{"id":623840,"date":"2023-08-31T11:10:44","date_gmt":"2023-08-31T11:10:44","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=623840"},"modified":"2023-08-31T11:17:43","modified_gmt":"2023-08-31T11:17:43","slug":"kc-sinha-exercise-4-4-mathematics-solution-class-9-chapter-4-algebraic-identities","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-4-4-mathematics-solution-class-9-chapter-4-algebraic-identities\/","title":{"rendered":"KC Sinha: Exercise 4.4 – Mathematics Solution Class 9 Chapter 4 Algebraic identities"},"content":{"rendered":"\n\n\n\n\n

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

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

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

Factorize the following if :<\/strong>
(i) (x+1) is a factor of 2x3<\/sup> – 3x2<\/sup> – 8x – 3<\/strong>
Sol :
Step-1\u21d22x3<\/sup>-3x2<\/sup>-8x-3
Step-2\u21d22x3<\/sup>+2x2<\/sup>-5x2<\/sup>-5x-3x-3
Step-3\u21d22x2<\/sup>(x+1)-5x(x+1)-3(x+1)
[After writing step 1 , jumped to 3 and written 3 times (x+1) and multiply it with suitable algebric expression or number so as to get the expression of step 1 . Now step 2 can be written down from step 3]
[Step 4 = Now taking common (x+1) ]
Step-4\u21d2(x+1)(2x2<\/sup>-5x-3)
Step-5\u21d2(x+1)(2x2<\/sup>+x-6x-3) [middle term split]
\u21d2(x+1)[x(2x+1)-3(2x+1)]
\u21d2(x+1)(x-3)(2x+1)<\/p>\n\n\n\n

\n\n\n\n

(ii) (2x+3) is a factor of 4x3<\/sup>+20x2<\/sup>+33x+18<\/strong>
Sol :
\u21d24x3<\/sup>+20x2<\/sup>+33x+18
\u21d24x3<\/sup>+6x2<\/sup>+14x2<\/sup>+21x+12x+18
\u21d22x2<\/sup>(2x+3)+7x(2x+3)+6(2x+3)
[taking common (2x+3)]
\u21d2(2x+3)(2x2<\/sup>+7x+6)
[mid term split]
\u21d2(2x+3)(2x2<\/sup>+4x+3x+6)
\u21d2(2x+3)[2x(x+2)+3(x+2)]
\u21d2(x+2)(2x+3)(2x+3)<\/p>\n\n\n\n

\n\n\n\n

(iii)  (x+9) is a factor of x3<\/sup>+13x2<\/sup>+31x-45<\/strong>
Sol :
\u21d2x3<\/sup>+13x2<\/sup>+31x-45
\u21d2x3<\/sup>+9x2<\/sup>+4x2<\/sup>+36x-5x-45
\u21d2x2<\/sup>(x+9)+4x(x+9)-5(x+9)
[taking common (x+9)]
\u21d2(x+9)[x2<\/sup>+4x-5]
\u21d2(x+9)[x2<\/sup>+4x-5]
\u21d2(x+9)[x2<\/sup>-x+5x-5]
\u21d2(x+9)[x(x-1)+5(x-1)]
\u21d2(x+9)(x+5)(x-1)<\/p>\n\n\n\n

\n\n\n\n

(iv) (x-1) is a factor of 2x3<\/sup>-5x2<\/sup>+x+2<\/strong>
Sol :
\u21d22x3<\/sup>-5x2<\/sup>+x+2
\u21d22x3<\/sup>-2x2<\/sup>-3x2<\/sup>+3x-2x+2
\u21d22x2<\/sup>(x-1)-3x(x-1)-2(x-1)
[taking common (x-1)]
\u21d2(x-1)(2x2<\/sup>-3x-2)
[mid-term split]
\u21d2(x-1)(2x2<\/sup>-4x+x-2)
\u21d2(x-1)[2x(x-2)+1(x-2)]
\u21d2(x-1)(2x+1)(x-2)<\/p>\n\n\n\n

\n\n\n\n

Q2 | Ex-4.4 | Class 9 |Algebraic Identities | KC SINHA Mathematics | myhelper<\/strong><\/p>\n\n\n\n

Page 4.27<\/p>\n\n\n\n

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

Find factors of the following :<\/strong>
(i) x3<\/sup>+2x2<\/sup>-6x+3<\/strong>
Sol :
Here c=3 and factors are \u00b11 , \u00b13
Also , p(1)=13<\/sup>+2\u00d712<\/sup>-6\u00d71+3
=1+2-6+3 = 0
\u2234(x-1) is factor of p(x)
\u21d2x3<\/sup>+2x2<\/sup>-6x+3
\u21d2x3<\/sup>-x2<\/sup>+3x2<\/sup>-3x-3x+3
\u21d2x2<\/sup>(x-1)+3x(x-1)-3(x-1)
[Taking common (x-1)]
\u21d2(x-1)(x2<\/sup>+3x-3)<\/p>\n\n\n\n

\n\n\n\n

(ii) x3<\/sup>-7x+6<\/strong>
Sol :
Here c=6 and factors are \u00b11 , \u00b12 and \u00b13
Also , p(1)=13<\/sup>-7\u00d71+6=0
\u2234(x-1) is factor of p(x)
\u21d2x3<\/sup>-7x+6
\u21d2x3<\/sup>-x2<\/sup>+x2<\/sup>-x-6x+6
\u21d2x2<\/sup>(x-1)+x(x-1)-6(x-1)
[Taking common (x-1)]
\u21d2(x-1)(x2<\/sup>+x-6)
[mid term split]
\u21d2(x-1)(x2<\/sup>-2x+3x-6)
\u21d2(x-1)[x(x-2)+3(x-2)]
\u21d2(x-1)(x+3)(x-2)<\/p>\n\n\n\n

\n\n\n\n

(iii) x3<\/sup>-2x2<\/sup>-x+2<\/strong>
Sol :
Here c=2
Also , the factors are \u00b11 , \u00b12
p(1)=13<\/sup>-2\u00d712<\/sup>-1+2
=1-2-1+2=0
\u2234(x-1) is factor of p(x)
\u21d2x3<\/sup>-2x2<\/sup>-x+2
\u21d2x3<\/sup>-x2<\/sup>-x2<\/sup>+x-2x+2
\u21d2x2<\/sup>(x-1)-x(x-1)-2(x-1)
[Taking common (x-1)]
\u21d2(x-1)[x2<\/sup>-x-2]
\u21d2(x-1)[x2<\/sup>+x-2x-2]
\u21d2(x-1)[x(x+1)-2(x+1)]
\u21d2(x-1)(x-2)(x+1)<\/p>\n\n\n\n

\n\n\n\n

(iv) x3<\/sup>+3x2<\/sup>-4<\/strong>
Sol :
Here c=-4 and numerical value is 4
Also , factors are \u00b11,\u00b12
p(1)=13<\/sup>+3\u00d712<\/sup>-4
=1+3-4=0
\u2234(x-1) is factor of p(x)
\u21d2x3<\/sup>+3x2<\/sup>-4
\u21d2x3<\/sup>-x2<\/sup>+4x2<\/sup>-4x+4x-4
\u21d2x2<\/sup>(x-1)+4x(x-1)+4(x-1)
[Taking common (x-1)]
\u21d2(x-1)[x2<\/sup>+4x+4]
[mid term split]
\u21d2(x-1)[x2<\/sup>+2x+2x+4]
\u21d2(x-1)[x(x+2)+2(x+2)]
\u21d2(x-1)(x+2)(x+2)<\/p>\n\n\n\n

\n\n\n\n

(v) x3<\/sup>-6x+4<\/strong>
Sol :
Here c=4 and factors are \u00b11 , \u00b12
Also , p(2)=23<\/sup>-6\u00d72+4
=8-12+4=0
\u2234(x-2) is factor of p(x)
\u21d2x3<\/sup>-6x+4
\u21d2x3<\/sup>-2x2<\/sup>+2x2<\/sup>-4x-2x+4
\u21d2x2<\/sup>(x-2)+2x(x-2)-2(x-2)
[Taking common (x-2)]
\u21d2(x-2)(x2<\/sup>+2x-2)<\/p>\n\n\n\n

\n\n\n\n

(vi) x3<\/sup>-13x-12<\/strong>
Sol :
Here c=-12 and numerical value is 12
factors are \u00b11,\u00b12,\u00b13,\u00b14,\u00b16,\u00b112
Also , p(-1)=13<\/sup>-13\u00d7-1-12
=-1+13-12=0
\u2234(x+1) factor of p(x)
\u21d2x3<\/sup>-13x-12
\u21d2x3<\/sup>+x2<\/sup>-x2<\/sup>-x-12x-12
\u21d2x2<\/sup>(x+1)-x(x+1)-12(x+1)
[Taking common (x-1)]
\u21d2(x+1)(x2<\/sup>-x-12)
[mid term split]
\u21d2(x+1)(x2<\/sup>+3x-4x-12)
\u21d2(x+1)[x(x+3)-4(x+3)]
\u21d2(x+1)(x+3)(x-4)<\/p>\n\n\n\n

\n\n\n\n

(vii) x3<\/sup>-8x2<\/sup>+17x-10<\/strong>
Sol :
Here c=-10 and numerical value is 10 and factors are \u00b11 , \u00b12 and \u00b15
Clearly , p(1)=13<\/sup>-8\u00d712<\/sup>+17\u00d71-10=0
\u2234(x-1) is a factor of p(x)
Also ,
\u21d2x3<\/sup>-8x2<\/sup>+17x-10
\u21d2x3<\/sup>-x2<\/sup>-7x2<\/sup>+7x+10x-10
\u21d2x2<\/sup>(x-1)-7x(x-1)+10(x-1)
[Taking common (x-1)]
\u21d2(x-1)(x2<\/sup>-7x+10)
[mid term split]
\u21d2(x-1)(x2<\/sup>-5x-2x+10)
\u21d2(x-1)[x(x-5)-2(x-5)]
\u21d2(x-1)(x-2)(x-5)<\/p>\n\n\n\n

\n\n\n\n

(viii) x3<\/sup>+13x2<\/sup>+31x-45<\/strong>
Sol :
Here c=-45 and numerical value is 45 and factors are \u00b11 , \u00b15 and \u00b13
Clearly , p(1)=13<\/sup>+13\u00d712<\/sup>+31\u00d71-45=0
\u2234(x-1) is a factor of p(x)
\u21d2x3<\/sup>+13x2<\/sup>+31x-45
\u21d2x3<\/sup>-x2<\/sup>+14x2<\/sup>-14x+45x-45
\u21d2x2<\/sup>(x-1)+14x(x-1)+45(x-1)
[Taking common (x-1)]
\u21d2(x-1)(x2<\/sup>+14x+45)
[mid term split]
\u21d2(x-1)(x2<\/sup>+9x+5x+45)
\u21d2(x-1)[x(x+9)+5(x+9)]
\u21d2(x-1)(x+5)(x+9)<\/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 Type 6 Question 1 Factorize the following if :(i) (x+1) is a factor of 2×3 – 3×2 – 8x – 3Sol :Step-1\u21d22×3-3×2-8x-3Step-2\u21d22×3+2×2-5×2-5x-3x-3Step-3\u21d22×2(x+1)-5x(x+1)-3(x+1)[After writing step 1 , jumped to 3 and written 3 times (x+1) and multiply it with suitable algebric expression or number so as to get the expression of step 1 . […]<\/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-623840","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-9","entry"],"yoast_head":"\nKC Sinha: Exercise 4.4 - Mathematics Solution Class 9 Chapter 4 Algebraic identities - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Exercise 4.3 Type 6 Question 1 Factorize the following if :(i) (x+1) is a factor of 2x3 - 3x2 - 8x - 3Sol\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-4-4-mathematics-solution-class-9-chapter-4-algebraic-identities\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"KC Sinha: Exercise 4.4 - 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