Found 225 Articles for Class 8

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

Given:The given expression is $9(a-b)^2-100(x-y)^2$.To do:We have to factorize the expression $9(a-b)^2-100(x-y)^2$.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.$9(a-b)^2-100(x-y)^2$ can be written as, $9(a-b)^2-100(x-y)^2=[3(a-b)]^2-[10(x-y)]^2$            [Since $9=3^2, 100=(10)^2$]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, $9(a-b)^2-100(x-y)^2=[3(a-b)]^2-[10(x-y)]^2$$9(a-b)^2-100(x-y)^2=[3(a-b)+10(x-y)][3(a-b)-10(x-y)]$$9(a-b)^2-100(x-y)^2=(3a-3b+10x-10y)(3a-3b-10x+10y)$Hence, the given expression can be factorized as $(3a-3b+10x-10y)(3a-3b-10x+10y)$.Read More

Given:The given algebraic expression is $(x-4y)^2-625$.To do:We have to factorize the expression $(x-4y)^2-625$.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.$(x-4y)^2-625$ can be written as, $(x-4y)^2-625=(x-4y)^2-(25)^2$             [Since $625=(25)^2$]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-4y)^2-625=(x-4y)^2-(25)^2$$(x-4y)^2-625=(x-4y+25)(x-4y-25)$Hence, the given expression can be factorized as $(x-4y+25)(x-4y-25)$.Read More

Given:The given expression is $x^3-144x$.To do:We have to factorize the expression $x^3-144x$.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.$x^3-144x$ can be written as, $x^3-144x=x(x^2-144)$$x^3-144x=x[(x)^2-(12)^2]$             [Since $144=(12)^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $x^3-144x=x[(x)^2-(12)^2]$$x^4-144x=x(x+12)(x-12)$Hence, the given expression can be factorized as $x(x+12)(x-12)$.Read More

Given:The given algebraic expression is $a^4-\frac{1}{b^4}$.To do:We have to factorize the expression $a^4-\frac{1}{b^4}$.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.$a^4-\frac{1}{b^4}$ can be written as, $a^4-\frac{1}{b^4}=(a^2)^2-(\frac{1}{b^2})^2$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-\frac{1}{b^4}=(a^2)^2-(\frac{1}{b^2})^2$$a^4-\frac{1}{b^4}=(a^2+\frac{1}{b^2})(a^2-\frac{1}{b^2})$Now, $(a^2-\frac{1}{b^2})$ can be written as, $(a^2-\frac{1}{b^2})=a^2-(\frac{1}{b})^2$Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize $(a^2-(\frac{1}{b})^2)$.$a^2-(\frac{1}{b})^2=(a+\frac{1}{b})(a-\frac{1}{b})$.............(I)Therefore, $a^4-\frac{1}{b^4}=(a^2+\frac{1}{b^2})(a+\frac{1}{b})(a-\frac{1}{b})$              ... Read More

Given:The given expression is $25x^4y^4-1$.To do:We have to factorize the expression $25x^4y^4-1$.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.$25x^4y^4-1$ can be written as, $25x^4y^4-1=(5x^2y^2)^2-(1)^2$             [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, $25x^4y^4-1=(5x^2y^2)^2-(1)^2$$25x^4y^4-1=(5x^2y^2+1)(5x^2y^2-1)$Hence, the given expression can be factorized as $(5x^2y^2+1)(5x^2y^2-1)$.Read More