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

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

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

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

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

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