Given:The given expression is $p^2q^2-p^4q^4$.To do:We have to factorize the expression $p^2q^2-p^4q^4$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$p^2q^2-p^4q^4$ can be written as, $p^2q^2-p^4q^4=p^2q^2[1-p^2q^2]$               (Taking $p^2q^2$ common)$p^2q^2-p^4q^4=p^2q^2[1^2-(pq)^2]$             [Since $p^2q^2=(pq)^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $p^2q^2-p^4q^4=p^2q^2[1^2-(pq)^2]$$p^2q^2-p^4q^4=p^2q^2(1+pq)(1-pq)$Hence, the given expression can ... Read More

Given:The given algebraic expression is $(3x+4y)^4-x^4$.To do:We have to factorize the expression $(3x+4y)^4-x^4$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression implies writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$(3x+4y)^4-x^4$ can be written as, $(3x+4y)^4-x^4=[(3x+4y)^2]^2-(x^2)^2$             [Since $(3x+4y)^4=[(3x+4y)^2]^2, x^4=(x^2)^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $(3x+4y)^4-x^4=[(3x+4y)^2]^2-(x^2)^2$$(3x+4y)^4-x^4=[(3x+4y)^2+x^2][(3x+4y)^2-x^2]$Now, Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize $(3x+4y)^2-x^2$.$(3x+4y)^2-x^2=(3x+4y+x)(3x+4y-x)$$(3x+4y)^2-x^2=(4x+4y)(2x+4y)$$(3x+4y)^2-x^2=4(x+y)2(x+2y)$$(3x+4y)^2-x^2=8(x+y)(x+2y)$.............(I)Therefore, $(3x+4y)^4-x^4=[(3x+4y)^2+x^2]8(x+y)(x+2y)$      ... Read More

Given:The given expression is $a^4-(2b+c)^4$.To do:We have to factorize the expression $a^4-(2b+c)^4$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$a^4-(2b+c)^4$ can be written as, $a^4-(2b+c)^4=(a^2)^2-[(2b+c)^2]^2$             [Since $a^4=(a^2)^2, (2b+c)^4=[(2b+c)^2]^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $a^4-(2b+c)^4=(a^2)^2-[(2b+c)^2]^2$$a^4-(2b+c)^4=[a^2+(2b+c)^2][a^2-(2b+c)^2]$Now, Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize $a^2-(2b+c)^2$.$a^2-(2b+c)^2=(a+2b+c)(a-2b-c)$.............(I)Therefore, $a^4-(2b+c)^4=[a^2+(2b+c)^2](a+2b+c)(a-2b-c)$        ... Read More

Given:The given expression is $256x^3-81x$.To do:We have to factorize the expression $256x^3-81x$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$256x^3-81x$ can be written as, $256x^3-81x=x(256x^2-81)$             (Taking $x$ common)$256x^3-81x=x[(16x)^2-(9)^2]$             [Since $256=(16)^2, 81=(9)^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $256x^3-81x=x[(16x)^2-(9)^2]$$256x^3-81x=x(16x+9)(16x-9)$Hence, the given expression can ... Read More

Given:The given algebraic expression is $\frac{50}{x^2}-\frac{2x^2}{81}$.To do:We have to factorize the expression $\frac{50}{x^2}-\frac{2x^2}{81}$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression implies writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$\frac{50}{x^2}-\frac{2x^2}{81}$ can be written as, $\frac{50}{x^2}-\frac{2x^2}{81}=2(\frac{25}{x^2}-\frac{x^2}{81})$             (Taking $2$ common)$\frac{50}{x^2}-\frac{2x^2}{81}=2[(\frac{5}{x})^2-(\frac{x}{9})^2]$             [Since $\frac{25}{x^2}=(\frac{5}{x})^2, \frac{x^2}{81}=(\frac{x}{9})^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $\frac{50}{x^2}-\frac{2x^2}{81}=2[(\frac{5}{x})^2-(\frac{x}{9})^2]$$\frac{50}{x^2}-\frac{2x^2}{81}=2(\frac{5}{x}+\frac{x}{9})(\frac{5}{x}-\frac{x}{9})$Hence, the given expression ... Read More

Given:The given algebraic expression is $x^5-16x^3$.To do:We have to factorize the expression $x^5-16x^3$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression implies writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$x^5-16x^3$ can be written as, $x^5-16x^3=x^3(x^2-16)$                    (Taking $x^3$ common)$x^5-16x^3=x^3[(x)^2-(4)^2]$             [Since $16=(4)^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $x^5-16x^3=x^3[(x)^2-(4)^2]$$x^5-16x^3=x^3(x+4)(x-4)$Hence, ... Read More

Given:The given expression is $75a^3b^2-108ab^4$.To do:We have to factorize the expression $75a^3b^2-108ab^4$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$75a^3b^2-108ab^4$ can be written as, $75a^3b^2-108ab^4=3ab^2(25a^2-36b^2)$           (Taking $3ab^2$ common)$75a^3b^2-108ab^4=3ab^2[(5a)^2-(6b)^2]$             [Since $25=5^2, 36=6^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $75a^3b^2-108ab^4=3ab^2[(5a)^2-(6b)^2]$$75a^3b^2-108ab^4=3ab^2(5a+6b)(5a-6b)$Hence, the given expression can be ... Read More

Given:The given algebraic expression is $\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2$.To do:We have to factorize the expression $\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression implies writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2$ can be written as, $\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2=(\frac{1}{4}xy)^2-(\frac{2}{7}yz)^2$             [Since $\frac{1}{16}=(\frac{1}{4})^2, \frac{4}{49}=(\frac{2}{7})^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2=(\frac{1}{4}xy)^2-(\frac{2}{7}yz)^2$$\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2=(\frac{1}{4}xy+\frac{2}{7}yz)(\frac{1}{4}xy-\frac{2}{7}yz)$$\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2=y(\frac{1}{4}x+\frac{2}{7}z)y(\frac{1}{4}x-\frac{2}{7}z)$$\frac{1}{16}x^2y^2-\frac{4}{49}y^2z^2=y^2(\frac{1}{4}x+\frac{2}{7}z)(\frac{1}{4}x-\frac{2}{7}z)$Hence, the given expression can be factorized as $y^2(\frac{1}{4}x+\frac{2}{7}z)(\frac{1}{4}x-\frac{2}{7}z)$.Read More

Given:The given expression is $(x+y)^2-(a-b)^2$.To do:We have to factorize the expression $(x+y)^2-(a-b)^2$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $(x+y)^2-(a-b)^2=[(x+y)+(a-b)][(x+y)-(a-b)]$$(x+y)^2-(a-b)^2=(x+y+a-b)(x+y-a+b)$Hence, the given expression can be factorized as $(x+y+a-b)(x+y-a+b)$.Read More

Given:The given algebraic expression is $(3+2a)^2-25a^2$.To do:We have to factorize the expression $(3+2a)^2-25a^2$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression implies writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$(3+2a)^2-25a^2$ can be written as, $(3+2a)^2-25a^2=(3+2a)^2-(5a)^2$             [Since $25=(5)^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $(3+2a)^2-25a^2=(3+2a)^2-(5a)^2$$(3+2a)^2-25a^2=(3+2a+5a)(3+2a-5a)$$(3+2a)^2-25a^2=(3+7a)(3-3a)$$(3+2a)^2-25a^2=(3+7a)3(1-a)$                    (Taking $3$ common)$(3+2a)^2-25a^2=3(3+7a)(1-a)$Hence, ... Read More