Found 225 Articles for Class 8

Given:The given expression is $a^4b^4-81c^4$.To do:We have to factorize the expression $a^4b^4-81c^4$.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.$a^4b^4-81c^4$ can be written as, $a^4b^4-81c^4=(a^2b^2)^2-(9c^2)^2$             [Since $a^4b^4=(a^2b^2)^2, 81c^4=(9c^2)^2$]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^4b^4-81c^4=(a^2b^2)^2-(9c^2)^2$$a^4b^4-81c^4=(a^2b^2+9c^2)(a^2b^2-9c^2)$Now, $a^2b^2-9c^2$ can be written as, $a^2b^2-9c^2=(ab)^2-(3c)^2$              ... Read More

Given:The given algebraic expression is $2a^5-32a$.To do:We have to factorize the expression $2a^5-32a$.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.$2a^5-32a$ can be written as, $2a^5-32a=2a(a^4-16)$             (Taking $2a$ common)$2a^5-32a=2a[(a^2)^2-4^2]$                  [Since $a^4=(a^2)^2, 16=4^2$]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^5-32a=2a[(a^2)^2-4^2]$$2a^5-32a=2a(a^2+4)(a^2-4)$Now, ... Read More

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

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

Given:The given expression is $a^2-b^2+a-b$.To do:We have to factorize the expression $a^2-b^2+a-b$.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, $a^2-b^2+a-b=(a+b)(a-b)+a-b$$a^2-b^2+a-b=(a-b)[(a+b)+1]$                                (Taking $(a-b)$ common)$a^2-b^2+a-b=(a-b)(a+b+1)$Hence, the given expression can be factorized as ... Read More

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

Given:The given expression is $(2x+1)^2-9x^4$.To do:We have to factorize the expression $(2x+1)^2-9x^4$.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.$(2x+1)^2-9x^4$ can be written as, $(2x+1)^2-9x^4=(2x+1)^2-(3x^2)^2$             [Since $9x^4=(3x^2)^2$]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, $(2x+1)^2-9x^4=(2x+1)^2-(3x^2)^2$$(2x+1)^2-9x^4=[2x+1+3x^2][2x+1-3x^2]$$(2x+1)^2-9x^4=(3x^2+2x+1)(-3x^2+2x+1)$Hence, the given expression can be factorized as $(3x^2+2x+1)(-3x^2+2x+1)$.Read More

Given:The given algebraic expression is $4(xy+1)^2-9(x-1)^2$.To do:We have to factorize the expression $4(xy+1)^2-9(x-1)^2$.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.$4(xy+1)^2-9(x-1)^2$ can be written as, $4(xy+1)^2-9(x-1)^2=[2(xy+1)]^2-[3(x-1)]^2$             [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, $4(xy+1)^2-9(x-1)^2=[2(xy+1)]^2-[3(x-1)]^2$$4(xy+1)^2-9(x-1)^2=[2(xy+1)+3(x-1)][2(xy+1)-3(x-1)]$$4(xy+1)^2-9(x-1)^2=[2xy+2+3x-3][2xy+2-3x+3]$$4(xy+1)^2-9(x-1)^2=(2xy+3x-1)(2xy-3x+5)$Hence, the given expression can be factorized as $(2xy+3x-1)(2xy-3x+5)$.Read More

Given:The given expression is $16(2x-1)^2-25y^2$.To do:We have to factorize the expression $16(2x-1)^2-25y^2$.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.$16(2x-1)^2-25y^2$ can be written as, $16(2x-1)^2-25y^2=[4(2x-1)]^2-(5y)^2$             [Since $16=4^2, 25=5^2$]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, $16(2x-1)^2-25y^2=[4(2x-1)]^2-(5y)^2$$16(2x-1)^2-25y^2=[4(2x-1)+5y][4(2x-1)-5y]$$16(2x-1)^2-25y^2=[4(2x)-4(1)+5y][4(2x)-4(1)-5y]$$16(2x-1)^2-25y^2=(8x-4+5y)(8x-4-5y)$$16(2x-1)^2-25y^2=(8x+5y-4)(8x-5y-4)$Hence, the given expression can be factorized as $(8x+5y-4)(8x-5y-4)$.Read More

Given:The given expression is $x-y-x^2+y^2$.To do:We have to factorize the expression $x-y-x^2+y^2$.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.$x-y-x^2+y^2$ can be written as, $x-y-x^2+y^2=x-y-(x^2-y^2)$Here, we can observe that $x^2-y^2$ 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, $x^2-y^2=(x+y)(x-y)$.............(I)This implies, $x-y-x^2+y^2=(x-y)-[(x+y)(x-y)]$            [Using (I)]$x-y-x^2+y^2=(x-y)[1-(x+y)]$                     (Taking $x-y$ common)$x-y-x^2+y^2=(x-y)(1-x-y)$Hence, the ... Read More