comparing with general formula

ax2+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

⇒-b2-4b+96

⇒-b2-12b+8b+96

⇒-b(b+12)+8(b+12)

⇒(12+b)(8-b)


Question 3

Factorize the following :

(i) 2x2+3x+1

Sol :

On comparing with general formula

ax2+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

⇒2x2+3x+1

⇒2x2+2x+x+1

⇒2x(x+1)+1(x+1)

⇒(2x+1)(x+1)


(ii) 3x2+4x+1

Sol :

On comparing with general formula

ax2+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

⇒3x2+4x+1

⇒3x2+3x+x+1

⇒3x(x+1)+1(x+1)

⇒(3x+1)(x+1)


(iii) 2x2+7x+3

Sol :

On comparing with general formula

ax2+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

⇒2x2+7x+3

⇒2x2+6x+x+3

⇒2x(x+3)+1(x+3)

⇒(2x+1)(x+3)


(iv) 6u2+17u+12

Sol :

On comparing with general formula

ax2+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

⇒6u2+17u+12

⇒6u2+9u+8u+12

⇒3u(2u+3)+4(2u+3)

⇒(2u+3)(3u+4)


(v) 3u2-10u+8

Sol :

On comparing with general formula

ax2+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

⇒3u2-10u+8

⇒3u2-6u-4u+8

⇒3u(u-2)-4(u-2)

⇒(u-2)(3u-4)


(vi) 12x2-25x+12

Sol :

On comparing with general formula

ax2+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

⇒12x2-25x+12

⇒12x2-16x-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) (p2+2p2q)

Sol :

Common factor p2

⇒p2(1+2q)


(ii) y5-5y2

Sol :

Common factor y2

⇒y2(y3-5)


(iii) a3b3-a2b2+2ab

Sol :

Common factor ab

⇒ab(a2b2-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-r2

Sol :

Common factors q , r

⇒pq+qr-pr-r2

⇒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) (a2-b2)c+(b2-c2)a

Sol :

⇒(a2-b2)c+(b2-c2)a

⇒ca2-cb2+ab2-ac2

⇒ca2+b2(-c+a)-ac2

⇒ca2-ac2+b2(-c+a)

⇒ac(a-c)+b2(a-c)

⇒(a-c)(ac+b2)


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