# Algebraic Expressions Solution of TS & AP Board Class 8 Mathematics

###### Exercise 11.1

**Question 1.**

Find the product of the following pairs:

6, 7k

**Answer:**

Given:6 and 7k

The product of 6 and 7k is = 6×7k = 42k

Hence the product of 6 and 7k is 42k.

**Question 2.**

Find the product of the following pairs:

−3l, −2m

**Answer:**

Given: -3l and -2m

The product of -3l and -2m is = -3l × -2m = 6lm

Hence the product of -3l and -2m is 6lm.

**Question 3.**

Find the product of the following pairs:

−5t^{2} −3t^{2}

**Answer:**

Given: -5t^{2} and -3t^{2}

The product of -5t^{2} and -3t^{2}

⇒ -5t^{2}×-3t^{2} = 15(t^{2}.t^{2})

⇒15t^{4}

Hence the product of -5t^{2} and -3t^{2} is 15t^{4}.

**Question 4.**

Find the product of the following pairs:

6n, 3m

**Answer:**

Given: 6n and 3m

The product of 6n and 3m

⇒ 6n × 3m = 18nm

Hence the product of 6n and 3m is 18nm.

**Question 5.**

Find the product of the following pairs:

−5p^{2}, −2p

**Answer:**

Given: -5p^{2} and -2p

The product of -5p^{2} and -2p

⇒-5p^{2}×-2p = 10(p^{2}.p)

⇒10p^{3}

Hence the product of -5p^{2} and -2p is 10p^{3}.

**Question 6.**

Complete the table of the products.

**Answer:**

Here the product is calculated in the following manner:

We multiply the terms on the x axis (horizontal)and y axis(vertical).

For example, when -2y^{2} is on the X axis (observe the 1^{st} row 3^{rd} column) and 3x is on the Y axis(2^{nd} row 1^{st} column) then ,the product

becomes -2y^{2}×3x = -6xy^{2}(observe the 2^{nd} row 3^{rd} column).

**Question 7.**

Find the volumes of rectangular boxes with given length, breadth and height in the following table.

**Answer:**

Here we have calculated the Volume V using the Length, Breadth and Height.

V = Length×Breadth×Height

**Question 8.**

Find the product of the following monomials

xy, x^{2}y, xy, x

**Answer:**

Here we have xy,x^{2}y,xy,x

Hence the product of the above terms is (xy)×(x^{2}y)×(x) =

(x.x^{2}.x) (y.y)

= x^{4}y^{2}

**Question 9.**

Find the product of the following monomials

a, b, ab, a^{3}b, ab^{3}

**Answer:**

Here we have a,b,ab,a^{3}b,ab^{3}

Hence the product of the above terms is (a)×(b)×(ab)×(a^{3}b)×(ab^{3})

= (a.a.a^{3}.a)(b.b.b.b^{3})

= a^{6}b^{6}

**Question 10.**

Find the product of the following monomials

kl, lm, km, klm

**Answer:**

Here we have kl,lm,km,klm

Hence the product of the above terms is (kl)×(lm)×(km)×(klm)

= (k.k.k)(l.l.l)(m.m.m)

= k^{3}l^{3}m^{3}

**Question 11.**

Find the product of the following monomials

pq ,pqr, r

**Answer:**

Here we have pq,pqr,r

Hence the product of the above terms is (pq)×(pqr)×(r)

= (p.p)(q.q)(r.r)

= p^{2}q^{2}r^{2}

**Question 12.**

Find the product of the following monomials

−3a, 4ab, −6c, d

**Answer:**

Here we have -3a,4ab,-6c,d

Hence the product of the above terms is (-3a)×(4ab)×(-6c)×(d)

= (-3×4×-6)(a.a)(b)(c)(d)

= 72a^{2}bcd

**Question 13.**

If A = xy, B = yz and C = zx, then find ABC = ...........

**Answer:**

Here we have A = xy ,B = yz and C = zx.

Therefore the product of A,B and C will be ABC = (xy)×(yz)×(zx) = x^{2}y^{2}z^{2}.

**Question 14.**

**Answer:**

Here we have P = 4x^{2} ,T = 5x and R = 5y.

Therefore the product of P,T and R will be PTR = (4x^{2})×(5x)×(5y) = 100x^{3}y

Hence is equal to i.e. x^{3}y.

**Question 15.**

Write some monomials of your own and find their products.

**Answer:**

The monomials are (i)5x,6y,7z and (ii)3x^{2}y,4xy^{2},7x^{3}y^{3}

(i) The given monomial is 5x,6y,7z .

Therefore the product of above terms is (5x)×(6y)×(7z) = 210xyz

(ii) The given monomial is 3x^{2}y,4xy^{2},7x^{3}y^{3} .

Therefore the product of above terms is (3x^{2}y)×(4xy^{2})×(7x^{3}y^{3}) = 84x^{6}y^{6}

###### Exercise 11.2

**Question 1.**

Complete the table:

**Answer:**

Here we have the First and Second expressions as 5q and p + q-2r respectively.

The process involved in the multiplication is :

Step1.Write the product of the expressions using multiplication symbol: 5q×(p + q-2r)

Step2.**Use distributive law** :Multiply the monomial by the first term of the polynomial then multiply the monomial by the second term and then multiply the monomial by the third term of the polynomial and add their products: 5pq + 5q^{2}-10qr

Step3.Simplify the terms: 5q^{2} + 5pq-10qr

2. Similarly ,following the above process we calculate

(kl + lm + mn)(3k)

= (3k)(kl) + (3k)(lm) + (3k)(mn)

= 3k^{2}l + 3klm + 3kmn

3. Similarly ,following the above process we calculate

(ab^{2})( a + b^{2} + c^{3}) = ab^{2}(a) + ab^{2} (b^{2}) + ab^{2}(c^{3}) = a^{2}b^{2} + ab^{4} + ab^{2}c^{3}

4.Similarly ,following the above process we calculate

(x-2y + 3z)(xyz) = ( x-2y + 3z)×(xyz) = x^{2}yz-2xy^{2}z + 3xyz^{2}

5. Similarly ,following the above process we calculate

(a^{2}bc + b^{2}cd-abd^{2})(a^{2}b^{2}c^{2}) = (a^{2}bc + b^{2}cd- abd^{2})×(a^{2}b^{2}c^{2}) = a^{4}b^{3}c^{3} + a^{2}b^{4}c^{3}d-a^{3}b^{3}c^{2}d^{2}

**Question 2.**

Simplify: 4y(3y + 4)

**Answer:**

Here we have 4y and 3y + 4.

Now the product of these terms is as follows:

The procedure involved in the multiplication is :

Step1.Write the product of the expressions using multiplication symbol: 4y(3y + 4).

Step2.**Use distributive law** :Multiply the monomial by the first term of the binomial then multiply the monomial by the second term of the binomial and then add their products: 4y(3y) + 4y(4)

Step3.Simplify the terms: 12y^{2} + 16y

Hence 4y(3y + 4) = 12y^{2} + 16y

**Question 3.**

Simplify x(2x^{2}−7x + 3) and find the values of it for (i) x = 1 and (ii) x = 0

**Answer:**

Here we have the monomial x and polynomial 2x^{2}-7x + 3

Now the product of these terms is as follows:

The process involved in the multiplication is :

Step1.Write the product of the expressions using multiplication symbol:

x(2x^{2}-7x + 3).

Step2.**Use distributive law** :Multiply the monomial by the first term of the trinomial then multiply the monomial by the second term of the trinomial and then the monomial by the third term of the trinomial and then add their products: x(2x^{2})-x(7x) + x(3)

Step3.Simplify the terms: 2x^{3}-7x^{2} + 3x

Hence x(2x^{2}−7x + 3) = 2x^{3}-7x^{2} + 3x

(i)When x = 1, 2x^{3}-7x^{2} + 3x = 2(1)^{3}-7(1)^{2} + 3(1) = -2

(ii)When x = 0, 2x^{3}-7x^{2} + 3x = 2(0)^{3}-7(0)^{2} + 3(0) = 0

**Question 4.**

Add the product: a(a−b), b(b−c), c(c−a)

**Answer:**

Here we have the expressions a(a-b),b(b-c) and c(c-a)

Using the **distributive law** a(a-b) = a^{2}-ab,b(b-c) = b^{2}-bc and c(c-a) = c^{2}-ca

Now we add the terms (a^{2}-ab) + (b^{2}-bc) + (c^{2}-ca)

= a^{2} + b^{2} + c^{2}-ab-bc-ca

Hence addition of the product of the above expressions is a^{2} + b^{2} + c^{2}-ab-bc-ca.

**Question 5.**

Add the product: x(x + y−r), y(x−y + r), z(x−y−z)

**Answer:**

Here we have the expressions x(x + y-r),y(x-y + r) and z(x-y-z)

Using the **distributive law** x(x + y-r)

= xy-y

^{2}+ yr and

z(x-y-z) = xz-yz-z^{2}

Now we add these terms (x^{2} + xy-xr) + (xy-y^{2} + yr) + (xz-yz-z^{2})

= x^{2}-y^{2}-z^{2} + 2xy-xr + yr + xz-yz

Hence addition of the product of the above expressions is x^{2}-y^{2}-z^{2} + 2xy-xr + yr + xz-yz.

**Question 6.**

Subtract the product of 2x(5x−y) from product of 3x(x + 2y)

**Answer:**

Here we have 2x(5x-y) and 3x(x + 2y)

Using the **distributive law** 2x(5x-y) = 2x(5x)-2x(y)

= 10x^{2}-2xy

and 3x(x + 2y)

= 3x(x) + 3x(2y)

= 3x^{2} + 6xy.

Now , we have to subtract 10x^{2}-2xy from 3x^{2} + 6xy.

i.e. (3x^{2} + 6xy)-(10x^{2}-2xy)

= 3x^{2} + 6xy-10x^{2} + 2xy

= -7x^{2} + 8xy

Hence after subtraction of the product of 2x(5x−y) from the product of 3x(x + 2y) is

-7x^{2} + 8xy.

**Question 7.**

Subtract 3k(5k−l + 3m) from 6k(2k + 3l−2m)

**Answer:**

Here we have 3k(5k−l + 3m) and 6k(2k + 3l−2m)

Using the **distributive law** 3k(5k−l + 3m) = 15k^{2}-3kl + 9km

and 6k(2k + 3l-2m) = 12k^{2} + 18kl-12km

Now , we have to subtract 15k^{2}-3kl + 9km from 12k^{2} + 18kl-12km.

i.e. (12k^{2} + 18kl-12km)-(15k^{2}-3kl + 9km) = -3k^{2} + 21kl-21k^{2}m^{2}

Hence after subtraction of the product of 3k(5k−l + 3m) from the product of 6k(2k + 3l−2m) is = -3k^{2} + 21kl-21k^{2}m^{2}

**Question 8.**

Simplify: a^{2}(a−b + c) + b^{2}(a + b−c)−c^{2}(a−b−c)

**Answer:**

Here we have the polynomials as a^{2}(a−b + c), b^{2}(a + b−c) and c^{2}(a−b−c)

Now using the **distributive law**, a^{2}(a−b + c) = a^{2}(a)-a^{2}(b) + a^{2}(c) = a^{3}-a^{2}b + a^{2}c

b^{2}(a + b−c) = b^{2}(a) + b^{2}(b)-b^{2}(c) = ab^{2} + b^{3}-b^{2}c and c^{2}(a−b−c) = c^{2}a-c^{2}b-c^{3}

Now we will simplify it, a^{2}(a−b + c) + b^{2}(a + b−c)−c^{2}(a−b−c) = (a^{3}-a^{2}b + a^{2}c) + (ab^{2} + b^{3}-b^{2}c)-(c^{2}a-c^{2}b-c^{3}) = a^{3}-a^{2}b + a^{2}c + ab^{2} + b^{3}-b^{2}c-c^{2}a + c^{2}b + c^{3} = a^{3} + b^{3} + c^{3}-a^{2}b + a^{2}c + ab^{2}-b^{2}c-c^{2}a + c^{2}b

Hence , a^{2}(a−b + c) + b^{2}(a + b−c)−c^{2}(a−b−c) = a^{2}b + a^{2}c + ab^{2}-b^{2}c-c^{2}a + c^{2}b

###### Exercise 11.3

**Question 1.**

Multiply the binomials:

2a−9 and 3a + 4

**Answer:**

Consider two binomials as 2a−9 and 3a + 4

Now,to get the product of two binomials 2a−9 and 3a + 4 we use the

**distributive law** i.e. (2a−9)×(3a + 4) = 2a(3a) + 2a(4)-9(3a)-9(4) = 6a^{2} + 8a-27a-36 = 6a^{2}-19a-36

Therefore ,the product of 2a−9 and 3a + 4 is 6a^{2}-19a-36.

**Question 2.**

Multiply the binomials:

x−2y and 2x−y

**Answer:**

Consider two binomials as x−2y and 2x−y

Now,to get the product of two binomials x−2y and 2x−y we use the

**distributive law** i.e. (x−2y)×( 2x−y) = x(2x)-x(y)-2y(2x)-2y(-y) = 2x^{2}-xy-4xy + 4y^{2}

Therefore ,the product of x−2y and 2x−y is 2x^{2} + 4y^{2}-5xy.

**Question 3.**

Multiply the binomials:

kl + lm and k−l

**Answer:**

Consider two binomials as kl + lm and k-l

Now,to get the product of two binomials kl + lm and k-l we use the

**distributive law** i.e. (kl + lm)×( k-l) = kl(k) + kl(-l) + lm(k) + lm(-l) = k^{2}l-kl^{2} + klm-l^{2}m

Therefore , the product of (kl + lm) and ( k-l) is k^{2}l-kl^{2} + klm-l^{2}m

**Question 4.**

Multiply the binomials:

m^{2}−n^{2} and m + n

**Answer:**

Consider two binomials as m^{2}−n^{2} and m + n

Now, to get the product of two binomials ) m^{2}−n^{2} and m + n we use the

**distributive law** i.e. (m^{2}−n^{2})×( m + n) = m^{2}(m) + m^{2}(n)-n^{2}(m)-n^{2}(n) =

m^{3} + m^{2}n-mn^{2}-n^{3}

Therefore , the product of m^{2}−n^{2} and m + n is m^{3} + m^{2}n-mn^{2}-n^{3}.

**Question 5.**

Find the product:

(x + y)(2x−5y + 3xy)

**Answer:**

Consider the binomial x + y and the trinomial 2x−5y + 3xy.

Now, to get the product of above expressions x + y and 2x−5y + 3xy,we use the

**distributive law** i.e (x + y)×(2x−5y + 3xy) = x(2x-5y + 3xy) + y(2x-5y + 3xy)

= x(2x) + x(-5y) + x(3xy) + y(-5y) + y(3xy) = 2x^{2}-5xy + 3x^{2}y-5y^{2} + 3xy^{2}

Therefore , the product of x + y and 2x−5y + 3xy is 2x^{2}-5y^{2} + 3x^{2}y + 3xy^{2}-5xy

**Question 6.**

Find the product:

(mn−kl + km) (kl−lm)

**Answer:**

Consider the trinomial mn−kl + km and the binomial kl−lm.

Now, to get the product of mn−kl + km and kl−lm,we use the

**distributive law** i.e. (mn−kl + km)×( kl−lm) = mn(kl-lm)-kl(kl-lm) + km(-lm)

= mnkl-m^{2}nl-k^{2}l^{2} + kl^{2}m + kml-klm^{2}

Therefore , the product of mn−kl + km and kl−lm is mnkl-m^{2}nl-k^{2}l^{2} + kl^{2}m + kml-klm^{2}

**Question 7.**

Find the product:

(a−2b + 3c)(ab^{2}−a^{2}b)

**Answer:**

Consider the trinomial a−2b + 3c and the binomial ab^{2}−a^{2}b.

Now, to get the product of a−2b + 3c and ab^{2}−a^{2}b,we use the

**distributive law** i.e. (a−2b + 3c)×( ab^{2}−a^{2}b) = a(ab^{2}-a^{2}b)-2b(ab^{2}-a^{2}b) + 3c(ab^{2}-a^{2}b)

= a^{2}b^{2}-a^{3}b-2ab^{3} + 2a^{2}b^{2} + 3ab^{2}c-3a^{2}bc = 3a^{2}b^{2}-a^{3}b-2ab^{3} + 3ab^{2}c-3a^{2}bc

Therefore , the product of a−2b + 3c and ab^{2}−a^{2}b is 3a^{2}b^{2}-a^{3}b-2ab^{3} + 3ab^{2}c-3a^{2}bc

**Question 8.**

Find the product:

(p^{3} + q^{3})(p−5q + 6r)

**Answer:**

Consider the binomial p^{3} + q^{3} and the trinomial p−5q + 6r.

Now, to get the product of p^{3} + q^{3} and p−5q + 6r,we use the

**distributive law** i.e (p^{3} + q^{3})(p−5q + 6r) = p^{3}(p-5q + 6r) + q^{3}(p-5q + 6r) = (p^{3}.p-5p^{3}q + p^{3}.6r) + (p.q^{3}-5q.q^{3} + 6q^{3}r) = p^{4}-5p^{3}q + 6p^{3}r + pq^{3}-5q^{4} + 6q^{3}r

Therefore , the product of p^{3} + q^{3}and p−5q + 6r is p^{4}-5p^{3}q + 6p^{3}r + pq^{3}-5q^{4} + 6q^{3}r

**Question 9.**

Simplify the following:

(x−2y)(y−3x) + (x + y)(x−3y)−(y−3x)(4x−5y)

**Answer:**

Here we have (x−2y)(y−3x) + (x + y)(x−3y)−(y−3x)(4x−5y)

Re-arranging, we get,

(x - 2y)(y - 3x) - (y - 3x)(4x - 5y) + (x + y)(x - 3y)

= (y - 3x)[(x - 2y) - (4x - 5y)] + (x + y)(x - 3y)

= (y - 3x)(-3x + 3y) + (x + y)(x - 3y)

= - 3 (y - 3x)(x - y) + (x + y)(x - 3y)

= -3 (xy - y^{2} - 3x^{2} + 3xy) + x^{2} - 3xy + xy - 3y^{2}= 9x^{2} + 3y^{2} - 12xy + x^{2} - 3y^{2} - 2xy

= 10x^{2} - 14xy

**Question 10.**

Simplify the following:

(m + n)(m^{2}−mn + n^{2})

**Answer:**

We have the binomial m + n and trinomial m^{2}−mn + n^{2}

Here we will use the **distributive law** as follows:

(m + n)(m^{2}−mn + n^{2}) = m(m^{2}−mn + n^{2}) + n(m^{2}−mn + n^{2}) = (m.m^{2}-m.mn + m.n^{2}) + (n.m^{2}-mn.n + n.n^{2}) = m^{3}-m^{2}n + mn^{2} + m^{2}n-mn^{2} + n^{3} = m^{3} + n^{3}

Hence , (m + n)(m^{2}−mn + n^{2}) = m^{3} + n^{3}

**Question 11.**

Simplify the following:

(a−2b + 5c)(a−b)− (a−b−c)(2a + 3c) + (6a + b)(2c−3a−5b)

**Answer:**

We have (a−2b + 5c)(a−b)− (a−b−c)(2a + 3c) + (6a + b)(2c−3a−5b)

Here, (a−2b + 5c)(a−b) = (a-b)(a-2b + 5c)(using **Commutative law**)

= a(a-2b + 5c)-b(a-2b + 5c)(using **distributive law** )

= a^{2}-2ab + 5ac-ab + 2b^{2}-5bc = a^{2} + 2b^{2}-3ab + 5ac-5bc

(a−b−c)(2a + 3c) = (2a + 3c)(a-b-c) (using **Commutative law**)

= 2a(a-b-c) + 3c(a-b-c) (using **distributive law** )

= 2a^{2}-2ab-2ac + 3ac-3bc-3c^{2}

= 2a^{2} + ac-2ab-3bc-3c^{2}

and (6a + b)(2c−3a−5b) = 6a(2c−3a−5b) + b(2c−3a−5b) = 12ac-18a^{2}-30ab + 2bc-3ab-5b^{2} = -18a^{2} + 5b^{2}-33ab + 2bc + 12ac

Therefore, a−2b + 5c)(a−b)− (a−b−c)(2a + 3c) + (6a + b)(2c−3a−5b) = a^{2} + 2b^{2}-3ab + 5ac-5bc-(2a^{2} + ac-2ab-3bc-3c^{2}) + ( -18a^{2} + 5b^{2}-33ab + 2bc + 12ac) =

a^{2} + 2b^{2}-3ab + 5ac-5bc-2a^{2}-2ac + 2ab + 3bc + 3c^{2}-18a^{2} + 5b^{2}-33ab + 2bc + 12ac =

= -19a^{2} + 7b^{2} + 3c^{2}-34ab + 15ac

**Question 12.**

Simplify the following:

(pq-qr + pr)(pq + qr)−(pr + pq)(p + q−r)

**Answer:**

We have , (pq-qr + pr)(pq + qr)−(pr + pq)(p + q−r)

(pq-qr + pr)(pq + qr) = (pq + qr)( pq-qr + pr) (using **Commutative law**)

= pq(pq-qr + pr) + qr( pq-qr + pr) (using **distributive law** )

= p^{2}q^{2}-pq^{2}r + p^{2}qr + pq^{2}r-q^{2}r^{2} + pqr^{2} = p^{2}q^{2} + p^{2}qr + pqr^{2}-q^{2}r^{2}

(pr + pq)(p + q−r) = pr(p + q-r) + pq(p + q−r) (using **distributive law** )

= p^{2}r + prq-pr^{2} + p^{2}q + pq^{2}-prq

= p^{2}r-pr^{2} + p^{2}q + pq^{2}

(pq-qr + pr)(pq + qr)−(pr + pq)(p + q−r) = p^{2}q^{2} + p^{2}qr + pqr^{2}-q^{2}r^{2}-( p^{2}r-pr^{2} + p^{2}q + pq^{2})

= p^{2}q^{2} + p^{2}qr + pqr^{2}-q^{2}r^{2}-p^{2}r + pr^{2}-p^{2}q-pq^{2}

Hence, (pq-qr + pr)(pq + qr)−(pr + pq)(p + q−r) = p^{2}q^{2} + p^{2}qr + pqr^{2}-q^{2}r^{2}-p^{2}r + pr^{2}-p^{2}q-pq^{2}

**Question 13.**

If a, b, c are positive real numbers such that, find the value of

**Answer:**

_{Given:}

, where a, b, c are positive real numbers

Here, we have

⇒b(a + b-c) = c(a-b + c)

⇒ab + b^{2}-bc = ac-bc + c^{2}(bc is on both the sides and they get subtracted)

⇒ab + b^{2} = ac + c^{2}

⇒ab-ac = c^{2}-b^{2}

⇒a(b-c) = (c-b)(c + b)

⇒-a(c-b) = (c + b)(c-b)

⇒-a = (c + b)[Here (c-b) is on both the sides,so when they get divided and the result is 1]

⇒(b + c) = -a(**commutative property**)

We have ,

⇒a(a-b + c) = b(-a + b + c)

⇒a^{2}-ab + ac = -ab + b^{2} + bc

⇒a^{2} + ac = b^{2} + bc

⇒a^{2}-b^{2} = bc-ac

⇒(a + b)(a-b) = -c(a-b)

⇒a + b = -c

We have ,

⇒a(a + b-c) = c(-a + b + c)

⇒a^{2} + ab-ac = -ac + bc + c^{2}

⇒a^{2} + ab = bc + c^{2}

⇒a^{2}-c^{2} = bc-ab

⇒(a + c)(a-c) = b(c-a)

⇒(a + c)(a-c) = -b(a-c)

⇒a + c = -b

Therefore, = -1(Here we have substituted the values of (a + b),(b + c),(c + a) from above evaluation)

###### Exercise 11.4

**Question 1.**

Select a suitable identity and find the following products

(3k + 4l) (3k + 4l)

**Answer:**

Given: (3k + 4l) (3k + 4l),it is a product of 2 binomial expressions which have the same terms 3k + 4l and 3k + 4l.

Now when we compare these expressions with the identities, we

find it in the form of (a + b)^{2},where a = 3k and b = 4l,

The identity (a + b)^{2} = a^{2} + 2ab + b^{2},

Hence, (3k + 4l) (3k + 4l) = (3k + 4l)^{2} = (3k)^{2} + 2(3k)(4l) + (4l)^{2} = 9k^{2} + 24kl + 16l^{2}

**Question 2.**

Select a suitable identity and find the following products

(ax^{2} + by^{2}) (ax^{2} + by^{2})

**Answer:**

Given: (ax^{2} + by^{2}) (ax^{2} + by^{2}),it is a product of 2 binomial expressions which have the same terms ax^{2} + by^{2} and ax^{2} + by^{2}.

Now when we compare these expressions with the identities, we

find it in the form of (a + b)^{2},where a = ax^{2} and b = by^{2},

The identity (a + b)^{2} = a^{2} + 2ab + b^{2},

Hence, (ax^{2} + by^{2}) (ax^{2} + by^{2}) = (ax^{2} + by^{2})^{2} = (ax^{2})^{2} + 2(ax^{2})(by^{2}) + (by^{2})^{2}

= a^{2}x^{4} + 2abx^{2}y^{2} + b^{2}y^{4}

**Question 3.**

Select a suitable identity and find the following products

(7d – 9e) (7d – 9e)

**Answer:**

Given: (7d – 9e) (7d – 9e),it is a product of 2 binomial expressions which have the same terms 7d – 9e and 7d – 9e.

Now when we compare these expressions with the identities, we

find it in the form of (a-b)^{2},where a = 7d and b = 9e,

The identity (a-b)^{2} = a^{2}-2ab + b^{2},

Hence, (7d – 9e) (7d – 9e) = (7d-9e)^{2} = (7d)^{2}-2(7d)(9e) + (9e)^{2} = 49d^{2}-126de + 81e^{2}

**Question 4.**

Select a suitable identity and find the following products

(m^{2} – n^{2}) (m^{2} + n^{2})

**Answer:**

Given:( m^{2} – n^{2}) (m^{2} + n^{2}),it is a product of 2 binomial expressions which have the terms (m^{2} – n^{2}) and (m^{2} + n^{2}).

Now when we compare these expressions with the identities, we

find it in the form of (a-b)(a + b),where a = m^{2} and b = n^{2},

The identity (a-b)(a + b) = a^{2}-b^{2},

Hence, (m^{2} – n^{2}) (m^{2} + n^{2}) = (m^{2})^{2}-(n^{2})^{2} = m^{4}-n^{4}

**Question 5.**

Select a suitable identity and find the following products

(3t + 9s) (3t – 9s)

**Answer:**

Given:(3t + 9s) (3t – 9s),it is a product of 2 binomial expressions which have the terms :(3t + 9s) and (3t – 9s).

Now when we compare these expressions with the identities, we

find it in the form of (a-b)(a + b),where a = 3t and b = 9s,

The identity (a-b)(a + b) = a^{2}-b^{2},

Hence, (3t + 9s) (3t – 9s) = (3t)^{2}-(9s)^{2} = 9t^{2}-81s^{2}

**Question 6.**

Select a suitable identity and find the following products

(kl – mn) (kl + mn)

**Answer:**

Given:( kl – mn) (kl + mn),it is a product of 2 binomial expressions which have the terms :( kl – mn) and (kl + mn).

Now when we compare these expressions with the identities, we

find it in the form of (a-b)(a + b),where a = kl and b = mn,

The identity (a-b)(a + b) = a^{2}-b^{2},

Hence, (kl – mn) (kl + mn) = (kl)^{2}-(mn)^{2}

**Question 7.**

Select a suitable identity and find the following products

(6x + 5) (6x + 6)

**Answer:**

Given:(6x + 5) (6x + 6),it is a product of 2 binomial expressions which have the terms (6x + 5) and (6x + 6).

Now when we compare these expressions with the identities, we

find it in the form of (x + a)(x + b),where x = 6x and a = 5 and b = 6,

Thus,(x + a)(x + b) = x^{2} + x(b + a) + ab

Hence, (6x + 5) (6x + 6) = (6x)^{2} + 6x(5 + 6) + 30 = 12x^{2} + 66x + 30

**Question 8.**

Select a suitable identity and find the following products

(2b – a) (2b + c)

**Answer:**

Given: (2b – a) (2b + c),it is a product of 2 binomial expressions which have the terms (2b – a) (2b + c).

Now when we compare these expressions with the identities, we

find it in the form of (x-a)(x + b),where x = 2b and a = a and b = c,

Thus,(x-a)(x + b) = x^{2} + bx-ax-ab = x^{2} + x(b-a)-ab

Hence, (2b – a) (2b + c) = (2b)^{2} + 2b(c-a)-ac = 4b^{2} + 2b(c-a)-ac

**Question 9.**

Evaluate the following by using suitable identities:

304^{2}

**Answer:**

Given:304^{2} = (300 + 4)^{2} = (300)^{2} + 2(300)(4) + (4)^{2}

= 90000 + 2400 + 16, where a = 300 and b = 4 in the identity (a + b)^{2} = a^{2} + 2ab + b^{2}

= 92416

**Question 10.**

Evaluate the following by using suitable identities:

509^{2}

**Answer:**

Given: 509^{2} = (500 + 9)^{2} = (500)^{2} + 2(500)(9) + (9)^{2}

= 2500 + 9000 + 81, where a = 500 and b = 9 in the identity (a + b)^{2} = a^{2} + 2ab + b^{2}

= 11581

**Question 11.**

Evaluate the following by using suitable identities:

992^{2}

**Answer:**

Given: 992^{2} = (900 + 2)^{2} = (900)^{2} + 2(900)(2) + (2)^{2}

= 8100 + 3600 + 4, where a = 900 and b = 2 in the identity (a + b)^{2} = a^{2} + 2ab + b^{2}

= 11704

**Question 12.**

Evaluate the following by using suitable identities:

799^{2}

**Answer:**

Given: 799^{2} = (800-1)^{2} = (800)^{2}-2(800)(1) + (1)^{2}

= 6400 + 1600 + 1, where a = 800 and b = 1 in the identity (a-b)^{2} = a^{2}-2ab + b^{2}

= 8001

**Question 13.**

Evaluate the following by using suitable identities:

304 × 296

**Answer:**

Given: 304×296 = (300 + 4)(300-4)

= (300)^{2}-(4)^{2}, where a = 300 and b = 4 in identity (a + b)(a-b) = a^{2}-b^{2}

= 90000-16 = 89984

**Question 14.**

Evaluate the following by using suitable identities:

83 × 77

**Answer:**

Given:83×77 = (80 + 3)(80-3)

= (80)^{2}-(3)^{2},where a = 80 and b = 3 in identity (a + b)(a-b) = a^{2}-b^{2}

= 6400-9 = 6391

**Question 15.**

Evaluate the following by using suitable identities:

109×108

**Answer:**

Given:109×108 = (100 + 9)(100 + 8)

= (100)^{2} + (9 + 8)100 + (9×8),

where x = 100,a = 9 and b = 8 in identity (x + a)(x + b) = x^{2} + (a + b)x + ab

= 10000 + 1700 + 72

= 11772

**Question 16.**

Evaluate the following by using suitable identities:

204×206

**Answer:**

Given: (200 + 4)(200 + 6)

= (200)^{2} + (4 + 6)200 + (4×6),

where x = 200,a = 4,b = 6 in identity (x + a)(x + b) = x^{2} + (a + b)x + ab

= 40000 + 2000 + 24

= 42024

###### Exercise 11.5

**Question 1.**

Verify the identity (a + b)^{2} ≡ a^{2} + 2ab + b^{2} geometrically by taking

a = 2 units, b = 4 units

**Answer:**

(a + b)^{2}Îža^{2} + 2ab + b^{2}

Draw a square with the side a + b i.e.,2 + 4

L.H.S of the whole square = (2 + 4)^{2} = (6)^{2} = 36

R.H.S = Area of the square with 2 units + Area of the square with 4 units +

Area of the 2,4 units + Area of the square with 4 ,2 units = 2^{2} + 4^{2} + 2×4 + 2×4 = 4 + 16 + 8 + 8 = 36

L.H.S = R.H.S

Hence,the identity is verified.

**Question 2.**

Verify the identity (a + b)^{2} ≡ a^{2} + 2ab + b^{2} geometrically by taking

a = 3 units, b = 1 unit

**Answer:**

(a + b)^{2}Îža^{2} + 2ab + b^{2}

Draw a square with the side a + b i.e.,3 + 1

L.H.S of the whole square = (3 + 1)^{2} = (4)^{2} = 16

R.H.S = Area of the square with 3 units + Area of the square with 1 unit +

Area of the 3,1 unit + Area of the square with 1 ,3 units = 3^{2} + 1^{2} + 3×1 + 1×3 = 9 + 1 + 3 + 3 = 16

L.H.S = R.H.S

Hence, the identity is verified.

**Question 3.**

Verify the identity (a + b)^{2} ≡ a^{2} + 2ab + b^{2} geometrically by taking

a = 5 units, b = 2 unit

**Answer:**

(a + b)^{2}Îža^{2} + 2ab + b^{2}

Draw a square with the side a + b i.e.,5 + 2

L.H.S of the whole square = (5 + 2)^{2} = (7)^{2} = 49

R.H.S = Area of the square with 5 units + Area of the square with 2 units +

Area of the 5,2 units + Area of the square with 2 ,5 units = 5^{2} + 2^{2} + 5×2 + 2×5 = 49

L.H.S = R.H.S

Hence,the identity is verified.

**Question 4.**

Verify the identity (a – b)^{2} ≡ a^{2} - 2ab + b^{2}geometrically by taking

a = 3 units, b = 1 unit

**Answer:**

(a – b)^{2} ≡ a^{2} - 2ab + b^{2}

Consider a square with side a.i.e.a = 3

The square is divided into 4 regions.

It consists of 2 squares with sides a-b and b respectively and 2 rectangles with length and breadth as ‘a-b’ and ‘b’ respectively.

Here a = 3 and b = 1.Therefore the 2 squares consist of sides ‘3-1’ and ‘1’ respectively and 2 rectangles with length and breadth as ‘3-1’ and ‘1’ respectively.

Now area of figure I = Area of whole square with side ‘a’ i.e.3 units-Area of figure II

-Area of figure III –Area of figure IV

L.H.S of area of figure I = (3-1)(3-1) = 2(2) = 4 units

R.H.S = Area of whole square with side 3 units-Area of figure II with 1,(3-1)units

-Area of figure III with 1,(3-1) units –Area of figure IV with 1,1 unit = 3^{2}-(1×(3-1))-

(1×(3-1))-(1×1) = 4 units

L.H.S = R.H.S

Hence,the identity is verified.

**Question 5.**

Verify the identity (a – b)^{2} ≡ a^{2} - 2ab + b^{2}geometrically by taking

a = 5 units, b = 2 units

**Answer:**

(a – b)^{2} ≡ a^{2} - 2ab + b^{2}

Consider a square with side a.i.e.a = 5

The square is divided into 4 regions.

It consists of 2 squares with sides a-b and b respectively and 2 rectangles with length and breadth as ‘a-b’ and ‘b’ respectively.

Here a = 5 and b = 2.Therefore the 2 squares consist of sides ‘5-2’ and ‘2’ respectively and 2 rectangles with length and breadth as ‘5-2’ and ‘2’ respectively.

Now area of figure I = Area of whole square with side ‘a’ i.e.5 units-Area of figure II

-Area of figure III –Area of figure IV

L.H.S of area of figure I = (5-2)(5-2) = 3(3) = 9 units

R.H.S = Area of whole square with side 5 units-Area of figure II with 2,(5-2)units

-Area of figure III with 2,(5-2) units –Area of figure IV with 2,2 units = 25-(2×3)-

( 2×3)-2^{2} = 25-6-6-4 = 9 units

L.H.S = R.H.S

Hence,the identity is verified.

**Question 6.**

Verify the identity (a + b) (a – b) ≡ a^{2} – b^{2} geometrically by taking

a = 3 units, b = 2 units

**Answer:**

Remove square from this whose side is ‘b’.(b<a)

Consider a square with side ‘a’.

We get the above figure by removing the square with side ‘b’.It consists of 2 parts

I and II.

So a^{2}-b^{2} = Area of figure I + Area of figure II = a(a-b) + b(a-b) = (a-b)(a + b)

Thus a^{2}-b^{2} = (a-b)(a + b).

Here a = 3 units and b = 2 units

So,L.H.S = a^{2}-b^{2} = 3^{2}-2^{2} = 5 units

R.H.S = (a-b)(a + b) = (3-2)(3 + 2) = 1(5) = 5 units

Therefore L.H.S = R.H.S

Hence,the identity is verified.

**Question 7.**

Verify the identity (a + b) (a – b) ≡ a^{2} – b^{2} geometrically by taking

a = 2 units, b = 1 unit

**Answer:**

Refer the above figure.

Here a^{2}-b^{2} = Area of figure I + Area of figure II = a(a-b) + b(a-b) = (a-b)(a + b)

Thus a^{2}-b^{2} = (a-b)(a + b).

Here a = 2 units and b = 1 units

So, L.H.S = a^{2}-b^{2} = 2^{2}-1^{2} = 3 units

R.H.S = (a-b)(a + b) = (2-1)(2 + 1) = 3 units

Therefore L.H.S = R.H.S

Hence,the identity is verified.