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

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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)

Q2 | Ex-4.3 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

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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)
⇒(b+12)(8-b)

Q3 | Ex-4.3 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

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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).

Q4 | Ex-4.3 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

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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

Q5 | Ex-4.3 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper

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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)