Found 466 Articles for Mathematics

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

Given:The given expression is $abx^2+(ay-b)x-y$.To do:We have to factorize the expression $abx^2+(ay-b)x-y$.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 $abx^2+(ay-b)x-y$ by grouping similar terms and taking out the common factors.We can write $abx^2+(ay-b)x-y$ as, $abx^2+(ay-b)x-y=abx^2+axy-bx-y$The terms in the given expression are $abx^2, ayx, -bx$ and $-y$.We can group the given terms as $abx^2, -bx$ and $axy, -y$. Therefore, by taking $bx$ as common in $abx^2, -bx$ and $y$ ... Read More

Given:The given algebraic expression is $x^3-2x^2y+3xy^2-6y^3$.To do:We have to factorize the expression $x^3-2x^2y+3xy^2-6y^3$.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 $x^3-2x^2y+3xy^2-6y^3$ by grouping similar terms and taking out the common factors. The terms in the given expression are $x^3, -2x^2y, 3xy^2$ and $-6y^3$.We can group the given terms as $x^3, 3xy^2$ and $-2x^2y, -6y^3$. Therefore, by taking $x$ as common in $x^3, 3xy^2$ and $-2y$ as common in $-2x^2y, ... Read More

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

Given:The given algebraic expression is $6xy+6-9y-4x$.To do:We have to factorize the expression $6xy+6-9y-4x$.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 $6xy+6-9y-4x$ by grouping similar terms and taking out the common factors. The terms in the given expression are $6xy, 6, -9y$ and $-4x$.We can group the given terms as $6xy, -4x$ and $6, -9y$. Therefore, by taking $2x$ as common in $6xy, -4x$ and $-3$ as common in $6, ... Read More

Given:The given expression is $x^3-y^2+x-x^2y^2$.To do:We have to factorize the expression $x^3-y^2+x-x^2y^2$.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 $x^3-y^2+x-x^2y^2$ by grouping similar terms and taking out the common factors. The terms in the given expression are $x^3, -y^2, x$ and $-x^2y^2$.We can group the given terms as $x^3, x$ and $-y^2, -x^2y^2$. Therefore, by taking $x$ as common in $x^3, x$ and $-y^2$ as common in $-y^2, -x^2y^2$, ... Read More