{"id":1693,"date":"2020-09-07T09:06:16","date_gmt":"2020-09-07T03:36:16","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1693"},"modified":"2021-07-02T15:03:45","modified_gmt":"2021-07-02T09:33:45","slug":"kseeb-solutions-class-9-maths-chapter-4-ex-4-5","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-4-ex-4-5\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.5"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.5<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.5.<\/p>\n

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.5<\/h2>\n

Question 1.
\nUse suitable identities to find the following products :
\n(i) (x + 4) (x + 10)
\n(ii) (x + 8) (x – 10)
\n(iii) (3x + 4) (3x – 5)
\n(iv) \\(\\left(\\mathrm{y}^{2}+\\frac{3}{2}\\right)\\left(\\mathrm{y}^{2}-\\frac{3}{2}\\right)\\)
\n(v) (3 – 2x) (3 + 2x)
\nSolution:
\n(i) (x + 4) (x + 10)
\nSuitable identity is (x + a) (x + b) = x2<\/sup> + (a + b)x + ab
\n(x + 4) (x + 10) = x2<\/sup> + (4 + 10)x + (4 \u00d7 10)
\n= x2<\/sup> + 14x + 40<\/p>\n

(ii) (x + 8) (x- 10)
\nSuitable identity is (x + a) (x + b) = x2<\/sup> + (a + b)x + ab
\n(x + 8) (x – 10)= (x)2<\/sup> + (8 – 10)x + (8 \u00d7 -10)
\n= x2<\/sup> + (-2)x + (-80)
\n= x2<\/sup> – 2x – 80<\/p>\n

(iii) (3x + 4) (3x – 5)
\nSuitable identity is (x + a) (x + b) = x2<\/sup> + (a + b)x + ab
\n(3x + 4) (3x – 5) = (3x)2<\/sup> + (+4 – 5)3x + (4 x -5)
\n= 9x2<\/sup> + (-1)3x + (-20)
\n= 9x2<\/sup> – 3x – 20<\/p>\n

(iv) \\(\\left(\\mathrm{y}^{2}+\\frac{3}{2}\\right)\\left(\\mathrm{y}^{2}-\\frac{3}{2}\\right)\\)
\nSuitable identity is (a + b) (a – b) = a2<\/sup> – b2<\/sup>
\n\\(\\begin{aligned}\\left(y^{2}+\\frac{3}{2}\\right)\\left(y^{2}-\\frac{3}{2}\\right) &=\\left(y^{2}\\right)^{2}-\\left(\\frac{3}{2}\\right)^{2} \\\\ &=y^{4}-\\frac{9}{4} \\end{aligned}\\)<\/p>\n

(v) (3 – 2x) (3 + 2x)
\nSuitable identity is (a + b) (a – b) = a2<\/sup> – b2<\/sup>
\n(3 – 2x) (3 + 2x) = (3)2<\/sup> – (2x)2<\/sup>
\n= 9 – 4x2<\/sup>.<\/p>\n

Question 2.
\nEvaluate the following products without mulitplying directly :
\n(i) 103 \u00d7 107
\n(ii) 95 \u00d7 96
\n(iii) 104 \u00d7 96
\nSolution:
\n(i) 103 \u00d7 107
\n= (100 + 3) (100 – 3)
\nAs per Identity,
\n(a + b)(a – b) = a2<\/sup> – b2<\/sup>
\n(100 + 3)(100 – 3) = (100)2 <\/sup>– (3)2<\/sup>
\n= 10000 – 9
\n= 9991<\/p>\n

(ii) 95 \u00d7 96
\n= (100 – 5) (100 + 6)
\nAs per Identity.
\n(x + a) (x + b) = x2<\/sup> + (a + b)x + ab
\n(100 – 5) (100 + 6) = (100)2 <\/sup>+ (-5 + 6)100 + (-5 \u00d7 6)
\n= 10000 + (1)100 + (-30)
\n= 10000 + 100 – 30
\n= 10100 – 30
\n= 10070<\/p>\n

(iii) 104 \u00d7 96
\n= (100 + 4) (100 – 4)
\nAs per Identity,
\n(a + b) (a – b) = a2<\/sup> – b2<\/sup>
\n(100 + 4) (100 – 4) = (100)2<\/sup> – (4)2<\/sup>
\n= 10000 – 16
\n= 9984<\/p>\n

Question 3.
\nFactorise the following using appropriate identities :
\n(i) 9x2<\/sup> + 6xy + y2<\/sup>
\n(ii) 4y2<\/sup> – 4y + 1
\n(iii) \\(x^{2}-\\frac{y^{2}}{100}\\)
\nSolution:
\n(i) 9x2<\/sup> + 6xy + y2<\/sup>
\n= (3x)2<\/sup> + 2(3x)(y) + (y)2<\/sup>
\n= (3x + y)2<\/sup> [\u2235 x2<\/sup> + 2xy + y2<\/sup> = (x + y)2<\/sup>]<\/p>\n

(ii) 4y2<\/sup> – 4y + 1
\n= (2y)2<\/sup> – 2(2y)(1) + (1)2<\/sup>
\n= (2y – 1)2<\/sup> [\u2235 x2<\/sup> – 2xy + y2<\/sup> = (x – y)2<\/sup>]<\/p>\n

(iii) \\(x^{2}-\\frac{y^{2}}{100}\\)
\n\"KSEEB<\/p>\n

Question 4.
\nExpand each of the following, using suitable identities :
\n(i) (x + 2y + 4z)2<\/sup>
\n(ii) (2x – y + z)2<\/sup>
\n(iii) (-2x + 3y + 2z)2<\/sup>
\n(iv) (3a – 7b – c)2<\/sup>
\n(v) (-2x + 5y – 3z)2<\/sup>
\n(vi) \\(\\left(\\frac{1}{4} a-\\frac{1}{2} b+1\\right)^{2}\\)
\nSolution:
\n(i) (x + 2y + 4z)2<\/sup>
\n(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca
\n(x + 2y + 4z)2<\/sup> =(x)2<\/sup> + (2y)2<\/sup> + (4z)2<\/sup> + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x)
\n= x2<\/sup> + 4y2<\/sup> + 16z2<\/sup> + 4xy + 16yz + 8zx<\/p>\n

(ii) (2x – y + z)2<\/sup>
\n(x + y + z)2<\/sup> = x2<\/sup> + y2<\/sup> + z2<\/sup> + 2xy + 2yz + 2zx (2x – y + z)2<\/sup>= (2x)2<\/sup> + (-y)2<\/sup> + (z)2<\/sup> + 2(2x)(-y) + 2(-y)(z) + 2(z)(2x)
\n= 4x2<\/sup> + y2<\/sup> + z2<\/sup> – 4xy – 2yz + 4zx.<\/p>\n

(iii) (-2x + 3y + 2z)2<\/sup>
\n(x + y + z)2<\/sup> = x2<\/sup> + y2<\/sup> + z2<\/sup> + 2xy + 2yz + 2zx (-2x + 3y + 2z)2
\n<\/sup>= (-2x)2<\/sup> + (3y)2<\/sup>2 + (2z)2<\/sup> + 2(-2x)(3y) + 2(3y)(2z) + 2(2z)(-2x)
\n= 4x2<\/sup> + 9x2<\/sup> + 4z2<\/sup> – 12xy + 12yz – 8zx<\/p>\n

(iv) (3a – 7b – c)2<\/sup>
\n(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> + 2ab + 2bc + 2ca (3a – 7b – c)2<\/sup> =(3a)2<\/sup> + (-7b)2<\/sup> + (-c)2<\/sup> + 2(3a)(-7b)
\n+ 2(-7b)(-c) + 2(-c)(3a) = 9a2<\/sup> + 49b2<\/sup> + c2<\/sup> – 42ab – 14bc – 6ca.<\/p>\n

(v) (-2x + 5y – 3z)2<\/sup>
\n(x + y + z)2<\/sup> = x2<\/sup> + y2<\/sup> + z2<\/sup> + 2xy + 2yz + 2zx (-2x + 5y – 3z)2<\/sup> = (-2x)2<\/sup> + (5y)2<\/sup> + (-3z)2<\/sup> + 2(-2x)(5y) + 2(5y)(-3z) + 2(-3z)(-2x) = 4x2<\/sup> + 25y2<\/sup> + 9z2<\/sup> – 20xy – 30yz + 12zx<\/p>\n

(vi) \\(\\left(\\frac{1}{4} a-\\frac{1}{2} b+1\\right)^{2}\\)
\n(a + b + c)2<\/sup> = a2<\/sup> + b2<\/sup> + c2<\/sup> +2ab + 2bc + 2ca
\n\"KSEEB<\/p>\n

Question 5.
\nFactorise :
\n(i) 4x2<\/sup> + 9y2<\/sup> + 16z2<\/sup> + 12xy – 24yz – 16xz
\n(ii) 2x2<\/sup> + y2<\/sup> + 8z2<\/sup> – 2\\(\\sqrt{2}\\)xy + 4\\(\\sqrt{2}\\)yz- 8xz
\nSolution:
\n(i) 4x2<\/sup> + 9y2<\/sup> + 16z2<\/sup>+ 12xy – 24yz – 16xz
\n= (2x)2<\/sup>+ (3y)2<\/sup> + (-4z)2<\/sup> + 2(2x)(3y) + 2(3y)(-4z) + 2(-4z)(2x)
\n= (2x + 3y – 4z)2<\/sup>.<\/p>\n

(ii) 2x2<\/sup> + y2<\/sup> + 8z2<\/sup> – 2\\(\\sqrt{2}\\)xy + 4\\(\\sqrt{2}\\)yz- 8xz
\n\"KSEEB<\/p>\n

Question 6.
\nWrite the following cubes in expanded form:
\n(i) (2x + 1)3<\/sup>
\n(ii) (2a – 3b)3<\/sup>
\n(iii) \\(\\left[\\frac{3}{2} x+1\\right]^{3}\\)
\n(iv) \\(\\left[\\mathrm{x}-\\frac{2}{3} \\mathrm{y}\\right]^{3}\\)
\nSolution:
\n(i) (2x + 1)3<\/sup>
\nAs per Identity,
\n(a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3ab(a + b)
\n(2x + 1)3<\/sup> = (2x)3<\/sup> + (1)3<\/sup> + 3(2x)(1)(2x + 1)
\n= 8x3<\/sup> + 1 + 6x(2x + 1)
\n= 8x3<\/sup> + 1 + 12x2<\/sup> + 6x
\n(2x + 1)3<\/sup> = 8x3<\/sup> + 12x2<\/sup> + 6x + 1.<\/p>\n

(ii) (2a – 3b)3<\/sup>
\nAs per Identity,
\n(a – b)3<\/sup> = a3<\/sup> – b3<\/sup> – 3ab(a – b)
\n(2a – 3b)3<\/sup>= (2a)3<\/sup>– (3b)3<\/sup> – 3(2a)(3b)(2a-3b)
\n= 8a3<\/sup> – 27b3<\/sup> – 18ab(2a – 3b)
\n(2a – 3b)3<\/sup>= 8a3<\/sup> – 27b3<\/sup> – 36a2<\/sup>b + 54ab2<\/sup><\/p>\n

(iii) \\(\\left[\\frac{3}{2} x+1\\right]^{3}\\)
\nAs per Identity,
\n(a + b)3<\/sup> = a3<\/sup> – b3<\/sup> – 3ab(a + b)
\n\"KSEEB<\/p>\n

(iv) \\(\\left[\\mathrm{x}-\\frac{2}{3} \\mathrm{y}\\right]^{3}\\)
\nAs per Identity,
\n(a – b)3<\/sup> = a3<\/sup> – b3<\/sup> – 3ab(a – b)
\n\"KSEEB<\/p>\n

Question 7.
\nEvaluate the following using suitable identities :
\n(i) (99)3<\/sup>
\n(ii) (102)3<\/sup>
\n(iii) (998)3<\/sup>
\nSolution:
\n(i) (99)3<\/sup> = (100 – 1)3<\/sup>
\nAs per Identity,
\n(a – b)3<\/sup> = a3<\/sup> – b3<\/sup> – 3ab(a – b)
\n(100 – 1)3<\/sup> = (100)3<\/sup> – (1)3<\/sup> – 3(100)(1)(100- 1)
\n(99)3<\/sup> = 1000000 – 1 – 300(100 – 1)
\n= 1000000 – 1 – 30000 + 300
\n= 970299
\n\u2234 (99)3<\/sup> = 970299<\/p>\n

(ii) (102)3<\/sup> = (100 + 2)3<\/sup>
\nAs per Identity,
\n(a + b)3<\/sup> = a3<\/sup> + b3<\/sup> + 3ab(a + b)
\n(100 + 2)3<\/sup> = (100)3<\/sup> + (2)3<\/sup> + 3( 100)(2)( 100 + 2)
\n= 1000000 + 8 + 600( 100 + 2)
\n= 1000000 + 8 + 60000 + 1200
\n(100 + 2)3<\/sup> = 1061208
\n\u2234 (102)3<\/sup>= 1061208<\/p>\n

(iii) (998)3<\/sup> = (1000 – 2)3<\/sup>
\nAs per Identity,
\n(a – b)3<\/sup> = a3<\/sup> – b3<\/sup> – 3ab(a – b)
\n(1000 – 2)3<\/sup> = (1000)3<\/sup> – (2)3 – 3( 1000) (2)(1000 – 2)
\n= 1000000000 – 8 – 6000(1000 – 1)
\n= 1000000000 – 8 – 6000000 + 6000
\n= 994005992
\n\u2234 (998)3<\/sup> = 994005992<\/p>\n

Question 8.
\nFactorise each of the following :
\n(i) 8a3<\/sup> + b3<\/sup> + 12a2<\/sup>b + 6ab2<\/sup>
\n(ii) 8a3<\/sup> – b3<\/sup> – 12a2<\/sup>b + 6ab2<\/sup>
\n(iii) 27- 125a3<\/sup>– 135a + 225a2<\/sup>
\n(iv) 64a3<\/sup> – 27b3<\/sup> – 144a2<\/sup>b + 108ab2<\/sup>
\n(v) \\(27 p^{3}-\\frac{1}{216}-\\frac{9}{2} p^{2}+\\frac{1}{4} p\\)
\nSolution:
\n(i) 8a3<\/sup> + b3<\/sup> + 12a2<\/sup>b + 6ab2<\/sup>
\n= (2a)3<\/sup> + (b)3<\/sup> + 3(2a)(b)(2a + b)
\n= (2a + b)3<\/sup> \u2235 a3<\/sup> + b3<\/sup> + 3ab(a + b) = (a + b)3<\/sup><\/p>\n

(ii) 8a3<\/sup> – b3<\/sup> – 12ab2<\/sup> + 6ab3<\/sup>
\n= (2a)3<\/sup> + (-b)3<\/sup> + 3(2a)(-b)(2a – b)
\n= (2a – b)3<\/sup> \u2235 a3<\/sup> – b3<\/sup> – 3ab(a – b) = (a – b)3<\/sup><\/p>\n

(iii) 27 – 125a3<\/sup> – 135a + 225a2<\/sup>
\nOR 125a3<\/sup> – 225a2<\/sup> + 135a – 27
\n= (5a)3<\/sup> + (-3)3<\/sup> + 3(5a)(-3)(5a – 3)
\n= (5a – 3)3<\/sup> \u2235 a3<\/sup> – b3<\/sup> – 3ab(a – b) = (a – b)3<\/sup><\/p>\n

(iv) 64a3<\/sup> – 27b3<\/sup> – 144a2<\/sup>b + 108ab2<\/sup>
\n= (4a)3<\/sup> + (-3b)3<\/sup> + 3(4a)(-3b)(4a – 3b)
\n= (4a-3b)3<\/sup> \u2235 a3<\/sup> – b3<\/sup> – 3ab(a – b) = (a – b)3<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 9.
\nVerify :
\n(i) x3<\/sup> + y3<\/sup> = (x + y) (x2<\/sup> – xy + y2<\/sup>)
\n(ii) x3<\/sup> – y3<\/sup> = (x – y) (x2<\/sup> + xy + y2<\/sup>)
\nSolution:
\n(i) x3<\/sup> + y3<\/sup> = (x + y)(x2<\/sup> – xy + y2<\/sup>)
\nL.H.S. = x3<\/sup> + y3<\/sup>
\nR.H.S. = (x + y)(x2<\/sup> – xy + y2<\/sup>)
\n= x(x2<\/sup> – xy + y2<\/sup>) + y(x2<\/sup> – xy + y2<\/sup>)
\n= x3<\/sup> – x2<\/sup>y + xy2<\/sup> + x2<\/sup>y – xy2<\/sup> + y3<\/sup>
\nR.H.S. = x3<\/sup> + y3<\/sup>
\n\u2234 L.H.S. = R.H.S.
\n\u2234 x3<\/sup> + y3<\/sup> = (x + y)(x2<\/sup> – xy + y2<\/sup>)<\/p>\n

(ii) x3<\/sup> – y3<\/sup> = (x – y)(x2<\/sup> + xy + y2<\/sup>)
\nL.H.S. = x3<\/sup> – y3<\/sup>
\nR.H.S. = (x – y)(x2<\/sup> + xy + y2<\/sup>)
\n= x(x2<\/sup> + xy + y2<\/sup>) – y(x2<\/sup> + xy + y2<\/sup>)
\n= x3<\/sup> + x2<\/sup>y + xy2<\/sup> – x2<\/sup>y – xy3<\/sup> – y3<\/sup>
\nR.H.S. = x3<\/sup> – y3<\/sup>
\n\u2234L.H.S. = R.H.S.
\n\u2234 x3<\/sup> – y3<\/sup> = (x – y)(x2<\/sup> + xy + y2<\/sup>)<\/p>\n

Question 10.
\nFactorise each of the following :
\n(i) 27y3<\/sup> + 125z3<\/sup>
\n(ii) 64m3<\/sup> – 343n3<\/sup>
\n[Hint: See Question 9).
\nSolution:
\n(i) 27y3<\/sup> + 125z3<\/sup>
\n= (3y)3<\/sup> + (5z)3<\/sup>
\n= (3y + 5z) {(3y)2<\/sup> – (3y)(5z) + (5z)2<\/sup>}
\n= (3y + 5z) (9y2<\/sup> – 15yz + 25z2<\/sup>)<\/p>\n

(ii) 64m3<\/sup> – 343n3<\/sup>
\n= (4m)3<\/sup> – (7n)3<\/sup>
\n= (4m – 7n) {(4m)2<\/sup> + (4m) (7n) + (7n)2<\/sup>}
\n= (4m – 7n) (17m2<\/sup> + 28mn + 49n2<\/sup>)<\/p>\n

Question 11.
\nFactorise : 27x3<\/sup> + y3<\/sup> + z3<\/sup> – 9xyz
\nSolution:
\n27x3<\/sup> + y3<\/sup> + z3<\/sup> – 9xyz
\n= (3x)3<\/sup> + (y)33<\/sup> + (z)3<\/sup> – 3(3x)(y)(z)
\n= (3x + y + z) {(3x)2<\/sup> + (y)2<\/sup> + (z)2<\/sup> – (3x)(y) – (y)(z) – (z)(3x)}
\n= (3x + y + z) (9x2<\/sup> + y2<\/sup> + z2<\/sup> – 3xy – yz – 3zx)<\/p>\n

Question 12.
\nVerify that x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = \\(\\frac{1}{2}\\) (x + y + z)[(x – y)2<\/sup> + (y – z)2<\/sup> + (z – x)2<\/sup>]
\nSolution:
\nL.H.S.= x3<\/sup> + y3<\/sup>+ z3<\/sup> – 3xyz
\nR.H.S. = \\(\\frac{1}{2}\\) (x+y+z) [(x – y)2<\/sup> + (y – z)2<\/sup> + (z – x)2<\/sup>]
\n= \\(\\frac{1}{2}\\) (x+y+z) [x2<\/sup> 2xy + y2<\/sup>+ y2<\/sup> – 2yz + z2<\/sup> + z2<\/sup> -2zx + x2<\/sup>]
\n= \\(\\frac{1}{2}\\) (x+y+z) [2x2<\/sup>+ 2y2<\/sup>+ 2z2<\/sup>– 2xy – 2yz – 2zx]
\n= \\(\\frac{1}{2}\\) (x+y+z) \u00d7 (x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\n= (x + y + z)(x2<\/sup> + y2<\/sup> + z2<\/sup> – xy – yz – zx)
\nR.H.S.= x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz
\n\u2234 L.H.S. = R.H.S.
\n\u2234 x3<\/sup> + y3<\/sup> + z3<\/sup> – 3xyz = \\(\\frac{1}{2}\\) (x + y + z)[(x – y)2<\/sup> + (y – z)2<\/sup> + (z – x)3<\/sup>]<\/p>\n

Question 13.
\nIf x + y + z = 0, show that x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz.
\nSolution:
\nx + y + z = 0
\nx + y = -z
\nby cubing on both sides,
\n(x + y)3<\/sup> = (-z)3
\nx3<\/sup>+ y3<\/sup> + 3xy(x + y) = -z3<\/sup>
\nx3<\/sup> + y3<\/sup> + 3xy (-z) = -z3<\/sup>
\n(\u2235 x + y =-z data given)
\nx3<\/sup> + y3<\/sup> – 3xyz = -z3<\/sup>
\n\u2234 x3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz.<\/p>\n

Question 14.
\nWithout actually calculating the cubes, find the value of each of the following:
\n(i) (-12)3<\/sup> + (7)3<\/sup> + (5)3<\/sup>
\n(ii) (28)3<\/sup> + (-15)3<\/sup> + (-13)3<\/sup>
\nSolution:
\n(i) (-12)3<\/sup> + (7)3<\/sup> + (5)3<\/sup>
\nLet -12 = x, 7 = y, 5 = z, then
\nx3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz.
\n\u2234 (-12)3<\/sup> + (7)3<\/sup> + (5)3
\n<\/sup>= 3(-12)(7)(5)
\n= -36 \u00d7 35
\n= -1260.<\/p>\n

(ii) (28)3<\/sup> + (-15)3<\/sup> + (-13)3<\/sup>
\nLet x = 28, y = -15, z = -13, then
\nx3<\/sup> + y3<\/sup> + z3<\/sup> = 3xyz
\n\u2234 (28)3<\/sup> + (-15)3<\/sup> + (-13)3
\n<\/sup>= 3(28)(- 15)(-13)
\n= 84 \u00d7 195
\n= 16380<\/p>\n

Question 15.
\nGive possible expressions for the length and breadth of each of the following rectangles, in which their areas are given :
\n(i) Area: 25a2<\/sup> – 35a +12
\n(ii) Area : 35y2<\/sup> + 13y – 12
\nSolution:
\n(i) Area: 25a2<\/sup> – 35a + 12
\n= 25a2<\/sup> – 20a – 15a + 12
\n= 5a(5a – 4) – 3(5a – 4)
\n= (5a -4) (5a – 3)
\nlength \u00d7 breadth = Area of rectangle
\n(5a – 4) (5a – 3) = 25a2<\/sup> – 35a + 12
\n\u2234 Length = (5a – 4)
\nBreadth = (5a – 3)<\/p>\n

(ii) Area : 35y2<\/sup> + 13y – 12
\nArea of Rectangle – Length \u00d7 Breadth
\n35y2<\/sup> + 13y – 12
\n35y2<\/sup> + 28y – 15y
\n7y(5y + 4) – 3(5y + 4)
\n(7y – 3) (5y + 4)
\n\u2234 Length = (7y – 3)
\nBreadth = (5y + 4)<\/p>\n

Question 16.
\nWhat are the possible expressions for the dimensions of the cuboids whose volumes are given below :
\n(i) Volume: 3x2<\/sup> – 12x
\n(ii) Volume: 12ky2<\/sup> + 8ky – 20k
\nSolution:
\n(i) length \u00d7 breadth \u00d7 height = Volume of Cuboid
\n= 3x2<\/sup> – 12x
\n= 3x2<\/sup> – 12x
\n= 3x(x – 4)
\n= 3 \u00d7 x \u00d7 (x – 4)
\nl \u00d7 b \u00d7 h
\n\u2234 Length = 3
\nBreadth = x
\nHeight = (x – 4)<\/p>\n

(ii) length \u00d7 breadth \u00d7 height = Volume of Cuboid
\n= 12ky2<\/sup> + 8ky – 20k
\n12ky2<\/sup> + 8ky – 20k
\n= 4k(3y2<\/sup> + 2y – 5)
\n= 4k (3y2<\/sup> + 5y – 3y – 5)
\n= 4k (y(3y + 5) – 1 (3y + 5))
\n= 4k \u00d7 (3y + 5) \u00d7 (y – 1)
\n= l \u00d7 b \u00d7 h
\n\u2234 Length = 4k
\nBreadth = (3y + 5)
\nHeight = (y – 1)<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.5 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.5, 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 4.5 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.5. Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.5 Question 1. Use suitable identities to find the following products : …<\/p>\n","protected":false},"author":6,"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\/1693"}],"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\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1693"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1693\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1693"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1693"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1693"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}