Found 225 Articles for Class 8

Given:The given algebraic expression is $36l^2-(m+n)^2$.To do:We have to factorize the expression $36l^2-(m+n)^2$.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.$36l^2-(m+n)^2$ can be written as, $36l^2-(m+n)^2=(6l)^2-(m+n)^2$             [Since $36=6^2$]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, $36l^2-(m+n)^2=[6l+(m+n)][6l-(m+n)]$$36l^2-(m+n)^2=(6l+m+n)(6l-m-n)$Hence, the given expression can be factorized as $(6l+m+n)(6l-m-n)$.Read More

Given:The given expression is $64-(a+1)^2$.To do:We have to factorize the expression $64-(a+1)^2$.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.$64-(a+1)^2$ can be written as, $64-(a+1)^2=(8)^2-(a+1)^2$             [Since $64=8^2$]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, $64-(a+1)^2=(8)^2-(a+1)^2$$64-(a+1)^2=(8+a+1)[(8)-(a+1)]$$64-(a+1)^2=(9+a)(8-a-1)$$64-(a+1)^2=(9+a)(7-a)$Hence, the given expression can be factorized as $(9+a)(7-a)$.Read More

Given:The given algebraic expression is $x^8-1$.To do:We have to factorize the expression $x^8-1$.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.$x^8-1$ can be written as, $x^8-1=(x^4)^2-(1)^2$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^8-1=(x^4)^2-(1)^2$$x^8-1=(x^4+1)(x^4-1)$Now, $(x^4-1)$ can be written as, $(x^4-1)=(x^2)^2-(1)^2$                    [Since $1=1^2$]Using the formula $a^2-b^2=(a+b)(a-b)$, ... Read More

Given:The given expression is $a^4-16b^4$.To do:We have to factorize the expression $a^4-16b^4$.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.$a^4-16b^4$ can be written as, $a^4-16b^4=(a^2)^2-(4b^2)^2$             [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, $a^4-16b^4=(a^2)^2-(4b^2)^2$$a^4-16b^4=(a^2+4b^2)(a^2-4b^2)$Now, $(a^2-4b^2)$ can be written as, $(a^2-4b^2)=a^2-(2b)^2$                ... Read More

Given:The given algebraic expression is $3a^5-48a^3$.To do:We have to factorize the expression $3a^5-48a^3$.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.$3a^5-48a^3$ can be written as, $3a^5-48a^3=3a^3(a^2-16)$               (Taking $3a^3$ common from both the terms)$3a^5-48a^3=3a^3(a^2-4^2)$             [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, ... Read More

Given:The given expression is $(x+2y)^2-4(2x-y)^2$.To do:We have to factorize the expression $(x+2y)^2-4(2x-y)^2$.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.$(x+2y)^2-4(2x-y)^2$ can be written as, $(x+2y)^2-4(2x-y)^2=(x+2y)^2-[2(2x-y)]^2$             [Since $4=2^2$]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, $(x+2y)^2-4(2x-y)^2=(x+2y)^2-[2(2x-y)]^2$$(x+2y)^2-4(2x-y)^2=[(x+2y)+2(2x-y)][(x+2y)-2(2x-y)]$$(x+2y)^2-4(2x-y)^2=[(x+2y)+2(2x)-2(y)][(x+2y)-2(2x)+2(y)]$$(x+2y)^2-4(2x-y)^2=(x+2y+4x-2y)(x+2y-4x+2y)$$(x+2y)^2-4(2x-y)^2=(5x)(4y-3x)$Hence, the given expression can be factorized as $5x(4y-3x)$.Read More

Given:The given algebraic expression is $(2a-b)^2-16c^2$.To do:We have to factorize the expression $(2a-b)^2-16c^2$.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.$(2a-b)^2-16c^2$ can be written as, $(2a-b)^2-16c^2=(2a-b)^2-(4c)^2]$             [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, $(2a-b)^2-16c^2=(2a-b)^2-(4c)^2$$(2a-b)^2-16c^2=(2a-b+4c)(2a-b-4c)$Hence, the given expression can be factorized as $(2a-b+4c)(2a-b-4c)$.Read More

Given:The given expression is $144a^2-169b^2$.To do:We have to factorize the expression $144a^2-169b^2$.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.$144a^2-169b^2$ can be written as, $144a^2-169b^2=(12a)^2-(13b)^2]$             [Since $144=(12)^2, 169=(13)^2$]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, $144a^2-169b^2=(12a)^2-(13b)^2$$144a^2-169b^2=(12a+13b)(12a-13b)$Hence, the given expression can be factorized as $(12a+13b)(12a-13b)$.Read More

Given:The given algebraic expression is $125x^2-45y^2$.To do:We have to factorize the expression $125x^2-45y^2$.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.$125x^2-45y^2$ can be written as, $125x^2-45y^2=5[25x^2-9y^2]$                (Taking $5$ as common)$125x^2-45y^2=5[(5x)^2-(3y)^2]$             [Since $25=5^2, 9=3^2$]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, $125x^2-45y^2=5[(5x)^2-(3y)^2]$$125x^2-45y^2=5(5x+3y)(5x-3y)$Hence, ... Read More

Given:The given expression is $12m^2-27$.To do:We have to factorize the expression $12m^2-27$.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.$12m^2-27$ can be written as, $12m^2-27=3[4m^2-9]$                (Taking $3$ as common)$12m^2-27=3[(2m)^2-(3)^2]$             [Since $4=2^2, 9=3^2$]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, $12m^2-27=3[(2m)^2-(3)^2]$$12m^2-27=3(2m+3)(2m-3)$Hence, the ... Read More