{"id":623838,"date":"2023-08-31T11:16:01","date_gmt":"2023-08-31T11:16:01","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=623838"},"modified":"2023-08-31T11:16:06","modified_gmt":"2023-08-31T11:16:06","slug":"kc-sinha-exercise-4-3-mathematics-solution-class-9-chapter-4-algebraic-identities-2","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-4-3-mathematics-solution-class-9-chapter-4-algebraic-identities-2\/","title":{"rendered":"KC Sinha: Exercise 4.3 – Mathematics Solution Class 9 Chapter 4 Algebraic identities"},"content":{"rendered":"\n\n\n\n\n

Exercise 4.3<\/h2>\n\n\n\n

Question 1<\/h4>\n\n\n\n

Factorize the following :<\/strong>
(i) x2<\/sup>+14x+45<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=14,c=45
where k=ac=1\u00d745=45(positive number) and b is also positive
factors of 45=5\u00d79 Also b=+5+9=14
\u21d2x2<\/sup>+14x+45
\u21d2x2<\/sup>+5x+9x+45
\u21d2x(x+5)+9(x+5)
\u21d2(x+9)(x+5)<\/p>\n\n\n\n

\n\n\n\n

(ii) x2<\/sup>+11x+24<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=11,c=24
where k=ac=1\u00d724=24(positive number) and b is also positive
factors of 24=3\u00d78 Also b=3+8=11
\u21d2x2<\/sup>+11x+24
\u21d2x2<\/sup>+3x+8x+24
\u21d2x(x+3)+8(x+3)
\u21d2(x+3)(x+8)<\/p>\n\n\n\n

\n\n\n\n

(iii) x2<\/sup>-5x+6<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=-5,c=6
where k=ac=1\u00d76=6(positive number) and b is negative
factors of 6=2\u00d73 Also b=-3-2=-5
\u21d2x2<\/sup>-5x+6
\u21d2x2<\/sup>-3x-2x+6
\u21d2x(x-3)-2(x-3)
\u21d2(x-2)(x-3)<\/p>\n\n\n\n

\n\n\n\n

(iv) x2<\/sup>-22x+120<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=-22,c=120
where k=ac=1\u00d7120=120(positive number) and b is negative
factors of 120=12\u00d710 Also b=-12-10=-22
\u21d2x2<\/sup>-22x+120
\u21d2x2<\/sup>-12x-10x+120
\u21d2x(x-12)-10(x-12)
\u21d2(x-10)(x-12)<\/p>\n\n\n\n

\n\n\n\n

(v) x2<\/sup>+6x-40<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=+6,c=-40
where k=ac=1\u00d7-40=-40(negative) and b is positive
factors of 40=4\u00d710 Also b=+10-4=6
\u21d2x2<\/sup>+6x-40
\u21d2x2<\/sup>+10x-4x-40
\u21d2x(x+10)-4(x+10)
\u21d2(x+10)(x-4)<\/p>\n\n\n\n

\n\n\n\n

(vi) x2<\/sup>+8x-48<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=8,c=-48
where k=ac=1\u00d7-48=-48(negative) and b is positive
factors of 48=4\u00d712 Also b=+12-4=8
\u21d2x2<\/sup>+8x-48
\u21d2x2<\/sup>+12x-4x-48
\u21d2x(x+12)-4(x+12)
\u21d2(x+12)(x-4)<\/p>\n\n\n\n

\n\n\n\n

(vii) y2<\/sup>-4y-21<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=-4,c=-21
where k=ac=1\u00d7-21=-21(negative) and b is negative
factors of 21=3\u00d77 Also b=-7+3=-4
\u21d2y2<\/sup>-4y-21
\u21d2y2<\/sup>-7y+3y-21
\u21d2y(y-7)+3(y-7)
\u21d2(y-7)(y+3)<\/p>\n\n\n\n

\n\n\n\n

(viii) x2<\/sup>-5x-14<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=-5,c=-14
where k=ac=1\u00d7-14=-14(negative) and b is negative
factors of 14=2\u00d77 Also b=-7+2=-5
\u21d2x2<\/sup>-5x-14
\u21d2x2<\/sup>-7x+2x-14
\u21d2x(x-7)+2(x-7)
\u21d2(x-7)(x+2)<\/p>\n\n\n\n

\n\n\n\n

Q2 | Ex-4.3 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 2<\/h4>\n\n\n\n

Factorize the following :<\/strong>
(i) p2<\/sup>-8p-65<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=-8,c=-65
where k=ac=1\u00d7-65=-65(negative) and b is negative
factors of 65=5\u00d713 also b=-13+5=-8
\u21d2p2<\/sup>-8p-65
\u21d2p2<\/sup>-13p+5p-65
\u21d2p(p-13)+5(p-13)
\u21d2(p-13)(p+5)<\/p>\n\n\n\n

\n\n\n\n

(ii) x2<\/sup>-x-132<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=-1,c=-132
where k=ac=1\u00d7-132=-132(negative) and b is negative
factors of 132=11\u00d712 also b=-12+11=-1
\u21d2x2<\/sup>-x-132
\u21d2x2<\/sup>-12x+11x-132
\u21d2x(x-12)+11(x-12)
\u21d2(x-12)(x+11)<\/p>\n\n\n\n

\n\n\n\n

(iii) p2<\/sup>+3p-108<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=1,b=3,c=-108
where k=ac=1\u00d7-108=-108(negative) and b is positive
factors of 108=12\u00d79 also b=+12-9=+3
\u21d2p2<\/sup>+3p-108
\u21d2p2<\/sup>+12p-9p-108
\u21d2p(p+12)-9(p+12)
\u21d2(p+12)(p-9)<\/p>\n\n\n\n

\n\n\n\n

(iv) 14-3a-5a2<\/sup><\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=-5,b=-3,c=+14
where k=ac=-5\u00d714=-70(negative) and b is negative
factors of 70=7\u00d710 also b=-10+7=-3
\u21d214-3a-5a2<\/sup>
\u21d2-5a2<\/sup>-3a+14
\u21d2-5a2<\/sup>-10a+7a+14
\u21d2-5a(a+2)+7(a+2)
\u21d2(a+2)(7-5a)<\/p>\n\n\n\n

\n\n\n\n

(v) 35-2b-b2<\/sup><\/strong>
Sol :
\u21d2-b2<\/sup>-2b+35
On comparing with general formula
ax2<\/sup>+bx+c
we get a=-1,b=-2,c=+35
where k=ac=-1\u00d735=-35(negative) and b is negative
factors of 35=7\u00d75 also b=-7+5=-2
\u21d2-b2<\/sup>-2b+35
\u21d2-b2<\/sup>-7b+5b+35
\u21d2-b(b+7)+5(b+7)
\u21d2(5-b)(7+b)<\/p>\n\n\n\n

\n\n\n\n

(vi) 96-4b-b2<\/sup><\/strong>
Sol :
\u21d2-b2<\/sup>-4b+96
On comparing with general formula
ax2<\/sup>+bx+c
we get a=-1,b=-4,c=+96
where k=ac=-1\u00d796=-96(negative) and b is negative
factors of 96=12\u00d78 also b=-12+8=-4
\u21d2-b2<\/sup>-4b+96
\u21d2-b2<\/sup>-12b+8b+96
\u21d2-b(b+12)+8(b+12)
\u21d2(b+12)(8-b)<\/p>\n\n\n\n

\n\n\n\n

Q3 | Ex-4.3 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 3<\/h4>\n\n\n\n

Factorize the following :<\/strong>
(i) 2x2<\/sup>+3x+1<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=2,b=3,c=1
where k=ac=2\u00d71=2(positive) and b is positive
factors of 2=1\u00d72 also b=+2+1=3
\u21d22x2<\/sup>+3x+1
\u21d22x2<\/sup>+2x+x+1
\u21d22x(x+1)+1(x+1)
\u21d2(2x+1)(x+1)<\/p>\n\n\n\n

\n\n\n\n

(ii) 3x2<\/sup>+4x+1<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=3,b=4,c=1
where k=ac=3\u00d71=3(positive) and b is positive
factors of 3=1\u00d73 also b=+3+1=4
\u21d23x2<\/sup>+4x+1
\u21d23x2<\/sup>+3x+x+1
\u21d23x(x+1)+1(x+1)
\u21d2(3x+1)(x+1)<\/p>\n\n\n\n

\n\n\n\n

(iii) 2x2<\/sup>+7x+3<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=2,b=7,c=3
where k=ac=2\u00d73=6(positive) and b is positive
factors of 6=6\u00d71 also b=+6+1=7
\u21d22x2<\/sup>+7x+3
\u21d22x2<\/sup>+6x+x+3
\u21d22x(x+3)+1(x+3)
\u21d2(2x+1)(x+3)<\/p>\n\n\n\n

\n\n\n\n

(iv) 6u2<\/sup>+17u+12<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=6,b=17,c=12
where k=ac=6\u00d712=72(positive) and b is positive
factors of 72=9\u00d78 also b=+9+8=17
\u21d26u2<\/sup>+17u+12
\u21d26u2<\/sup>+9u+8u+12
\u21d23u(2u+3)+4(2u+3)
\u21d2(2u+3)(3u+4)<\/p>\n\n\n\n

\n\n\n\n

(v) 3u2<\/sup>-10u+8<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=3,b=-10,c=8
where k=ac=3\u00d78=24(positive) and b is negative
factors of 24=6\u00d74 also b=-6-4=-10
\u21d23u2<\/sup>-10u+8
\u21d23u2<\/sup>-6u-4u+8
\u21d23u(u-2)-4(u-2)
\u21d2(u-2)(3u-4)<\/p>\n\n\n\n

\n\n\n\n

(vi) 12x2<\/sup>-25x+12<\/strong>
Sol :
On comparing with general formula
ax2<\/sup>+bx+c
we get a=12,b=-25,c=12
where k=ac=12\u00d712=144(positive) and b is negative
factors of 144=16\u00d79 also b=-16-9=-25
\u21d212x2<\/sup>-25x+12
\u21d212x2<\/sup>-16x-9x+12
\u21d24x(3x-4)-3(3x-4)
\u21d2(3x-4)(4x-3)<\/p>\n\n\n\n

\n\n\n\n

Type 5
Problems based on factorization by taking out common factor from each term of given expression.<\/strong>
WORKING RULE:<\/strong>
1. Observe the given expression attentively and f\u200cind out, which number or algebraic term or expression is a common factor in each term of the given expression.<\/strong>
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).<\/strong><\/p>\n\n\n\n

\n\n\n\n

Q4 | Ex-4.3 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 4<\/h4>\n\n\n\n

Factorize :<\/strong>
(i) (p2<\/sup>+2p2<\/sup>q)<\/strong>
Sol :
Common factor p2<\/sup>
\u21d2p2<\/sup>(1+2q)<\/p>\n\n\n\n

\n\n\n\n

(ii) y5<\/sup>-5y2<\/sup><\/strong>
Sol :
Common factor y2<\/sup>
\u21d2y2<\/sup>(y3<\/sup>-5)<\/p>\n\n\n\n

\n\n\n\n

(iii) a3<\/sup>b3<\/sup>-a2<\/sup>b2<\/sup>+2ab<\/strong>
Sol :
Common factor ab
\u21d2ab(a2<\/sup>b2<\/sup>-ab+2)<\/p>\n\n\n\n

\n\n\n\n

Type 6
<\/strong><\/p>\n\n\n\n

Q5 | Ex-4.3 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 5<\/h4>\n\n\n\n

Factorize :<\/strong>
(i) ax-bx-ay+by<\/strong>
Sol :
Common factors x , y
\u21d2x(a-b)-y(a-b)
\u21d2(a-b)(x-y)<\/p>\n\n\n\n

\n\n\n\n

(ii) pq+qr-pr-r2<\/sup><\/strong>
Sol :
Common factors q , r
\u21d2pq+qr-pr-r2<\/sup>
\u21d2q(p+r)-r(p+r)
\u21d2(p+r)(q-r)<\/p>\n\n\n\n

\n\n\n\n

(iii) (x-y)2<\/sup>+2x-2y<\/strong>
Sol :
\u21d2(x-y)2<\/sup>+2x-2y
\u21d2(x-y)2<\/sup>+2(x-y)
Common factor (x-y)
\u21d2(x-y)(x-y+2)<\/p>\n\n\n\n

\n\n\n\n

(iv) (a2<\/sup>-b2<\/sup>)c+(b2<\/sup>-c2<\/sup>)a<\/strong>
Sol :
\u21d2(a2<\/sup>-b2<\/sup>)c+(b2<\/sup>-c2<\/sup>)a
\u21d2ca2<\/sup>-cb2<\/sup>+ab2<\/sup>-ac2<\/sup>
\u21d2ca2<\/sup>+b2<\/sup>(-c+a)-ac2<\/sup>
\u21d2ca2<\/sup>-ac2<\/sup>+b2<\/sup>(-c+a)
\u21d2ac(a-c)+b2<\/sup>(a-c)
\u21d2(a-c)(ac+b2<\/sup>)<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 9 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

  Exercise 4.3 Question 1 Factorize the following :(i) x2+14x+45Sol :On comparing with general formulaax2+bx+cwe get a=1,b=14,c=45where k=ac=1\u00d745=45(positive number) and b is also positivefactors of 45=5\u00d79 Also b=+5+9=14\u21d2x2+14x+45\u21d2x2+5x+9x+45\u21d2x(x+5)+9(x+5)\u21d2(x+9)(x+5) (ii) x2+11x+24Sol :On comparing with general formulaax2+bx+cwe get a=1,b=11,c=24where k=ac=1\u00d724=24(positive number) and b is also positivefactors of 24=3\u00d78 Also b=3+8=11\u21d2x2+11x+24\u21d2x2+3x+8x+24\u21d2x(x+3)+8(x+3)\u21d2(x+3)(x+8) (iii) x2-5x+6Sol :On comparing with general formulaax2+bx+cwe […]<\/p>\n","protected":false},"author":302,"featured_media":623835,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[921],"tags":[],"boards":[],"class_list":["post-623838","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-9","entry"],"yoast_head":"\nKC Sinha: Exercise 4.3 - 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