{"id":658,"date":"2020-09-07T11:52:56","date_gmt":"2020-09-07T06:22:56","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=658"},"modified":"2021-07-02T15:03:45","modified_gmt":"2021-07-02T09:33:45","slug":"kseeb-solutions-class-9-maths-chapter-4-ex-4-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-4-ex-4-4\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.4.<\/p>\n

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.4.<\/h2>\n

Question 1.
\nDetermine which of the following polynomials has (x+1) a factor :
\ni) x3<\/sup> + x2<\/sup> + x + 1
\nii) x4<\/sup> + x3<\/sup> + x2<\/sup> + x + 1
\niii) x4<\/sup> + 3x3<\/sup> + 3x2<\/sup> + x + 1
\niv) x3<\/sup> – x2<\/sup> – (2 + \\(\\sqrt{2}\\) )x + \\(\\sqrt{2}\\)
\nAnswer:
\ni) x- 1 is a factor of p(x)
\nx + 1 = x – a
\na = -1
\nFor the value of p(a),
\nvalue of r(x) = 0.
\n\u2234 p(x) = x3<\/sup> + x2<\/sup> + x + 1
\np(-1) = (-1)3<\/sup> + (-1)2<\/sup> + (-1) + 1
\n= -1 + 1 -1 + 1
\np(-1) = 0
\nHere, p(a) = r(x) = 0
\n\u2234 x + 1 is a factor.<\/p>\n

ii) If x + 1 = x – a, then a = -1
\np(x) = x4<\/sup> + x3<\/sup> + x2<\/sup> + x + 1
\np(-1)= (-1)4<\/sup> + (-1)3<\/sup> +(-1)2<\/sup> + (-1)+ 1
\n= 1 – 1 + 1 – 1 + 1
\n= 3 – 2 p(-1)= 1
\nHere, r(x) = p(a)= 1 Reminder is not zero.
\n\u2234 x+1 is not a factor.<\/p>\n

iii) If x + 1 = x – a then
\na = -1
\np(x) = x4<\/sup> + 3x3<\/sup>\u00a0+ 3x2<\/sup> + x + 1
\np(-1) = (-1)4<\/sup> + 3(-1 )3<\/sup> +3(-1 )2<\/sup> + (-1) + 1
\n= 1 + 3(-1) + 3(1) + 1(-1) + 1
\n= 1- 3 + 3 – 1 + 1
\n= 5 – 4
\nP(-1)= 1
\nHere, r(x) = p(a)=l Remainder is not zero
\n\u2234 x+1 is not a factor.<\/p>\n

iv) If x + 1 = x – a then,
\na = -1
\np(x) = x3<\/sup> – x2<\/sup> – (2 + \\(\\sqrt{2}\\))x+ \\(\\sqrt{2}\\)
\np(-1) = (-1)3<\/sup> – (-1)2<\/sup> -(2 + \\(\\sqrt{2}\\))(-1) + \\(\\sqrt{2}\\)
\n= -1 -(+1) – (2 – \\(\\sqrt{2}\\))+ \\(\\sqrt{2}\\)
\n= -1 – 1 + 2 + \\(\\sqrt{2}\\) + \\(\\sqrt{2}\\)
\n= -2 + 2 + \\(2 \\sqrt{2}\\)
\n= = + \\(2 \\sqrt{2}\\)
\np(-1) = \\(2 \\sqrt{2}\\)
\nHere, r(x) = p(a) = \\(2 \\sqrt{2}\\) Value of remainder r(x) is not zero.
\n\u2234 x + 1 is not a factor.<\/p>\n

Question 2.
\nUse the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases :
\ni) p(x) = 2x3<\/sup> + x2<\/sup> – 2x – 1, g(x) = x+1
\nii) p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1, g(x) = x + 2
\niii) p(x) = x3<\/sup> – 4x2<\/sup> + x + 6, g(x) = x – 3
\nAnswer:
\ni) p(x) = 2x3<\/sup> + x2<\/sup> – 2x – 1
\ng(x) = x + 1
\nIf x + 1 = 0, then x = -1
\np(x) = 2x3<\/sup> + x2<\/sup>\u00a0– 2x – 1
\np(-1)= 2(-1)3<\/sup> + (-1)2<\/sup> -2(- 1) – 1
\n= 2(-1) + (1) + 2 – 1
\n= -2 + 1 + 2 – 1
\nP(-1)= 0
\nHere, r(x) = p(a) = 0,
\n\u2234 g(x) is the a factor f(x)<\/p>\n

ii) p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\ng(x) = x + 2
\nIf x + 2 = 0, then
\nx = -2
\n\u2234 p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\np(-2) = (-2)3<\/sup> + 3(-2)2<\/sup> + 3 (-2) + 1
\n= -8 + 3(4) + 3(-2) + 1
\n= -8 + 12 – 6 + 1
\n= 13 – 24
\np(-2)= -11
\nHere we have r(x) = p(a) =-11.
\nValue of r(x) is not equal to zero.
\n\u2234 g(x) is not a factor of f(x).<\/p>\n

iii) p(x) = x3<\/sup> – 4x2<\/sup> + x + 6
\ng(x) = x – 3
\nLet, x – 3 = 0, then
\nx = 3
\np(x) = x3<\/sup> – 4x2<\/sup> + x + 6
\np(3) = (3)2<\/sup> – 4(3)2<\/sup> + 3 + 6
\n= 27 – 4(9) + 3 + 6
\n= 27 – 36 + 3 + 6
\n= 36 – 36
\np(3) = 0
\nHere, we have r(x) = p(a) = 0
\n\u2234 (x – 3) is the factor of p(x).<\/p>\n

Question 3.
\nFind the value of k, if x – 1 is a factor of p(x) in each of the following cases :
\ni) p(x) = x2<\/sup> + x + k
\nii) p(x) = 2x2<\/sup> + kx + \\(\\sqrt{2}\\)
\niii) p(x) = kx2<\/sup> – \\(\\sqrt{2} \\mathrm{x}\\) + 1
\niv) p(x) = kx2<\/sup> – 3x + k
\nAnswer:
\ni) p(x) = x2<\/sup> + x + k
\ng(x) = x – 1
\nk = ?
\nIf x – 1 = 0, then
\nx = 1
\np(x) = x2<\/sup> + x + k
\np(1) = (1)2<\/sup> + 1 + k
\np( 1) = 1 + 1 + k
\np( 1) = 2 + k
\nIf g(x) is a factor, then we have r(x) = 0
\n\u2234 p(1) = 0
\n2 + k= 0
\n\u2234 k = 0 – 2
\nk = -2<\/p>\n

ii) p(x) = 2x2<\/sup> + kx + \\(\\sqrt{2}\\)
\ng(x) = x – 1 k = ?
\nIf x – 1 = 0, then x = 1
\np(x) = 2x2<\/sup> + kx + \\(\\sqrt{2}\\)
\np(1) = 2(1)2<\/sup> + k(1) + \\(\\sqrt{2}\\)
\n= 2(1) + k(l) + \\(\\sqrt{2}\\)
\np(1) = 2 + k + \\(\\sqrt{2}\\)
\nIf (x – 1) is the factor of p(x), then we have p(1) = 0.
\n\u2234 2 + k + \\(\\sqrt{2}\\) = 0
\nk = -2 – \\(\\sqrt{2}\\)<\/p>\n

iii) p(x) = kx2<\/sup> – \\(\\sqrt{2} x\\) + 1
\ng(x) = x – 1 k = ?
\nIf x – 1 = 0, then
\nx = 1
\np(x) = kx2<\/sup> – \\(\\sqrt{2} x\\) + 1
\np(1) = k(1)2<\/sup>– \\(\\sqrt{2}\\)(1) + 1
\n= k( 1) – \\(\\sqrt{2}\\) + 1
\np(1) = k- \\(\\sqrt{2}\\) + 1
\nIf (x – 1) is the factor of p(x) then we have p(1) = 0.
\n\u2234 p(1) = k – \\(\\sqrt{2}\\) + 1 = 0
\n\u2234 k= \\(\\sqrt{2}\\) – 1<\/p>\n

iv) p(x) = kx2<\/sup> – 3x + k
\ng(x) = x – 1 k = ?
\nIf x – 1 = 0, then x – 1
\np(x) = kx2<\/sup> – 3x + k
\np(1) = k(1)2<\/sup> – 3(1) + k
\n= k(1) – 3(1) + k
\n= k – 3 + k
\np(1) = 2k – 3
\nIf (x – 1) is the factor of p(x), then we have p(1) = 0.<\/p>\n

Question 4.
\nFactorise :
\ni) 12x2<\/sup> – 7x + 1
\nii) 2x2<\/sup> + 7x + 3
\niii) 6x2<\/sup> + 5x – 6
\niv) 3x2<\/sup> – x – 4
\nAnswer:
\ni) 12x2<\/sup> – 7x + 1
\n\"KSEEB
\n= 12x2<\/sup> – 4x – 3x + 1
\n= 4x (3x – 1) – 1(3x – 1)
\n= (3x – 1) (4x – 1)<\/p>\n

ii) 2x2<\/sup> + 7x + 3
\n\"KSEEB
\n= 2x2<\/sup> + 6x + x + 3
\n= 2x(x + 3) + 1 (x + 3)
\n= (x + 3) (2x + 1)<\/p>\n

iii) 6x2<\/sup> + 5x – 6
\n\"KSEEB
\n= 6x2<\/sup> + 9x – 4x – 6
\n= 3x(2x + 3) – 2(2x + 3)
\n= (2x + 3) (3x – 2)<\/p>\n

iv) 3x2<\/sup> – x – 4
\n\"KSEEB
\n= 3x2<\/sup> – 4x + 3x – 4
\n=x(3x – 4) + 1 (3x – 4)
\n= (3x – 4) (x + 1)<\/p>\n

Question 5.
\nFactorise :
\ni) x2<\/sup> – 2x2<\/sup> – x + 2
\nii) x2<\/sup> – 3x2<\/sup> – 9x – 5
\niii) x3<\/sup> + 13x2<\/sup> + 32x + 20
\niv) 2y3<\/sup> + y2<\/sup> – 2y – 1
\nAnswer:
\ni) x3<\/sup> – 2x2<\/sup> – x + 2
\n= x2<\/sup>(x – 2) – 1 (x – 2)
\n= (x – 2) (x2<\/sup> – 1)
\n= (x – 2) (x + 1) (x- 1)<\/p>\n

ii) x3<\/sup> – 3x2<\/sup> – 9x – 5
\n= x3<\/sup> – 5x2<\/sup> + 2x2<\/sup> – 10x + x – 5
\n= x2<\/sup>(x – 5) + 2x(x – 5) + 1 (x – 5)
\n= (x – 5) (x2<\/sup> + 2x + 1)
\n= (x – 5) {x2<\/sup> + x + x + 1}
\n= (x – 5) (x(x + 1) + 1(x + 1)}
\n= (x- 5)(x + 1) (x + 1)<\/p>\n

iii) x3<\/sup> + 13x2<\/sup> + 32x + 20
\n= x3<\/sup> + 10x2<\/sup> + 3x2<\/sup> + 30x + 2x + 26
\n= x2<\/sup>(x + 10) + 3x(x + 10) + 2(x + 10)
\n= (x + 10) (x2<\/sup> + 3x + 2)
\n= (x + 10) {x2<\/sup> + 2x + x + 2)
\n= (x + 10) {x(x + 2) + 1 (x + 2))
\n= (x + 10) (x + 2) (x + 1)<\/p>\n

iv) 2y3<\/sup> + y2<\/sup> – 2y – 1
\n= y2<\/sup>(2y + 1) – 1(2y + 1)
\n= (2y + 1) (y2<\/sup>– 1)
\n= (2y + 1) {(y)2<\/sup> – (1)2<\/sup>}
\n= (2y+ 1) (y + 1) (y- 1)<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.4, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.4 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.4. Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.4. Question 1. Determine which of the following polynomials has (x+1) a …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/658"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=658"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/658\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=658"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=658"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=658"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}