Given:The given expressions are:(i) $49-x^2-y^2+2xy$(ii) $a^2+4b^2-4ab-4c^2$(iii) $x^2-y^2-4xz+4z^2$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 $49-x^2-y^2+2xy$.$49-x^2-y^2+2xy$ can be written as, $49-x^2-y^2+2xy=49-(x^2+y^2-2xy)$$49-x^2-y^2+2xy=7^2-[(x)^2-2(x)(y)+(y)^2]$            [Since $49=7^2$ and $2xy=2(x)(y)$]Here, we can observe that the given expression is of the form $m^2-2mn+n^2$. So, by using the formula $(m-n)^2=m^2-2mn+n^2$, we can factorize the given expression.Here, $m=x$ and $n=y$ Therefore, $49-x^2-y^2+2xy=7^2-[(x)^2-2(x)(y)+(y)^2]$$49-x^2-y^2+2xy=7^2-(x-y)^2$Now, Using the formula ... Read More

Given:The given expressions are:(i) $25x^2-10x+1-36y^2$(ii) $a^2-b^2+2bc-c^2$(iii) $a^2+2ab+b^2-c^2$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 $25x^2-10x+1-36y^2$.$25x^2-10x+1-36y^2$ can be written as, $25x^2-10x+1-36y^2=[(5x)^2-2(5x)(1)+(1)^2]-(6y)^2$                [Since $25x^2=(5x)^2, 10x=2(5x)(1)$ and $36y^2=(6y)^2$]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, $m=5x$ and $n=1$ Therefore, $25x^2-10x+1-36y^2=[(5x)^2-2(5x)(1)+(1)^2]-(6y)^2$$25x^2-10x+1-36y^2=(5x-1)^2-(6y)^2$Now, ... Read More

Given:The given expressions are:(i) $a^2-8ab+16b^2-25c^2$(ii) $x^2-y^2+6y-9$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 $a^2-8ab+16b^2-25c^2$.$a^2-8ab+16b^2-25c^2$ can be written as, $a^2-8ab+16b^2-25c^2=[(a)^2-2(a)(4b)+(4b)^2]-(5c)^2$          [Since $8ab=2(a)(4b), 16b^2=(4b)^2$ and $25c^2=(5c)^2$]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, $m=a$ and $n=4b$ Therefore, $a^2-8ab+16b^2-25c^2=[(a)^2-2(a)(4b)+(4b)^2]-(5c)^2$$a^2-8ab+16b^2-25c^2=(a-4b)^2-(5c)^2$Now, Using the formula ... Read More

Given:The given expressions are:(i) $25-p^2-q^2-2pq$(ii) $x^2+9y^2-6xy-25a^2$(iii) $49-a^2+8ab-16b^2$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 $25-p^2-q^2-2pq$.$25-p^2-q^2-2pq$ can be written as, $25-p^2-q^2-2pq=25-[p^2+2pq+q^2]$$25-p^2-q^2-2pq=5^2-[(p)^2+2(p)(q)+(q)^2]$             [Since $25=5^2$ and $2pq=2(p)(q)$]Here, we can observe that the given expression is of the form $m^2+2mn+n^2$. So, by using the formula $(m+n)^2=m^2+2mn+n^2$, we can factorize the given expression.Here, $m=p$ and $n=q$ Therefore, $25-p^2-q^2-2pq=5^2-[(p)^2+2(p)(q)+(q)^2]$$25-p^2-q^2-2pq=5^2-(p+q)^2$Now, Using the formula ... Read More

Given:The given expressions are:(i) $4x^4+1$.(ii) $4x^4+y^4$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 $4x^4+1$.$4x^4+1$ can be written as, $4x^4+1=4x^4+1+4x^2-4x^2$                    (Add and subtract $4x^2$)$4x^4+1=[(2x^2)^2+2(2x^2)(1)+1^2]-4x^2$             [Since $4x^4=(2x^2)^2, 1=(1)^2$ and $4x^2=2(2x^2)(1)$]Here, we can observe that the given expression is of the form $m^2+2mn+n^2$. So, by using the formula ... Read More

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

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

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

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

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