Found 225 Articles for Class 8

Given:The given algebraic expression is $49(a-b)^2-25(a+b)^2$.To do:We have to factorize the expression $49(a-b)^2-25(a+b)^2$.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.$49(a-b)^2-25(a+b)^2$ can be written as, $49(a-b)^2-25(a+b)^2=[7(a-b)]^2-[5(a+b)]^2$             [Since $49=(7)^2, 25=5^2$]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, $49(a-b)^2-25(a+b)^2=[7(a-b)]^2-[5(a+b)]^2$$49(a-b)^2-25(a+b)^2=[7(a-b)+5(a+b)][7(a-b)-5(a+b)]$$49(a-b)^2-25(a+b)^2=(7a-7b+5a+5b)(7a-7b-5a-5b)$$49(a-b)^2-25(a+b)^2=(12a-2b)(2a-12b)$$49(a-b)^2-25(a+b)^2=2(6a-b)2(a-6b)$$49(a-b)^2-25(a+b)^2=4(6a-b)(a-6b)$Hence, the given expression can be factorized as $4(6a-b)(a-6b)$.Read More

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

Given:The given algebraic expression is $x^4-625$.To do:We have to factorize the expression $x^4-625$.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.$x^4-625$ can be written as, $x^4-625=(x^2)^2-(25)^2$             [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^4-625=(x^2)^2-(25)^2$$x^4-625=(x^2+25)(x^2-25)$Now, $(x^2-25)$ can be written as, $(x^2-25)=x^2-5^2$Using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize ... Read More

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

Given:The given algebraic expression is $3x^3y-243xy^3$.To do:We have to factorize the expression $3x^3y-243xy^3$.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.$3x^3y-243xy^3$ can be written as, $3x^3y-243xy^3=3xy(x^2-81y^2)$              (Taking $3xy$ common)$3x^3y-243xy^3=3xy[(x)^2-(9y)^2]$             [Since $81=(9)^2$]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, $3x^3y-243xy^3=3xy[(x)^2-(9y)^2]$$3x^3y-243xy^3=3xy(x+9y)(x-9y)$Hence, the given expression ... Read More

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