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

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