NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.4 - Polynomials

Q1 (i) Use suitable identities to find the following product: $(x+4)(x+10)$

Answer:

We will use identity

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

Put $a=4andb=10$

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

$=x^2+14x+40$

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

Q1 (ii) Use suitable identities to find the following product: $(x+8)(x-10)$

Answer:

We will use identity

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

Put $a=8andb=-10$

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

$=x^2-2x-80$

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

Q1 (iii) Use suitable identities to find the following product: $(3x+4)(3x-5)$

Answer:

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

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

We will use identity

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

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

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

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

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

Q1 (iv) Use suitable identities to find the following product: $(y^2+\frac{3}{2})(y^2-\frac{3}{2})$

Answer:

We will use identity

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

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

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

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

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

Q1 (v) Use suitable identities to find the following product: $(3-2x)(3+2x)$

Answer:

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

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

We will use identity

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

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

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

$=9-4x^2$

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

Q2 (i) Evaluate the following product without multiplying directly: $103\times 107$

Answer:

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 $x=100,a=3andb=7$

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

$=10000+1000+21=11021$

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

Q2 (ii) Evaluate the following product without multiplying directly: $95\times 96$

Answer:

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 $x=100,a=-5andb=-4$

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

$=10000-900+20=9120$

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

Q2 (iii) Evaluate the following product without multiplying directly: $104\times 96$

Answer:

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 $x=100anda=4$

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

$=10000-16=9984$

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

Q3 (i) Factorise the following using appropriate identities: $9x^2+6xy+y^2$

Answer:

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, $a=3xandb=y$

Therefore,

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

Q3 (ii) Factorise the following using appropriate identities: $4y^2-4y+1$

Answer:

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

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

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

Here, $a=2yandb=1$

Therefore,

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

Q3 (iii) Factorise the following using appropriate identities: $x^2-\frac{y^2}{100}$

Answer:

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, $a=xandb=\frac{y}{10}$

Therefore,

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

Q4 (i) Expand each of the following, using suitable identities: $(x+2y+4z{)}^2$

Answer:

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

We will Use identity

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

Here, $a=x,b=2yandc=4z$

Therefore,

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

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

Q4 (ii) Expand each of the following, using suitable identities: $(2x-y+z{)}^2$

Answer:

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

We will Use identity

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

Here, $a=2x,b=-yandc=z$

Therefore,

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

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

Q4 (iii) Expand each of the following, using suitable identities: $(-2x+3y+2z{)}^2$

Answer:

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

We will Use identity

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

Here, $a=-2x,b=3yandc=2z$

Therefore,

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

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

Q4 (iv) Expand each of the following, using suitable identities: $(3a-7b-c{)}^2$

Answer:

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

We will Use identity

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

Here, $x=3a,y=-7bandz=-c$

Therefore,

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

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

Q4 (v) Expand each of the following, using suitable identities: $(-2x+5y-3z{)}^2$

Answer:

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

We will Use identity

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

Here, $a=-2x,b=5yandc=-3z$

Therefore,

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

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

Q4 (vi) Expand each of the following, using suitable identities: ${[\frac{1}{4}a-\frac{1}{2}b+1]}^2$

Answer:

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

We will Use identity

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

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

Therefore,

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

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

Q5 (i) Factorise: $4x^2+9y^2+16z^2+12xy-24yz-16xz$

Answer:

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, $a=2x,b=3yandc=-4z$

Therefore,

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

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

Q5 (ii) Factorise: $2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz$

Answer:

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

$\Rightarrow 2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz$ $=(-\sqrt{2}x{)}^2+(y{)}^2+(2\sqrt{2}z{)}^2+2.(-\sqrt{2}).y+2.y.2\sqrt{2}z+2.(-\sqrt{2}x).2\sqrt{2}z$

We will Use identity

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

Here, $a=-\sqrt{2}x,b=yandc=2\sqrt{2}z$

Therefore,

$2x^2+y^2+8z^2-2\sqrt{2}xy+4\sqrt{2}yz-8xz=(-\sqrt{2}x+y+2\sqrt{2}z{)}^2$

$=(-\sqrt{2}x+y+2\sqrt{2}z)(-\sqrt{2}x+y+2\sqrt{2}z)$

Q6 (i) Write the following cubes in expanded form: $(2x+1{)}^3$

Answer:

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

We will use identity

$(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2$

Here, $a=2xandb=1$

Therefore,

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

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

Q6 (ii) Write the following cube in expanded form: $(2a-3b{)}^3$

Answer:

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

We will use identity

$(x-y{)}^3=x^3-y^3-3x^{2}y+3x{y}^2$

Here, $x=2aandy=3b$

Therefore,

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

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

Q6 (iii) Write the following cube in expanded form: ${[\frac{3}{2}x+1]}^3$

Answer:

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

We will use identity

$(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2$

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

Therefore,

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

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

Q6 (iv) Write the following cube in expanded form: ${[x-\frac{2}{3}y]}^3$

Answer:

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

We will use identity

$(a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2$

Here, $a=xandb=\frac{2y}{3}$

Therefore,

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

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

Q7 (i) Evaluate the following using suitable identities: $(99{)}^3$

Answer:

We can rewrite $(99{)}^3$ as

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

We will use identity

$(a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2$

Here, $a=100andb=1$

Therefore,

$(100-1{)}^1=(100{)}^3-(1{)}^3-3.(100{)}^2.1+3.100{.1}^2$

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

Q7 (ii) Evaluate the following using suitable identities: $(102{)}^3$

Answer:

We can rewrite $(102{)}^3$ as

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

We will use identity

$(a+b{)}^3=a^3+b^3+3a^{2}b+3a{b}^2$

Here, $a=100andb=2$

Therefore,

$(100+2{)}^1=(100{)}^3+(2{)}^3+3.(100{)}^2.2+3.100{.2}^2$

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

Q7 (iii) Evaluate the following using suitable identities: $(998{)}^3$

Answer:

We can rewrite $(998{)}^3$ as

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

We will use identity

$(a-b{)}^3=a^3-b^3-3a^{2}b+3a{b}^2$

Here, $a=1000andb=2$

Therefore,

$(1000-2{)}^1=(1000{)}^3-(2{)}^3-3.(0100{)}^2.2+3.1000{.2}^2$

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

Q8 (i) Factorise the following: $8a^3+b^3+12a^{2}b+6a{b}^2$

Answer:

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

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

We will use identity

$(x+y{)}^3=x^3+y^3+3x^{2}y+3x{y}^2$

Here, $x=2aandy=b$

Therefore,

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

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

Q8 (ii) Factorise the following: $8a^3-b^3-12a^{2}b+6a{b}^2$

Answer:

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

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

We will use identity

$(x-y{)}^3=x^3-y^3-3x^{2}y+3x{y}^2$

Here, $x=2aandy=b$

Therefore,

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

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

Q8 (iii) Factorise the following: $27-125a^3-135a+225a^2$

Answer:

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

$=(3{)}^3-(25a{)}^3-3.(3{)}^2.5a+3.3.(5a{)}^2$

We will use identity

$(x-y{)}^3=x^3-y^3-3x^{2}y+3x{y}^2$

Here, $x=3andy=5a$

Therefore,

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

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

Q8 (iv) Factorise the following: $64a^3-27b^3-144a^{2}b+108a{b}^2$

Answer:

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

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

We will use identity

$(x-y{)}^3=x^3-y^3-3x^{2}y+3x{y}^2$

Here, $x=4aandy=3b$

Therefore,

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

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

Q8 (v) Factorise the following: $27p^3-\frac{1}{216}-\frac{9}{2}{p}^2+\frac{1}{4}p$

Answer:

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

$(x-y{)}^3=x^3-y^3-3x^{2}y+3x{y}^2$

Here, $x=3pandy=\frac{1}{6}$

Therefore,

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

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

Q9 (i) Verify: $x^3+y^3=(x+y)(x^2-xy+y^2)$

Answer:

We know that

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

Now,

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

$\Rightarrow x^3+y^3=(x+y)((x+y{)}^2-3xy)$

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

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

Hence proved.

Q9 (ii) Verify: $x^3-y^3=(x-y)(x^2+xy+y^2)$

Answer:

We know that

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

Now,

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

$\Rightarrow x^3-y^3=(x-y)((x-y{)}^2+3xy)$

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

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

Hence proved.

Q10 (i) Factorise the following: $27y^3+125z^3$

Answer:

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, $a=3yandb=5z$

Therefore,

$27y^3+125z^3=(3y+5z)((3y{)}^2+(5z{)}^2-3y.5z)$

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

Q10 (ii) Factorise the following: $64m^3-343n^3$

Answer:

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, $a=4mandb=7n$

Therefore,

$64m^3-343n^3=(4m-7n)((4m{)}^2+(7n{)}^2+4m.7n)$

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

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

Answer:

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, $a=3x,b=yandc=z$

Therefore,

$27x^3+y^3+z^3-9xyz$ $=(3x+y+z)((3x{)}^2+(y{)}^2+(z{)}^2-3x.y-y.z-z.3x)$

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