Found 225 Articles for Class 8

Given:The given algebraic expression is $144a^2-289b^2$.To do:We have to factorize the expression $144a^2-289b^2$.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.$144a^2-289b^2$ can be written as, $144a^2-289b^2=(12a)^2-(17b)^2]$             [Since $144=(12)^2, 289=(17)^2$]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^2-289b^2=(12a)^2-(17b)^2$$144a^2-289b^2=(12a+17b)(12a-17b)$Hence, the given expression can be factorized as $(12a+17b)(12a-17b)$.Read More

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

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

Given:The given algebraic expression is $ab-a-b+1$.To do:We have to factorize the expression $ab-a-b+1$.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.Here, we can factorize the expression $ab-a-b+1$ by grouping similar terms and taking out the common factors. The terms in the given expression are $ab, -a, -b$ and $1$.We can group the given terms as $ab, -a$ and $-b, 1$. Therefore, by taking $a$ as common in $ab, -a$ and $-1$ as common in $-b, ... Read More

Given:The given expression is $x^2-11xy-x+11y$.To do:We have to factorize the expression $x^2-11xy-x+11y$.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 factorize the expression $x^2-11xy-x+11y$ by grouping similar terms and taking out the common factors. The terms in the given expression are $x^2, -11xy, -x$ and $11y$.We can group the given terms as $x^2, -11xy$ and $-x, 11y$. Therefore, by taking $x$ as common in $x^2, -11xy$ and $-1$ as common in $-x, 11y$, ... Read More

Given:The given algebraic expression is $a(a+b-c)-bc$.To do:We have to factorize the expression $a(a+b-c)-bc$.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 factorize the expression $a(a+b-c)-bc$ by grouping similar terms and taking out the common factors. $a(a+b-c)-bc$ can be written as, $a(a+b-c)-bc=a(a)+a(b)-a(c)-bc$$a(a+b-c)-bc=a^2+ab-ac-bc$The terms in the given expression are $a^2, ab, -ac$ and $-bc$.We can group the given terms as $a^2, ab$ and $-ac, -bc$. Therefore, by taking $a$ as common in $a^2, ab$ and ... Read More

Given:The given expression is $a(a-2b-c)+2bc$.To do:We have to factorize the expression $a(a-2b-c)+2bc$.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.Here, we can factorize the expression $a(a-2b-c)+2bc$ by grouping similar terms and taking out the common factors. $a(a-2b-c)+2bc=a(a)-a(2b)-a(c)+2bc$$a(a-2b-c)+2bc=a^2-2ab-ac+2bc$The terms in the given expression are $a^2, -2ab, -ac$ and $2bc$.We can group the given terms as $a^2, -2ab$ and $-ac, 2bc$. Therefore, by taking $a$ as common in $a^2, -2ab$ and $-c$ as common in $-ac, 2bc$, ... Read More

Given:The given expression is $ab(x^2+1)+x(a^2+b^2)$.To do:We have to factorize the expression $ab(x^2+1)+x(a^2+b^2)$.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.Here, we can factorize the expression $ab(x^2+1)+x(a^2+b^2)$ by grouping similar terms and taking out the common factors. $ab(x^2+1)+x(a^2+b^2)$ can be written as, $ab(x^2+1)+x(a^2+b^2)=ab(x^2)+ab(1)+x(a^2)+x(b^2)$$ab(x^2+1)+x(a^2+b^2)=abx^2+ab+a^2x+b^2x$The terms in the given expression are $abx^2, ab, a^2x$ and $b^2x$.We can group the given terms as $abx^2, a^2x$ and $ab, b^2x$. Therefore, by taking $ax$ as common in $abx^2, a^2x$ ... Read More

Given:The given expression is $16(a-b)^3-24(a-b)^2$.To do:We have to factorize the expression $16(a-b)^3-24(a-b)^2$.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.Here, we can factorize the expression $16(a-b)^3-24(a-b)^2$ by grouping similar terms and taking out the common factors. The terms in the given expression are $16(a-b)^3$ and $-24(a-b)^2$.Therefore, by taking $(a-b)^2$ as common, we get, $16(a-b)^3-24(a-b)^2=(a-b)^2[16(a-b)-24]$Now, taking $8$ common in $[16(a-b)-24]$, we get, $16(a-b)^3-24(a-b)^2=(a-b)^28[2(a-b)-3]$$16(a-b)^3-24(a-b)^2=8(a-b)^2[2(a)-2(b)-3]$$16(a-b)^3-24(a-b)^2=8(a-b)^2(2a-2b-3)$Hence, the given expression can be factorized as $8(a-b)^2(2a-2b-3)$.Read More

Given:The given algebraic expression is $(ax+by)^2+(bx-ay)^2$.To do:We have to factorize the expression $(ax+by)^2+(bx-ay)^2$.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 factorize the expression $(ax+by)^2+(bx-ay)^2$ by grouping similar terms and taking out the common factors. We can write $(ax+by)^2+(bx-ay)^2$ as, $(ax+by)^2+(bx-ay)^2=(ax)^2+2(ax)(by)+(by)^2+(bx)^2-2(bx)(ay)+(ay)^2$                     [Since $(m+n)^2=m^2+2mn+n^2$ and $(m-n)^2=m^2-2mn+n^2$]$(ax+by)^2+(bx-ay)^2=a^2x^2+2abxy+b^2y^2+b^2x^2-2abxy+a^2y^2$ $(ax+by)^2+(bx-ay)^2=a^2x^2+b^2y^2+b^2x^2+a^2y^2$The terms in the given expression are $a^2x^2, b^2y^2, b^2x^2$ and $a^2y^2$.We can group the given terms as $a^2x^2, ... Read More