NCERT Solutions for Exercise 2.5 Class 9 Maths Chapter 2 Polynomials

# NCERT Solutions for Exercise 2.5 Class 9 Maths Chapter 2 Polynomials

Edited By Vishal kumar | Updated on Oct 30, 2023 11:23 AM IST

## NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.5 - Download Free PDF

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials: Exercise 2.5 - NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.5 introduces us to many identities which are covered in the whole exercise. An algebraic identity is an algebraic equation that is true for all values of the variables occurring in it. NCERT solutions for Class 9 Maths chapter 2 exercise 2.5 includes a variety of problems related to the application of all the algebraic identities in the question. In exercise 2.5 Class 9 Maths a lot of problems which also includes real-life application.

Overall NCERT solutions for Class 9 Maths chapter 2 exercise 2.5 is the most important exercise as it is the base for the most important branch of mathematics called algebra. Exercise 2.5 Class 9 Maths includes the identities with two variables as well as three variables.

The 9th class maths exercise 2.5 answers have been meticulously prepared by subject experts and are presented in a comprehensive and easily understandable manner. This exercise consists of a total of sixteen questions, each containing multiple parts. Students can readily access these class 9 maths chapter 2 exercise 2.5 in PDF format, allowing them to use them offline without requiring an internet connection, and they are made available free of charge. Additionally, along with exercise 2.5 class 9 maths, you'll also find Exercise 2.5 and other related exercises in the curriculum.

## Access Polynomials Class 9 Chapter 2 Exercise: 2.5

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put $a = 4 \ \ and \ \ b = 10$

$(x+4)(x+10)= x^2+(10+4)x+10\times 4$

$= x^2+14x+40$

Therefore, $(x + 4) ( x + 10)$ is equal to $x^2+14x+40$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put $a = 8 \ \ and \ \ b = -10$

$(x+8)(x-10)= x^2+(-10+8)x+8\times (-10)$

$= x^2-2x-80$

Therefore, $(x+8)(x-10)$ is equal to $x^2-2x-80$

We can write $(3x+4)(3x - 5)$ as

$(3x+4)(3x - 5)= 9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put $a = \frac{4}{3} \ \ and \ \ b = -\frac{5}{3}$

$\dpi{100} 9\left ( x+\frac{4}{3} \right )\left ( x-\frac{5}{3} \right )= 9\left ( x^2+\left ( \frac{4}{3}-\frac{5}{3} \right )x+\frac{4}{3} \times \left ( -\frac{5}{3} \right ) \right )$

$=9x^2-3x-20$

Therefore, $(3x+4)(3x - 5)$ is equal to $9x^2-3x-20$

We will use identity

$(x+a)(x-a)=x^2-a^2$

Put $x=y^2 \ \ and \ \ a = \frac{3}{2}$

$(y^2 + \frac{3}{2})(y^2 - \frac{3}{2}) = \left ( y^2 \right )^2-\left(\frac{3}{2} \right )^2$

$= y^4-\frac{9}{4}$

Therefore, $(y^2 + \frac{3}{2})(y^2 - \frac{3}{2})$ is equal to $y^4-\frac{9}{4}$

We can write $(3 - 2x) (3 + 2x)$ as

$(3 - 2x) (3 + 2x)=-4\left ( x-\frac{3}{2} \right )\left(x+\frac{3}{2} \right )$

We will use identity

$(x+a)(x-a)=x^2-a^2$

Put $a = \frac{3}{2}$

$-4(x + \frac{3}{2})(x- \frac{3}{2}) =-4\left ( \left ( x \right )^2-\left(\frac{3}{2} \right )^2 \right )$

$=9-4x^2$

Therefore, $(3 - 2x) (3 + 2x)$ is equal to $9-4x^2$

We can rewrite $103 \times 107$ as

$\Rightarrow 103 \times 107= (100+3)\times (100+7)$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put $x =100 , a=3 \ \ and \ \ b = 7$

$(100+3)\times (100+7)= (100)^2+(3+7)100+3\times 7$

$=10000+1000+21= 11021$

Therefore, value of $103 \times 107$ is $11021$

We can rewrite $95 \times 96$ as

$\Rightarrow 95 \times 96= (100-5)\times (100-4)$

We will use identity

$(x+a)(x+b)=x^2+(a+b)x+ab$

Put $x =100 , a=-5 \ \ and \ \ b = -4$

$(100-5)\times (100-4)= (100)^2+(-5-4)100+(-5)\times (-4)$

$=10000-900+20= 9120$

Therefore, value of $95 \times 96$ is $9120$

We can rewrite $104 \times 96$ as

$\Rightarrow 104 \times 96= (100+4)\times (100-4)$

We will use identity

$(x+a)(x-a)=x^2-a^2$

Put $x =100 \ \ and \ \ a=4$

$(100+4)\times (100-4)= (100)^2-(4)^2$

$=10000-16= 9984$

Therefore, value of $104 \times 96$ is $9984$

We can rewrite $9x^2 + 6xy + y^2$ as

$\Rightarrow 9x^2 + 6xy + y^2 = (3x)^2+2\times 3x\times y +(y)^2$

Using identity $\Rightarrow (a+b)^2 = (a)^2+2\times a\times b +(b)^2$

Here, $a= 3x \ \ and \ \ b = y$

Therefore,

$9x^2+6xy+y^2 = (3x+y)^2 = (3x+y)(3x+y)$

We can rewrite $4y^2 - 4y + 1$ as

$\Rightarrow 4y^2 - 4y + 1=(2y)^2-2\times2y\times 1+(1)^2$

Using identity $\Rightarrow (a-b)^2 = (a)^2-2\times a\times b +(b)^2$

Here, $a= 2y \ \ and \ \ b = 1$

Therefore,

$4y^2 - 4y + 1=(2y-1)^2=(2y-1)(2y-1)$

We can rewrite $x^2 - \frac{y^2}{100}$ as

$\Rightarrow x^2 - \frac{y^2}{100} = (x)^2-\left(\frac{y}{10} \right )^2$

Using identity $\Rightarrow a^2-b^2 = (a-b)(a+b)$

Here, $a= x \ \ and \ \ b = \frac{y}{10}$

Therefore,

$x^2 - \frac{y^2}{100} = \left ( x-\frac{y}{10} \right )\left ( x+\frac{y}{10} \right )$

Given is $(x + 2y+4z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = x , b = 2y \ \ and \ \ c = 4z$

Therefore,

$(x + 2y+4z)^2 = (x)^2+(2y)^2+(4z)^2+2.x.2y+2.2y.4z+2.4z.x$

$= x^2+4y^2+16z^2+4xy+16yz+8zx$

Given is $(2x - y + z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = 2x , b = -y \ \ and \ \ c = z$

Therefore,

$(2x -y+z)^2 = (2x)^2+(-y)^2+(z)^2+2.2x.(-y)+2.(-y).z+2.z.2x$

$= 4x^2+y^2+z^2-4xy-2yz+4zx$

Given is $(-2x + 3y + 2z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a =- 2x , b = 3y \ \ and \ \ c = 2z$

Therefore,

$(-2x +3y+2z)^2 = (-2x)^2+(3y)^2+(2z)^2+2.(-2x).3y+2.3y.2z+2.z.(-2x)$

$= 4x^2+9y^2+4z^2-12xy+12yz-8zx$

Given is $(3a - 7b - c)^2$

We will Use identity

$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$

Here, $x =3a , y = -7b \ \ and \ \ z = -c$

Therefore,

$(3a - 7b - c)^2=(3a)^2+(-7b)^2+(-c)^2+2.3a.(-7b)+2.(-7b).(-c)+2.(-c)$ $.3a$

$= 9a^2+49b^2+c^2-42ab+14bc-6ca$

Given is $(-2x + 5y -3z)^2$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a =- 2x , b = 5y \ \ and \ \ c = -3z$

Therefore,

$(-2x +5y-3z)^2$ $= (-2x)^2+(5y)^2+(-3z)^2+2.(-2x).5y+2.5y.(-3z)+2.(-3z).(-2x)$

$= 4x^2+25y^2+9z^2-20xy-30yz+12zx$

Given is $\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$

We will Use identity

$(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx$

Here, $x =\frac{a}{4} , y = -\frac{b}{2} \ \ and \ \ z = 1$

Therefore,

$\left [\frac{1}{4}a - \frac{1}{2}b + 1\right ]^2$ $=\left(\frac{a}{4} \right )^2+\left(-\frac{b}{2} \right )^2+(1)^2+2.\left(\frac{a}{4} \right ). \left(-\frac{b}{2} \right )+2. \left(-\frac{b}{2} \right ).1+2.1.\left(\frac{a}{4} \right )$

$= \frac{a^2}{16}+\frac{b^2}{4}+1-\frac{ab}{4}-b+\frac{a}{2}$

We can rewrite $4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$ as

$\Rightarrow 4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz$ $= (2x)^2+(3y)^2+(-4z)^2+2.2x.3y+2.3y.(-4z)+2.(-4z).2x$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = 2x , b = 3y \ \ and \ \ c = -4z$

Therefore,

$4x^2 +9y^2 +16z^2 + 12xy -24yz - 16xz = (2x+3y-4z)^2$

$= (2x+3y-4z)(2x+3y-4z)$

We can rewrite $2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$ as

$\Rightarrow 2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz$ $= (-\sqrt2x)^2+(y)^2+(2\sqrt2z)^2+2.(-\sqrt2).y+2.y.2\sqrt2z+2.(-\sqrt2x).2\sqrt2z$

We will Use identity

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$

Here, $a = -\sqrt2x , b = y \ \ and \ \ c = 2\sqrt2z$

Therefore,

$2x^2 + y^2 +8z^2 - 2\sqrt2 xy + 4\sqrt2 yz - 8xz=(-\sqrt2x+y+2\sqrt2z)^2$

$=(-\sqrt2x+y+2\sqrt2z)(-\sqrt2x+y+2\sqrt2z)$

Given is $(2x + 1)^3$

We will use identity

$(a+b)^3=a^3+b^3+3a^2b+3ab^2$

Here, $a= 2x \ \ and \ \ b= 1$

Therefore,

$(2x+1)^3=(2x)^3+(1)^3+3.(2x)^2.1+3.2x.(1)^2$

$= 8x^3+1+12x^2+6x$

Given is $(2a-3b)^3$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here, $x= 2a \ \ and \ \ y= 3b$

Therefore,

$(2a-3b)^3=(2a)^3-(3b)^3-3.(2a)^2.3b+3.2a.(3b)^2$

$= 8a^3-9b^3-36a^2b+54ab^2$

Given is $\left[\frac{3}{2}x + 1\right ]^3$

We will use identity

$(a+b)^3=a^3+b^3+3a^2b+3ab^2$

Here, $a= \frac{3x}{2} \ \ and \ \ b= 1$

Therefore,

$\left(\frac{3x}{2}+1 \right )^3= \left(\frac{3x}{2} \right )^3+(1)^3+3.\left(\frac{3x}{2} \right )^2.1+3.\frac{3x}{2}.(1)^2$

$= \frac{27x^3}{8}+1+\frac{27x^2}{4}+\frac{9x}{2}$

Given is $\left[x - \frac{2}{3} y\right ]^3$

We will use identity

$(a-b)^3=a^3-b^3-3a^2b+3ab^2$

Here, $a=x \ and \ \ b= \frac{2y}{3}$

Therefore,

$\left[x - \frac{2}{3} y\right ]^3 = x^3-\left(\frac{2y}{3} \right )^3-3.x^2.\frac{2y}{3}+3.x.\left(\frac{2y}{3} \right )^2$

$= x^3-\frac{8y^3}{27}-2x^2y+\frac{4xy^2}{3}$

We can rewrite $(99)^3$ as

$\Rightarrow (99)^3=(100-1)^3$

We will use identity

$(a-b)^3=a^3-b^3-3a^2b+3ab^2$

Here, $a=100 \ and \ \ b= 1$

Therefore,

$(100-1)^1=(100)^3-(1)^3-3.(100)^2.1+3.100.1^2$

$= 1000000-1-30000+300= 970299$

We can rewrite $(102)^3$ as

$\Rightarrow (102)^3=(100+2)^3$

We will use identity

$(a+b)^3=a^3+b^3+3a^2b+3ab^2$

Here, $a=100 \ and \ \ b= 2$

Therefore,

$(100+2)^1=(100)^3+(2)^3+3.(100)^2.2+3.100.2^2$

$= 1000000+8+60000+1200= 1061208$

We can rewrite $(998)^3$ as

$\Rightarrow (998)^3=(1000-2)^3$

We will use identity

$(a-b)^3=a^3-b^3-3a^2b+3ab^2$

Here, $a=1000 \ and \ \ b= 2$

Therefore,

$(1000-2)^1=(1000)^3-(2)^3-3.(0100)^2.2+3.1000.2^2$

$= 1000000000-8-6000000+12000= 994011992$

We can rewrite $8a^3 + b^3 + 12a^2 b + 6ab^2$ as

$\Rightarrow 8a^3 + b^3 + 12a^2 b + 6ab^2$ $= (2a)^3+(b)^3+3.(2a)^2.b+3.2a.(b)^2$

We will use identity

$(x+y)^3=x^3+y^3+3x^2y+3xy^2$

Here, $x=2a \ \ and \ \ y= b$

Therefore,

$8a^3 + b^3 + 12a^2 b + 6ab^2 = (2a+b)^3$

$=(2a+b)(2a+b)(2a+b)$

We can rewrite $8a ^3 - b^3 - 12a^2 b + 6ab^2$ as

$\Rightarrow 8a ^3 - b^3 - 12a^2 b + 6ab^2$ $= (2a)^3-(b)^3-3.(2a)^2.b+3.2a.(b)^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here, $x=2a \ \ and \ \ y= b$

Therefore,

$8a ^3 - b^3 - 12a^2 b + 6ab^2 =(2a-b)^3$

$=(2a-b)(2a-b)(2a-b)$

We can rewrite $27 - 125a^ 3 - 135a + 225a^2$ as

$\Rightarrow 27 - 125a^ 3 - 135a + 225a^2^{}$ $= (3)^3-(25a)^3-3.(3)^2.5a+3.3.(5a)^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here, $x=3 \ \ and \ \ y= 5a$

Therefore,

$27 - 125a^ 3 - 135a + 225a^2 = (3-5a)^3$

$=(3-5a)(3-5a)(3-5a)$

We can rewrite $64a^3 - 27b^3 - 144a^2 b + 108ab^2$ as

$\Rightarrow 64a^3 - 27b^3 - 144a^2 b + 108ab^2$ $= (4a)^3-(3b)^3-3.(4a)^2.3b+3.4a.(3b)^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here, $x=4a \ \ and \ \ y= 3b$

Therefore,

$64a^3 - 27b^3 - 144a^2 b + 108ab^2=(4a-3b)^2$

$=(4a-3b)(4a-3b)(4a-3b)$

We can rewrite $27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$ as

$\Rightarrow 27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p$ $= (3p)^3-\left(\frac{1}{6} \right )^3-3.(3p)^2.\frac{1}{6}+3.3p.\left(\frac{1}{6} \right )^2$

We will use identity

$(x-y)^3=x^3-y^3-3x^2y+3xy^2$

Here, $x=3p \ \ and \ \ y= \frac{1}{6}$

Therefore,

$27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4} p = \left ( 3p-\frac{1}{6} \right )^3$

$= \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right ) \left ( 3p-\frac{1}{6} \right )$

We know that

$(x+y)^3=x^3+y^3+3xy(x+y)$

Now,

$\Rightarrow x^3+y^3=(x+y)^3-3xy(x+y)$

$\Rightarrow x^3+y^3=(x+y)\left((x+y)^2-3xy \right )$

$\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2+2xy-3xy \right )$ $(\because (a+b)^2=a^2+b^2+2ab)$

$\Rightarrow x^3+y^3=(x+y)\left(x^2+y^2-xy \right )$

Hence proved.

We know that

$(x-y)^3=x^3-y^3-3xy(x-y)$

Now,

$\Rightarrow x^3-y^3=(x-y)^3+3xy(x-y)$

$\Rightarrow x^3-y^3=(x-y)\left((x-y)^2+3xy \right )$

$\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2-2xy+3xy \right )$ $(\because (a-b)^2=a^2+b^2-2ab)$

$\Rightarrow x^3-y^3=(x-y)\left(x^2+y^2+xy \right )$

Hence proved.

We know that

$a^3+b^3=(a+b)(a^2+b^2-ab)$

Now, we can write $27y^3 + 125z^3$ as

$\Rightarrow 27y^3 + 125z^3 = (3y)^3+(5z)^3$

Here, $a = 3y \ \ and \ \ b = 5z$

Therefore,

$27y^3+125z^3= (3y+5z)\left((3y)^2+(5z)^2-3y.5z \right )$

$27y^3+125z^3= (3y+5z)\left(9y^2+25z^2-15yz \right )$

We know that

$a^3-b^3=(a-b)(a^2+b^2+ab)$

Now, we can write $64m^3 - 343n^3$ as

$\Rightarrow 64m^3 - 343n^3 = (4m)^3-(7n)^3$

Here, $a = 4m \ \ and \ \ b = 7n$

Therefore,

$64m^3-343n^3= (4m-7n)\left((4m)^2+(7n)^2+4m.7n \right )$

$64m^3-343n^3= (4m-7n)\left(16m^2+49n^2+28mn \right )$

Q11 Factorise: $27x^3 + y^3 + z^3 - 9xyz$

Given is $27x^3 + y^3 + z^3 - 9xyz$

Now, we know that

$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Now, we can write $27x^3 + y^3 + z^3 - 9xyz$ as

$\Rightarrow 27x^3 + y^3 + z^3 - 9xyz$ $=(3x)^3+(y)^3+(z)^3-3.3x.y.z$

Here, $a= 3x , b = y \ \ and \ \ c = z$

Therefore,

$27x^3 + y^3 + z^3 - 9xyz$ $=(3x+y+z)\left((3x)^2+(y)^2+(z)^2-3x.y-y.z-z.3x \right )$

$=(3x+y+z)\left(9x^2+y^2+z^2-3xy-yz-3zx \right )$

We know that

$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Now, multiply and divide the R.H.S. by 2

$x^3+y^3+z^3-3xyz = \frac{1}{2}(x+y+z)(2x^2+2y^2+2z^2-2xy-2yz-2zx)$

$= \frac{1}{2}(x+y+z)(x^2+y^2-2xy+x^2+z^2-2zx+y^2+z^2-2yz)$

$= \frac{1}{2}(x+y+z)\left((x-y)^2+(y-z)^2 +(z-x)^2\right )$ $\left(\because a^2+b^2-2ab=(a-b)^2 \right )$

Hence proved.

We know that

$x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Now, It is given that $x + y + z = 0$

Therefore,

$x^3+y^3+z^3-3xyz =0(x^2+y^2+z^2-xy-yz-zx)$

$x^3+y^3+z^3-3xyz =0$

$x^3+y^3+z^3=3xyz$

Hence proved.

Given is $(-12)^3 + (7)^3 + (5)^3$

We know that

If $x+y+z = 0$ then , $x^3+y^3+z^3 = 3xyz$

Here, $x = -12 , y = 7 \ \ an d \ \ z = 5$

$\Rightarrow x+y+z = -12+7+5 = 0$

Therefore,

$(-12)^3 + (7)^3 + (5)^3 = 3 \times (-12)\times 7 \times 5 = -1260$

Therefore, value of $(-12)^3 + (7)^3 + (5)^3$ is $-1260$

Given is $(28)^3 + (-15)^3 + (-13)^3$

We know that

If $x+y+z = 0$ then , $x^3+y^3+z^3 = 3xyz$

Here, $x = 28 , y = -15 \ \ an d \ \ z = -13$

$\Rightarrow x+y+z =28-15-13 = 0$

Therefore,

$(28)^3 + (-15)^3 + (-13)^3 = 3 \times (28)\times (-15) \times (-13) = 16380$

Therefore, value of $(28)^3 + (-15)^3 + (-13)^3$ is $16380$

 $25a^2 - 35a + 12$

We know that

Area of rectangle is = $length \times breadth$

It is given that area = $25a^2-35a+12$

Now, by splitting middle term method

$\Rightarrow 25a^2-35a+12 = 25a^2-20a-15a+12$

$= 5a(5a-4)-3(5a-4)$

$= (5a-3)(5a-4)$

case (i) :- Length = $(5a-4)$ and Breadth = $(5a-3)$

case (ii) :- Length = $(5a-3)$ and Breadth = $(5a-4)$

 $35y^2 + 13y- 12$

We know that

Area of rectangle is = $length \times breadth$

It is given that area = $35y^2 + 13y- 12$

Now, by splitting the middle term method

$\Rightarrow 35y^2 + 13y- 12 =35y^2+28y-15y-12$

$= 7y(5y+4)-3(5y+4)$

$= (7y-3)(5y+4)$

case (i) :- Length = $(5y+4)$ and Breadth = $(7y-3)$

case (ii) :- Length = $(7y-3)$ and Breadth = $(5y+4)$

 Volume : $3x^2 - 12x$

We know that

Volume of cuboid is = $length \times breadth \times height$

It is given that volume = $3x^2-12x$

Now,

$\Rightarrow 3x^2-12x=3\times x\times (x-4)$

Therefore,one of the possible answer is possible

Length = $3$ and Breadth = $x$ and Height = $(x-4)$

 Volume : $12ky^2 + 8ky - 20k$

We know that

Volume of cuboid is = $length \times breadth \times height$

It is given that volume = $12ky^2+8ky-20k$

Now,

$\Rightarrow 12ky^2+8ky-20k = k(12y^2+8y-20)$

$= k(12y^2+20y-12y-20)$

$= k\left(4y(3y+5)-4(3y+5) \right )$

$= k(3y+5)(4y-4)$

$= 4k(3y+5)(y-1)$

Therefore,one of the possible answer is possible

Length = $4k$ and Breadth = $(3y+5)$ and Height = $(y-1)$

## More About NCERT Solutions for Class 9 Maths Exercise 2.1

Class 9 Maths chapter 2 exercise 2 includes some of the basic problems in question one in which we have to apply the algebraic identities. Question two and question seven have problems based on splitting and applying identities. There are some problems based on identities in finding areas and volumes. Hence we can say that NCERT solutions for Class 9 Maths exercise 2.1 is a cluster of all types of questions from direct to hard. So this is the best source for practicing algebraic identities in order to make the base strong for whole algebra.

Also Read| Polynomials Class 9 Notes

## Benefits of NCERT Solutions for Class 9 Maths Exercise 2.5

• Class 9 Maths chapter 2 exercise 2.5 is the most important exercise of chapter 2

• NCERT Class 9 Maths chapter 2 exercise 2.5, will be useful in chapters of Class 10 such as chapter 2 polynomial, chapter 3 linear equation with two variable and Chapter 4 quadratic equation

• NCERT Class 9 Maths chapter 2 exercise 2.5, will be useful in chapters of class 11 such as chapter 5 complex number and quadratic equation and chapter 6 linear inequalities

• NCERT Class 9 Maths chapter 2 exercise 2.5, will be helpful in JEE Main as algebra is in the syllabus

## Key Features of Exercise 2.4 Class 9 Maths

1. Comprehensive Exercise: 9th class maths exercise 2.5 answers is a comprehensive exercise that covers various topics related to polynomials.

2. Conceptual Clarity: The primary objective of this class 9 maths chapter 2 exercise 2.5 is to help students develop a clear understanding of polynomial factorization and related theorems.

3. Diverse Problem Set: Exercise 2.5 class 9 maths offers a variety of problems with different levels of complexity, allowing students to enhance their skills in polynomial factorization.

4. Expert-Created Solutions: Class 9 maths ex 2.5 Solutions to the problems are typically provided in the exercise. These solutions are crafted by subject matter experts to ensure accuracy and clarity.

5. PDF Availability: Students can often download a PDF version of the ex 2.5 class 9 solutions, allowing them to access and use them offline. This resource is provided at no cost.

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## Subject Wise NCERT Exemplar Solutions

1. What exactly do you mean when you say algebraic identities?

An algebraic identity is a mathematical equation that holds true for all possible values of the variables in the equation.

2. Name all the algebraic identities used in NCERT solutions for Class 9 Maths exercise 2.5.

Identity I    : (x + y)²= x² + 2xy + y²

Identity II   : (x – y)² = x²  – 2xy + y²

Identity III  : x² – y²= (x + y) (x – y)

Identity IV  : (x + a) (x + b) = x²+ (a + b)x + ab

Identity V   : (x + y + z)² = x²+ y²+ z² + 2xy + 2yz + 2zx

Identity VI  : (x + y)³ = x³ + y³+ 3xy (x + y)

Identity VII : (x – y)³ = x³ – y³ – 3xy(x – y) = x³ – 3x²y + 3xy² – y³

Identity VIII : x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)

3. What are binomial algebraic identities?

(x + y)²= x² + 2xy + y²

(x – y)² = x²  – 2xy + y²

x² – y²= (x + y) (x – y)

(x + y)³ = x³ + y³+ 3xy (x + y)

(x – y)³ = x³ – y³ – 3xy(x – y) = x³ – 3x²y + 3xy² – y³

4. What are trinomial algebraic identities?

(x + a) (x + b) = x² + (a + b)x + ab

(x + y + z)² = x²+ y²+ z² + 2xy + 2yz + 2zx

x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)

5. In the NCERT Solutions for Class 9 Maths chapter 4 exercise 5, how many questions are covered?

There are 16 questions in Class 9 Maths chapter 4  exercise 9. 5

6. What is the number of solved examples covered before NCERT solutions for Class 9 Maths chapter 4 exercise 9.5?

There are 9  solved examples covered before NCERT solutions for Class 9 Maths chapter 4 exercise 9.5

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9