Polynomials: 9 Math
NCERT Exercise 2.5 Part-3: 9th math
NCERT Exercise 2.5 Polynomials class ninth math Question (10) Factorize each of the following
(i) 27y3 + 125z3
Solution
Given, 27y3 + 125z3
= (3y)3 + (5z)3
Using identity x3 + y3 = (x + y)(x2 – xy +y2), we get
(3y)3 + (5z)3
= (3y + 5z)[(3y)2 – 3y×5z + (5z)2]
= (3y + 5z)(9y2 – 15yz + 25z2) Answer
(ii) 64m3 – 343n3
Solution
Given, 64m3 – 343n3
= (4m)3 – (7n)3
Using identity x3 – y3 = (x – y)(x2 + xy) + y2), we get
(4m)3 – (7n)3
= (4m – 7n) [(4m)2 + 4m × 7n + (7n)2]
= (4m – 7n) (16m2 + 28mn + 49n2) Answer
NCERT Exercise 2.5 Polynomials class ninth math Question (11) Factorize: 27x3 + y3 + z3 – 9xyz
Solution
Given, 27x3 + y3 + z3 – 9xyz
= (3x)3 + y3 + z3 – 3(3x)yz
We know that, x3 + y3 + z3 – 3xyz
Thus, after replacing x = 3x, y = y and z = z, we ge
(3x + y + z)[(3x)2 +y2 + z2 – 3xy – yz – zx]
= (3x + y + z)(9x2 + z2 + z2 – 3zy – yz – zx) Answer
NCERT Exercise 2.5 Polynomials class ninth math Question (12) Verify that x3 + y3 + z3 – 3xyz =1/2(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
Solution
Given, x3 + y3 + z3 – 3xyz =1/2(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
RHS = 1/2(x + y + z) [(x – y)2 + (y – z)2 + (z – x)2]
= 1/2 ( x + y + z)(x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx)
= 1/2× [x(x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx) + y(x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx) + z(x2 + y2 – 2xy + y2 + z2 – 2yz + z2 + x2 – 2zx)]
= 1/2×(x3 + x3 + y3 + y3 + z3 + z3 – 2x2y + x2y +x2y + xy2 + xy2 – 2xy2 + xz2 + xz2 – 2xz2 – 2x2z + x2z + x2z + yz2 +yz2 – 2yz2 – 2y2z + y2z + y2z – 2xyz – 2xyz – 2xyz)
= 1/2(2x3 + 2y3 + 2z3 – 2x2y + 2x2y + 2xy2 – 2xy2 + 2yz2 – 2yz2 – 2y2z + 2y2z – 2xyz – 2xyz – 2xyz)
= 1/2(2x3 + 2y3 + 2z3 – 6xyz)
`= 1/2xx2(x^3+y^3+z^3– 3xyz)`
= x3 + y3 + z3 – 3xyz
= LHS Proved
NCERT Exercise 2.5 Polynomials class ninth math Question (13) If x + y + z = 0, show that x3 + y3 + z3 = 3xyz
Solution
Given, x + y + z = 0
Therefore, show that x3 + y3 + z3 = 3xyz
We know that, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
Now, after replacting (x + y + z) =0 [As given in the question], we get
x3 + y3 + z3 – 3xyz = 0 × (x2 + y2 + z2 – xy – yz – zx)
⇒ x3 + y3 + z3 – 3xyz = 0
⇒ x3 + y3 + z3 = 3xyz Proved
NCERT Exercise 2.5 Polynomials class ninth math Question (14) Without actually calculating the cubes, find the value of each of the following:
(i) (– 12)3 + (7)3 + (5)3
Solution
Given, (– 12)3 + (7)3 + (5)3
Let, x = – 12, y=7 and z = 5
We know that, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
Thus, after replacting the value of x = – 12, y=7 and z = 5, we get
(– 12)3 + (7)3 + (5)3 – 3(– 12)(7)(5) = (– 12 + 7 + 5) (x2 + y2 + z2 – xy – yz – zx)
⇒ (– 12)3 + (7)3 + (5)3 – 3(– 12)(7)(5) = 0 (x2 + y2 + z2 – xy – yz – zx)
⇒ (– 12)3 + (7)3 + (5)3 – 3(– 12)(35) = 0
⇒ (– 12)3 + (7)3 + (5)3 + 36(35) = 0
⇒ (– 12)3 + (7)3 + (5)3 + 1260 = 0
⇒ (– 12)3 + (7)3 + (5)3 = –1260
= –1260 Answer
(ii) (28)3 + (– 15)3 + (– 13)3
Solution
Given, (28)3 + (– 15)3 + (– 13)3
Here, let x = 28, y = – 15 and z = – 13
We know that, x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
Now, after substituting values of x = 28, y = – 15 and z = – 13, we get
(28)3 + (– 15)3 + (– 13)3 – 3(28)(– 13)(– 15) = [28 + (– 13) + (– 15)] (x2 + y2 + z2 – xy – yz – zx)
⇒ (28)3 + (– 15)3 + (– 13)3 – 84(– 13)(– 15) = [28 – 13 – 15] (x2 + y2 + z2 – xy – yz – zx)
⇒ (28)3 + (– 15)3 + (– 13)3 – 84(195) = 0 (x2 + y2 + z2 – xy – yz – zx)
⇒ (28)3 + (– 15)3 + (– 13)3 – 16380 = 0
⇒ (28)3 + (– 15)3 + (– 13)3 = 16380
= 16380 Answer
NCERT Exercise 2.5 Polynomials class ninth math Question (15) Given possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area: 25a2 – 35a + 12
Solution
Given, Area of a rectangle = 25a2 – 35a + 12
Thus, length and breadth = ?
Now, 25a2 – 35a + 12
= 25a2 – 15a + 20a + 12
= 5a(5a – 3) – 4(5a – 3)
= (5a – 3)(5a – 4)
Thus, Length = (5a – 3) and breadth = (5a – 4) Answer
(ii) Area : 35y2 + 13y – 12
Solution
Given, Area of a rectangle = 35y2 + 13y – 12
Thus, length and breadth = ?
Now, 35y2 + 13y – 12
= 35y2 + 28y – 15y – 12
= 7y(5y + 4) – 3(5y + 4)
= (5y + 4) (7y – 3)
Thus, possible length and breadth are (7y – 3) and (5y + 4) Answer
NCERT Exercise 2.5 Polynomials class ninth math Question (16) What are the possible expressions for the dimensions of the cuboids whose volume are given below:
(i) Volume: 3x2 – 12
Solution
Given, volume of a cuboid = 3x2 – 12x
Thus, possible dimenstions i.e. length, breadth and height = ?
Now, 3x2 – 12x
= 3x(4x – 3)
Thus, possible length, breadth and height are 3, x and (4x – 3) Answer
(ii) Volume: 12ky2 + 8 ky – 20k
Solution
Given, Volume of a cuboid = 12ky2 + 8 ky – 20k
Thus, possible dimensions = ?
Now, 12ky2 + 8 ky – 20k
= 4k (3y2 + 2y – 5)
= 4k (3y2 – 3y + 5y – 5)
= 4k [3y (y – 1) + 5(y – 1)]
= 4k (y – 1) (3y + 5)
Thus, possible dimensions of the given cuboid are 4k, (y – 1) and (3y + 5) Answer
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