Chapter 2 Polynomials Ex 2.5 Solutions
Question - 11 : - Factorise: 27x3+y3+z3–9xyz
Answer - 11 : -
The expression27x3+y3+z3–9xyz canbe written as (3x)3+y3+z3–3(3x)(y)(z)
27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z)
We know that, x3+y3+z3–3xyz =(x+y+z)(x2+y2+z2–xy –yz–zx)
27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z)
= (3x+y+z)[(3x)2+y2+z2–3xy–yz–3xz]
= (3x+y+z)(9x2+y2+z2–3xy–yz–3xz)
Question - 12 : - Verify that:
x3+y3+z3–3xyz= (1/2) (x+y+z)[(x–y)2+(y–z)2+(z–x)2]
Answer - 12 : -
We know that,
x3+y3+z3−3xyz = (x+y+z)(x2+y2+z2–xy–yz–xz)
⇒ x3+y3+z3–3xyz= (1/2)(x+y+z)[2(x2+y2+z2–xy–yz–xz)]
= (1/2)(x+y+z)(2x2+2y2+2z2–2xy–2yz–2xz)
= (1/2)(x+y+z)[(x2+y2−2xy)+(y2+z2–2yz)+(x2+z2–2xz)]
=(1/2)(x+y+z)[(x–y)2+(y–z)2+(z–x)2]
Question - 13 : - If x+y+z= 0, show that x3+y3+z3 = 3xyz.
Answer - 13 : -
We know that,
x3+y3+z3-3xyz = (x +y+z)(x2+y2+z2–xy–yz–xz)
Now, according to the question, let (x+y+z) = 0,
then, x3+y3+z3 -3xyz =(0)(x2+y2+z2–xy–yz–xz)
⇒ x3+y3+z3–3xyz= 0
⇒ x3+y3+z3 =3xyz
Hence Proved
Question - 14 : - Without actuallycalculating the cubes, find the value of each of the following:
(i) (−12)3+(7)3+(5)3
(ii) (28)3+(−15)3+(−13)3
Answer - 14 : -
(i) (−12)3+(7)3+(5)3
Let a = −12
b = 7
c = 5
We know that if x+y+z = 0, then x3+y3+z3=3xyz.
Here, −12+7+5=0
(−12)3+(7)3+(5)3 = 3xyz
= 3×-12×7×5
= -1260
(ii) (28)3+(−15)3+(−13)3
Solution:
(28)3+(−15)3+(−13)3
Let a = 28
b = −15
c = −13
We know that if x+y+z = 0, then x3+y3+z3 =3xyz.
Here, x+y+z = 28–15–13 = 0
(28)3+(−15)3+(−13)3 = 3xyz
= 0+3(28)(−15)(−13)
=16380
Question - 15 : - Give possibleexpressions for the length and breadth of each of the following rectangles, inwhich their areas are given:
(i) Area : 25a2–35a+12
(ii) Area: 35y2+13y–12
Answer - 15 : -
(i) Area : 25a2–35a+12
Using the splitting the middle term method,
We have to find a number whose sum = -35 and product =25×12=300
We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20=300]
25a2–35a+12 = 25a2–15a−20a+12
= 5a(5a–3)–4(5a–3)
= (5a–4)(5a–3)
Possible expression for length = 5a–4
Possible expression for breadth = 5a –3
(ii) Area : 35y2+13y–12
Using the splitting the middle term method,
We have to find a number whose sum = 13 and product = 35×-12 =420
We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=420]
35y2+13y–12 = 35y2–15y+28y–12
= 5y(7y–3)+4(7y–3)
= (5y+4)(7y–3)
Possible expression for length = (5y+4)
Possible expression for breadth = (7y–3)
Question - 16 : - What are thepossible expressions for the dimensions of the cuboids whose volumes are givenbelow?
(i) Volume: 3x2–12x
(ii) Volume: 12ky2+8ky–20k
Answer - 16 : -
(i) Volume : 3x2–12x
3x2–12x can be written as 3x(x–4) by taking 3xout of both the terms.
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x–4)
(ii) Volume:
12ky2+8ky–20k
12ky2+8ky–20k can be written as 4k(3y2+2y–5)by taking 4k out of both the terms.
12ky2+8ky–20k = 4k(3y2+2y–5)
[Here, 3y2+2y–5can be written as 3y2+5y–3y–5 using splitting the middle termmethod.]
= 4k(3y2+5y–3y–5)
= 4k[y(3y+5)–1(3y+5)]
= 4k(3y+5)(y–1)
Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y -1)