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Found 225 Articles for Class 8
93 Views
Given:The given expressions are:(i) $a^2+4ab+3b^2$.(ii) $96-4x-x^2$(iii) $a^4+3a^2+4$To do:We have to factorize the given algebraic expressions.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.(i) The given expression is $a^2+4ab+3b^2$.$a^2+4ab+3b^2$ can be written as, By splitting and grouping the terms we can factorize the given expression. $a^2+4ab+3b^2=a^2+ab+3ab+3b^2$ [Since $4ab=ab+3ab$]Therefore, $a^2+4ab+3b^2=a^2+ab+3ab+3b^2$$a^2+4ab+3b^2=a(a+b)+3b(a+b)$$a^2+4ab+3b^2=(a+b)(a+3b)$Hence, the given expression can be factorized as $(a+b)(a+3b)$.(ii) The given expression is $96-4x-x^2$.By splitting and grouping ... Read More
82 Views
Given:The given expressions are:(i) $16-a^6+4a^3b^3-4b^6$.(ii) $a^2-2ab+b^2-c^2$(iii) $x^2+2x+1-9y^2$To do:We have to factorize the given algebraic expressions.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.(i) The given expression is $16-a^6+4a^3b^3-4b^6$.$16-a^6+4a^3b^3-4b^6$ can be written as, $16-a^6+4a^3b^3-4b^6=16-[(a^3)^2-2(a^3)(2b^3)+(2b^3)^2$$16-a^6+4a^3b^3-4b^6=4^2-[(a^3)^2-2(a^3)(2b^3)+(2b^3)^2]$ [Since $16=4^2, a^6=(a^3)^2, 4b^6=(2b^3)^2$ and $4a^3b^3=2(a^3)(2b^3)$]Here, we can observe that the given expression is of the form $m^2-2mn+n^2$. So, by using the formula $(m-n)^2=m^2-2mn+n^2$, we can factorize the given expression.Here, $m=a^3$ and $n=2b^3$ Therefore, $16-a^6+4a^3b^3-4b^6=4^2-[(a^3)^2-2(a^3)(2b^3)+(2b^2)^3]$$16-a^6+4a^3b^3-4b^6=4^2-(a^3-2b^3)^2$Now, Using ... Read More
178 Views
Given:The given expression is $9a^4-24a^2b^2+16b^4-256$.To do:We have to factorize the algebraic expression $9a^4-24a^2b^2+16b^4-256$.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.$9a^4-24a^2b^2+16b^4-256$ can be written as, $9a^4-24a^2b^2+16b^4-256=(9a^4-24a^2b^2+16b^4)-256$$9a^4-24a^2b^2+16b^4-256=[(3a^2)^2-2(3a^2)(4b^2)+(4b^2)^2]-256$ [Since $9a^4=(3a^2)^2, 16b^4=(4b^2)^2$ and $24a^2b^2=2(3a^2)(4b^2)$]Here, we can observe that the given expression is of the form $m^2-2mn+n^2$. So, by using the formula $(m-n)^2=m^2-2mn+n^2$, we can factorize the given expression.Here, $m=3a^2$ and $n=4b^2$ Therefore, $9a^4-24a^2b^2+16b^4-256=[(3a^2)^2-2(3a^2)(4b^2)+(4b^2)^2]-256$$9a^4-24a^2b^2+16b^4-256=(3a^2-4b^2)^2-256$Now, $(3a^2-4b^2)^2-256$ can be written as, $(3a^2-4b^2)^2-256=(3a^2-4b^2)^2-(16)^2$ ... Read More
111 Views
Given:The given algebraic expression is $9z^2-x^2+4xy-4y^2$.To do:We have to factorize the expression $9z^2-x^2+4xy-4y^2$.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.$9z^2-x^2+4xy-4y^2$ can be written as, $9z^2-x^2+4xy-4y^2=9z^2-(x^2-4xy+4y^2)$$9z^2-x^2+4xy-4y^2=9z^2-[x^2-2(x)(2y)+(2y)^2]$ [Since $x^2=(x)^2, 4y^2=(2y)^2$ and $4xy=2(x)(2y)$]Here, we can observe that the given expression is of the form $m^2-2mn+n^2$. So, by using the formula $(m-n)^2=m^2-2mn+n^2$, we can factorize the given expression.Here, $m=x$ and $n=2y$ Therefore, $9z^2-x^2+4xy-4y^2=9z^2-[x^2-2(x)(2y)+(2y)^2]$$9z^2-x^2+4xy-4y^2=9z^2-[(x-2y)^2]$Now, $9z^2-[(x-2y)^2]$ can be written as, $9z^2-[(x-2y)^2]=(3z)^2-(x-2y)^2$ ... Read More
122 Views
Given:The given expression is $a^2+2ab+b^2-16$.To do:We have to factorize the algebraic expression $a^2+2ab+b^2-16$.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.$a^2+2ab+b^2-16$ can be written as, $a^2+2ab+b^2-16=a^2+2(a)(b)+(b)^2-16$ [Since $a^2=(a)^2, b^2=(b)^2$ and $2ab=2(a)(b)$]Here, we can observe that the given expression is of the form $m^2+2mn+n^2$. So, by using the formula $(m+n)^2=m^2+2mn+n^2$, we can factorize the given expression.Here, $m=a$ and $n=b$ Therefore, $a^2+2ab+b^2-16=(a)^2+2(a)(b)+(b)^2-16$$a^2+2ab+b^2-16=(a+b)^2-16$Now, $(a+b)^2-16$ can be written as, $(a+b)^2-16=(a+b)^2-4^2$ ... Read More
74 Views
Given:The given algebraic expression is $36a^2+36a+9$.To do:We have to factorize the expression $36a^2+36a+9$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$36a^2+36a+9$ can be written as, $36a^2+36a+9=(6a)^2+2(6a)(3)+(3)^2$ [Since $36a^2=(6a)^2, 9=(3)^2$ and $36a=2(6a)(3)$]Here, we can observe that the given expression is of the form $m^2+2mn+n^2$. So, by using the formula $(m+n)^2=m^2+2mn+n^2$, we can factorize the given expression.Here, $m=6a$ and $n=3$ Therefore, $36a^2+36a+9=(6a)^2+2(6a)(3)+(3)^2$$36a^2+36a+9=(6a+3)^2$$36a^2+36a+9=(6a+3)(6a+3)$Hence, the given expression can be factorized as ... Read More
97 Views
Given:The given expression is $p^2q^2-6pqr+9r^2$.To do:We have to factorize the algebraic expression $p^2q^2-6pqr+9r^2$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression implies writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$p^2q^2-6pqr+9r^2$ can be written as, $p^2q^2-6pqr+9r^2=(pq)^2-2(pq)(3r)+(3r)^2$ [Since $p^2q^2=(pq)^2, 9r^2=(3r)^2$ and $6pqr=2(9pq)(3r)$]Here, we can observe that the given expression is of the form $m^2-2mn+n^2$. So, by using the formula $(m-n)^2=m^2-2mn+n^2$, we can factorize the given expression.Here, $m=pq$ and $n=3r$ Therefore, $p^2q^2-6pqr+9r^2=(pq)^2-2(pq)(3r)+(3r)^2$$p^2q^2-6pqr+9r^2=(pq-3r)^2$$p^2q^2-6pqr+9r^2=(pq-3r)(pq-3r)$Hence, the given expression can be factorized as ... Read More
85 Views
Given:The given algebraic expression is $9a^2-24ab+16b^2$.To do:We have to factorize the expression $9a^2-24ab+16b^2$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$9a^2-24ab+16b^2$ can be written as, $9a^2-24ab+16b^2=(3a)^2-2(3a)(4b)+(4b)^2$ [Since $9a^2=(3a)^2, 16b^2=(4b)^2$ and $24ab=2(3a)(4b)$]Here, we can observe that the given expression is of the form $m^2-2mn+n^2$. So, by using the formula $(m-n)^2=m^2-2mn+n^2$, we can factorize the given expression.Here, $m=3a$ and $n=4b$ Therefore, $9a^2-24ab+16b^2=(3a)^2-2(3a)(4b)+(4b)^2$$9a^2-24ab+16b^2=(3a-4b)^2$$9a^2-24ab+16b^2=(3a-4b)(3a-4b)$Hence, the given expression can be factorized as ... Read More
148 Views
Given:The given expression is $18a^2x^2-32$.To do:We have to factorize the expression $18a^2x^2-32$.Solution:Factorizing algebraic expressions:Factorizing an algebraic expression means writing the expression as a product of two or more factors. Factorization is the reverse of distribution. An algebraic expression is factored completely when it is written as a product of prime factors.$18a^2x^2-32$ can be written as, $18a^2x^2-32=2(9a^2x^2-16)$ (Taking $2$ common)$18a^2x^2-32=2[(3ax)^2-(4)^2]$ [Since $9a^2x^2=(3ax)^2, 16=4^2$]Here, we can observe that the given expression is a difference of two squares. So, by using the formula $a^2-b^2=(a+b)(a-b)$, we can factorize the given expression. Therefore, $18a^2x^2-32=2[(3ax)^2-(4)^2]$$18a^2x^2-32=2(3ax+4)(3ax-4)$Hence, the given ... Read More
122 Views
Given:The given algebraic expression is $x^3-x$.To do:We have to factorize the expression $x^3-x$.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.$x^3-x$ can be written as, $x^3-x=x(x^2-1)$ (Taking $x$ common)$x^3-x=x(x^2-1^2)$ [Since $1=1^2$]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 ... Read More