NCERT Solutions For Class 8 Maths Chapter 8 Algebraic Expressions And Identities

Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Expressions

While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them; then handle the unlike terms. Note that the sum of x number of like terms is another like term whose coefficient is the sum of the coefficients of the like terms being added.

Addition And Subtraction Of Algebraic Expressions Exercise 8.1

Question 1. Add the following.

xb – bc, bc – cx, cx – xb

x – b + xb, b – c + be, c – x + xc

2p²q² – 3pq + 4, 5 + 7pq – 3p²q²

l² + m², m² + n² n² + l², 2lm + 2mn + 2nl

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Solution: 1. xb – bc, bc – cx, cx – xb

Thus, the sum of the given expressions is 0.

2. x-b + xb, b-c + bc, c-x+xc

Thus, the sum of the given expressions is ab + bc + cx

3. 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²

4. l² + m², m² + n², n² + l², 2lm + 2mn + 2nl

Thus, the sum of the given expressions is 2(l² + m² + n² + Im + mn + nl).

Question 2.

1. Subtract 4x – 7xb + 3b + 12 from 12x – 9xb + 5b – 3
2. Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + lOxyz
3. Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from IS – 3p – 11q + 5pq – 2pq² + 5p²q

Solution:

Thus, the required answer is 8a – 2ab + 26- 15

Thus, the required answer is 28 + bp- 18q + 8pq- Ipq² + p²q

8.2 Multiplication Of Algebraic Expressions: Introduction

There exist x number of situations when we need to multiply algebraic expressions. For example, in finding the xerox of x rectangle whose sides are given xs expressions

Question 1. Can you think of two more such situations, where we may need to multiply algebraic expressions?

[Hint: Think of speed and time; Think of interest to be paid, the principal and the rate of simple interest; etc.].

Solution:

Distance = Speed x Time

Simple Interest

Principal x Rate of simple interest

⇒ \(=\frac{\text { per annum } \times \text { Time in years }}{100}\)

Multiplying A Monomial By A Monomial

A monomial multiplied by an x monomial always gives an x monomial.

Multiplying Two Monomials

In the product of two monomials

Coefficient = coefficient of first monomial x coefficient of second monomial

Algebraic factor – algebraic factor of first monomial x algebraic factor of second monomial

Multiplying Three Or More Monomixls

We first multiply the first two monomials and then multiply the resulting monomial by the third monomial. This method can be extended to the product of any number of monomials.

Rules Of Signs

(1) The product of two factors is positive or negative accordingly xs the two factors have like signs or unlike signs. Note that

1. (+) x (+) = +
2. (+) X (-) = –
3. (-) x (+) = –
4. (-) x (-) = +

(2) If x is x vxrixble xnd p, q xre positive inteqers, then

x^p x x^q = x^p+q

Question 1. Find 4x x 5y x 7z. First, find 4x x 5y and. multiply it by 7z; or first, find 5y x 7z and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter?
Solution:

4x x 5y x 7z

4x x 5y = (4 x 5) x (x x y)

= 20 x (xy) = 20 xy

(4x x 5y) x 7z = 20xy x lz

= (20 x 7) x (xy x z)

= 140 x (XYZ)

= 140xyz …(1)

OR

5y x 7z = (5 x 7) x (y x z)

– 35 X (yz) = 35yz

4x X (5y X 7z) = 4x X 35yz

= (4 x 35) x (x x yz)

= 140 x (xyz)

= I40xyz…..(2)

We observe from Eqns. (1) and (2) that the result is the same. It shows that the product of monomials is associative, i.e., the order in which we multiply monomials does not matter.

Addition And Subtraction Of Algebraic Expressions Exercise 8.2

Question 1. Find the product of the following pairs of monomials:

1. 4, 7p
2. -4p, 7p
3. -4p, 7pq
4. 4p³, -3p
5. 4p, 0.

Solution:

(1)  4, 7p

4 x 7p = (4 x 7) x p

= 28 x p

= 28p

(2)  – 4p, Ip

(-4p)x(7p) = {(-4) x 7} x (P x P)

= (-28) x p2

= -28p2

(3)  -4p, 7pq

(- 4p) x (Ipq) = {(- 4) x 7} x (p x (pq)}

= (-28) x (p x p x q)

= (-28) x p²

= -28p²q

(4)  4p², – 3p

(4p³) x ( -3p) = {4 x (_ 3)} x (p³ Xp)

= (-12) X p⁴

= – 12p⁴

(5)  4p, 0

(4p) x 0 = (4×0) x p

0 x p = 0

Question 2. Find the press of rectangles with the following pairs of monomials xs their lengths and breadths respectively:

(p, q) ; (10m, 5n); (20x² 5y²); (4x, 3.x²); (3mn, 4np).

Solution:

1.  (p, q)

Length = p

Brexdth = q

Area of the rectangle

= Length x Brexdth

= P x q

= pq

2.  (10m, 5n)

Length = 10 m

Brexdth = 5 n

Area of the rectangle

= Length x Brexdth

= (10m) x (5n)

= (10 x 5) x (m x n)

= 50 x (mn)

= 50 mn

3. (2Ox² , 5y²)

Length = 20X²

Brexdth = 5y²

Area of the rectangle

= Length x Brexdth

= (20x²) x (5y²)

= (20 x 5) x (x² x y²)

= 100 x (x2/)

= 100x²y²

4.  (4x,3x²)

Length = 4x

Brexdth = 3X²

Area of the rectangle

= Length x Brexdth

= (4x) x (3x²)

= (4 x 3) x (x x x²)

= 12 x x³ = 12X³

5.  (3mn, 4np)

Length = 3mn

Brexdth = 4np

Area of the rectangle

= Length x Brexdth

= (Smn) x (4np)

= (3×4) x (mn) x (np)

= 12 x m x (n x n)xp

= 12 mn²p.

Question 3. Complete the table of products.

Solution:

We have.

1. 2x x -5y = {2 x (- 5)} x (x x y) = – lOxy

2x x 3X² = (2 x 3) x (x x x²) = 6X³

2x x (- 4xy) = {2 x (-4)} x x x (xy)

= -8 X (x X x) X y

= – 8 x²y

2x X 7x²y = (2 X 7) X (x X x²) x y = 14x²y

2x x (-9X²y²) = {2 x (-9)> x (x x x²) Xy²

= – 18x³y²

2. (-5y)x(2x)

= {(-5) x 2} X (y X x)

= – 10 30

(_5y) x (- 5p)

= {(-5) x (-5)} x y xy

= 25 y²

(~5y) x 3X2

= {(-5) x 3} x yx2

= — 15.x²y

(-5y) x (— 430’)

= {(-5) X (- 4)} x XX (y X y)

= 20 xy²

(-5y) x (7x2y)

= {(-5) X 7} X X2 X (yXy)

= -35 x²y²

(~5y) x (-qx2ÿ)

= {(-5) x (- 9)} x x2x(yxy2)

= 45 x²y³

3. 3x² x 2x = (3 x 2) x (x2 x x) = 6X²

3X² x 3x² = (3 x 3) x (x² x x²) = 9 x⁴

3x⁴ x (- 4 xy) = 3 x (- 4)} x (x2 x x) x y = – 12x³y

3x² x 7x²y

= (3 x 7) x (x2 x x1) xy

= 21×4;

3x² x (- 9X² y²)

= {3 x (- 9)} x (x² x x²)xy²

= – 27xy

4. (-4xy) x 2x

= {(_4) x 2} x {x x x) xy

= -8x²y

(-4xy) x (-5y)

= {(- 4) x (- 5)} X X x (y x y)

= 20xy²

(-4x²y) x 3X²

= {(- 4) x 3} x (ix x²) x y

= – 12x²y

(-430) x (- 4ry)

= {(-4) X (_4)}x (ixi)x(y Xy)

= 16x²y²

(-4xy) X Ixÿy

= {(-4) X 7} X (x X X2) X (y X y)

= -28x³y²

(- 4xy) X (- 9X2/)

={(-4) x (-9)} X (x X x2) X (y X y2)

= 36x³y³

5. 7x²y x 2x

= (7 x 2) x (x2 x x) x y

= 14x²y

7x²y x (- 5y)

= {7 x (-5)} x x² X (y Xy)

= – 35x²y²

7x²y X(3x²)

– (7 x 3) X (*2 x x2) Xy = 21x4y

7x²y x (- 4 xy)

= {7 X (-4)} X (x² x x) X (y X y)

= -28x³y²

7x2y x 7x2p

= (7 x 7) X (x² X X²) x (y X y)

= 49x⁴y²

7-X²y X (_ 9x²ÿ2)

= {7X(-9)} X (x²X X²) X (y x y²)

= -63x⁴y³

(- 9x²,y²) x 2x

= {(-9) x 2} x x( x x) x y2

= -18x³y²

(-9xV) x (-5y)

= {(-9) x (-5)} x x² x x² (y²x y)

– 45x²y³

(- 9x2y²) x (3X²)

= {(- 9) x 3} x (x² x x²) x y2

= – 27x⁴ y²

(- 9x²y) x (- 4xy)

= {(-9) x (-4)} x (x² x X) x (y2 x y)

= 36 x⁴

(~9x2y2) x (7x2y)

= {(-9)x7}x(x2Xx2)X(y2xy)

= – 63 xy

(-9x2y²) x (- 9xY)

= {(-9) x (-9)}x ft? xÿ x (y2 x y2)

= 81 x⁴y⁴

Question 4. Obtain the volume of rectangular boxes with the following length, breadth, and height respectively:

1. 5x, 3x², 7x.4
2. 2p, 4q, 8r
3. xy, 2x²y, 2xy²
4. x, 2b, 3c.

Solution:

1.  5x, 3x², 7×4

Length = 5x

Brexdth = 3x²

Height = 7×4

Volume of the rectxnqulxr box

= Lenqth x Brexdth x Heiqht

= (5x) x (3x²) x (7×4)

= (5 x 3 x 7) x (x x x² x x4)

= 105×7

2.   2p, xq, 8r

Lenqth = 2p

Brexdth = xq

Heiqht = 8r

Volume of the rectxnqulxr box

= Lenqth x Brexdth x Heiqht

= (2p) x (xq) x (8r)

= (2 x 4 x 8) x (p x q x r)

= 64 pqr

3.  xy, 2x²y, 2xy²

Length = xy

Brexdth = 2x²y

Height = 2xy²

The volume of the rectangular box

= Length x Breadth x Height

= (xy) x (2x²y) x (2xy²)

= (2 x 2) x (x x x² x x) x (yxyxy²)

= 4x4y4

4.  x, 2b, 3c

Length = x

Brexdth = 2b

Height = 3c

The volume of the rectangular box

= Length x Breadth x Height

= (x) x (2b) x (3c)

= (2 x 3) x (x x b x c)

= 6xbc

Question 5. Obtain the product of

1. xy, yz, zx
2. x, -x², x3
3. 2, 4y, 8y2, 16y³
4. x, 2b, 3c,6xbc
5. m, -mn, mn

Solution:

1.  xy, yz, zx

Required product

= (xy) x (yz) x (zx)

= (x x X) X (y X y) X (Z X z)

= x² x x ²z²

= x2²y²z²

2.  x, -x², x³

Required product

= (x) x (-x²) x (x²)

= – (x x x² x x³)

= -x⁶

3.  2, 4y, 8y² 16y³

Required product

= (2) x (4y) x (8y²) x (16y³)

= (2 x 4 x 8 x 16) x (y x y2 x y3)

= 1024y⁶

4.  x, 2b, 3c, 6xbc

Required product

= (x) x (26) x (3c) x (6xbc)

= (2 x 3 x 6) x (x x x) x (6 x 6) x(c x c)

= 36 x²b²c²

5.  m, – mn, mn

Required product

= (m) x (-mn) x (mnp)

= (- 1) x (m x m x m) x (n xn) xp

= -m²n²p

Multiplying A Monomial By A Polynomial

While multiplying x polynomial by x monomial, we multiply every term in the polynomial by the monomial.

Using the distributive law, we carry out the multiplication term by term.
It states that if P, Q, and R are three monomials, then

1. P x (Q + R) = (P x Q) + (p x R)
2.  (Q + R) x p = (Q x P) + (R x P)

Question. Find the product:

1.  2x(3x + 5xy) (H)x,2 (2xb – 5c).

Solution:

1.  2x (3x + 5ry)

2x (3x + 5xy) = (2x) x (3x) + (2x) x (5xy)

= (2 x 3) x(xxx) + (2 x 5) x (x x x) x y

= 6x² + 10 x2y

2.  a² (2xb-5c)

x2(2xb – 5c) = (x2) x (2xb) – (x2) x (5C)

= (1 x 2) x (x2 x x) X b -(1 X 5) X x2 X c

= 2a³b – 5a²c

8.4.2 Multiplying A Monomixl By A Trinomial

By using the distributive law, we carry out the multiplication term by term

Question. Find the product: (4p² + 5p + 7) x 3p.
Solution:

(4p² + 5p + 7) x 3p.

= (4p²x 3p) + (5p x 3p)+ (7 x 3p)

= (4 x 3) x (p² x P) + (5 x 3) x (p x p) + (7 x 3) x p

= 12p3+ 15p² + 21p

Addition And Subtraction Of Algebraic Expressions Exercise 8.3

Question 1. Carry out the. multiplication of the expressions in exch for the following pairs:

1. 4p, q + r
2. ab, a – b
3. a + b, 7a²b²
4. a²-9, 4a
5. PQ + qr + rp, 0

Solution:

1.   4p, q + r

(4p) x (q+ r) = (dp) x (q) + (4 p) x (r)

By distributive law

= (4 x 1) x p x q + (4 x 1) x p x r

= 4pq + 4pr

(2) xb, x-b

(ab) x (a-b) = (ab) x (a) – (ab) x (b)

By distributive law

= (1 x 1) x (a x a) xb

– (1 x 1) a x (b x b)

= a²b – ab²

3.  a+b, 7a²b²

(x +b) x (7a²b²)

= x x 7a² b² + 5 x 7×2 b2

By distributive law

= (1 x 7) x (a x x²) x b² + (1 x 7) x a² x (a x b2)

= 7a³b² + 7a²b³

a²– 9, 4a

(a²– 9) x (4a)

= a²x 4a – 9 x 4a

By distributive law

= ( 1 x 4) x (a² x a) – (9 x 4) x a

= 4a³-36a

PQ + qr + rp, 0

(PQ + qr+ rp) x (0)

= (pq) x 0 + (qr) x 0 + (rp) x 0.

By distributive law

= 0 + 0 + 0 = 0.

Question 2. Complete the table:

Solution:

We have,

(1) a x (b + c + d)

= a x b + a x c + a x d

= ab + ac + ad

(2) (X + y – 5) x 5xy

= pC x 5pc_y + y x 5*y + (-5) x 5xy

= (1 X 5) X (x x X) X y

+ (1 x 5) X X X (y X y)

+ {(-5) x 5}x xy

= 5x²y + 5xy² – 25xy

(3) p x (6p² – 7p + 5)

= p x 6p² + p x (- 7p) + p x 5

= (1 x 6) x (p xp²)

+ {1 x (- 7)} x (p x p) + 5 x p

= 6p³ – 7p² + 5p

(4) 4p²q² x (p² – q²)

= 4p²q² x p² + 4p²q² x (-q²)

= (4 x 1) x (p² x p²) x q²

+ {4 X (- 1)} X p² X (q² X q²)

= 4p⁴ q² – 4p² q⁴.

(5)(a + b + c) x abc

= a x abc + b x abc + c x abc

= (ax a) x b x c

+ a x (b x b) x c

+ a x b x (c x c)

= a² bc + ab² c + abc²

Question 3. Find the product:

(a²) x(2a²²) x (4a²⁶)

⇒ \(\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^2 y^2\right)\)

⇒ \(\left(-\frac{10}{3} p q^3\right) \times\left(\frac{6}{5} p^3 q\right)\)

x x x² x x³x x⁴

Sol. (i) (x2) x (2×22) x (4×26)

= (1 x 2 x 4) x (x2 x x22 x x26)

= 8 X a^2+22+26

Bylaws of exponents

= 8 x a⁵⁰ = 8×50

⇒ \(\left(\frac{2}{3} x y\right) \times\left(-\frac{9}{10} x^2 y^2\right)\)

⇒ \(\left\{\frac{2}{3} \times\left(-\frac{9}{10}\right)\right\} \times\left(x \times x^2\right) \times\left(y \times y^2\right)\)

⇒ \(-\frac{3}{5} x^3 y^3\) I by laws of exponents

3.   \(\left(-\frac{10}{3} p q^3\right) \times\left(\frac{6}{5} p^3 q\right)\)

⇒ \(=\left\{\left(-\frac{10}{3}\right) \times \frac{6}{5}\right\} \times\left(p \times p^3\right) \times\left(q^3 \times q\right)\)

= -4p⁴q⁴.    Bylaws of exponents

(4) X x x² X x³ X x⁴

X x x2 x x3 X x4 = (1 X 1 X 1 X 1) X X¹ X x² X x³ X x⁴

= (1) X x ^1+2+3+4

= X10.

Bylaws of exponents

Question 4. (x) Simplify: 3x (4x -5) +3 and find Us values for (i) x = 3, (ii) x =\(\frac{1}{2}\)

b) Simplify: x(x2 + x + 1) + 5 xnd find its vxlue for

(i) x = 0, (ii) x = l xnd (Hi) x=-l.

Solution :

1.  3x (4x-5) + 3

= (3x) (xx) – (3x) (5) + 3

= (3 x 4) x (x x x) – (3 x 5) x x + 3

= 12x² – 15x + 3

(i) Whenx = 3,

12x² – 15.x + 3

= 12(3)² – 15(3) + 3

= 12 x 9 – 45 + 3

= 108 – 45 + 3 = 66

2.  When x =\(\frac{1}{2}\)

12X² – 15.x + 3

⇒ \(12\left(\frac{1}{2}\right)^2-15\left(\frac{1}{2}\right)+3\)

⇒ \(12\left(\frac{1}{4}\right)-\frac{15}{2}+3\)

⇒ \(3-\frac{15}{2}+3\)

⇒ \(6-\frac{15}{2}\)

⇒ \(\frac{12-15}{2}\)

⇒ \(-\frac{3}{2}\)

(b) a(a² + x + 1) + 5

a x a² + a x a + (a x 1 + 5 )

= a³ + a² + a+ 5

1.  When a = 0

a³ + a²+ a + 5 = (0)3 + (0)2 + (0) + 5

=0+0+0+5=5

2.  When a =1

a³ + a² + a+ 5 = (l)3 + (l)2 + (1) + 5

=l+l+l+5=8

3.  When a=-l

a³ +a² + a+ 5

= (- 1)3 + (- l)2 + (- 1) + 5

= —1 + 1 — 1 + 5 = 4

Question  5.

1. add : p(p – q), q(q – r) and r(r-p)
2. add: 2x(z – x – y)and 2y (z -y- x)
3. Subtract: 31 (l – 4m. + 5n) from 41 (10n-3m + 21)
4. Subtract: 3x(x +b+c)-2b(x-b+c) from 4c(- x + b+ c).

Solution:

(1) First expression

= p(p-q)=pxp-pxq

= p² – pq

Second expression

= q(q-r)

= q x q – q x r

= q² – qr

Third expression

= r(r – p)

=r x r – r x p

= r² – rp

adding the three expressions,

image

First expression

= 2x (z – x – y)

= (2x) X (2)-(2x) X (x) – (2x) X (y)

= 2xz -2x² – 2xy

Second expression

= 2y(z-y-x)

= (2y) x (z) – (2y) x (y) – (2y)x(x)

= 2yz – 2y² – 2yx

adding the two expressions,

image

First expression

= 31(1 – 4m + 5n)

= (31) x (l) – (31) x (4m) + (31) x (bn)

= 312 – 12lm + 15In

Second expression

= 41 (10n – 3m + 2l)

= (4l) x (lOn) – (4l) x (3m) + (4l) x (2l)

= 40ln – 12lm + 8l²

Subtrading,

image

First expression

= 3x (x + 6 + c) – 26 (x – 6 + c)

= [(3x) x (x) + (3x) x (6) + (3x) x (c)]

-[(26) x (x) + (26) x (6) – (26) x (c)]

= [3a² + 3ab + 3ac]- [2ab- 2b² + 2bc]

= 3a² + 2b² + 3ab – 2ab- 2bc + 3ac

– 3a² + 2b² + ab- 2bc + 3ac

Second expression

= 4c (- x + 6 + c)

= 4c x (-x) + 4c x 6 + 4c x c

= -4xc + 46c + 4c2

Subtracting

image

8.5 Multiplying A Polynomial By A Polynomial

We multiply the term of one polynomial by each term of the other polynomial. Also, we combine the like terms in the product.

Multiplying A Binomial By A Binomial

We use distributive law and multiply each of the two terms of one binomial by exchanging the two terms of the other binomial and combining like terms in the product.

Thus, if P, Q, R, and S are four monomials, then

(P + Q) x (R + S) = P x (R + S) +Qx(R + S)

= (P x R + P x S) + (QXR + QXS)

= PR + PS + QR + QS.

8.5.2 Multiplying A Binomial By A Trinomial

We use distributive law and multiply each of the three terms in the trinomial by exchanging the two terms in the binomial and combining like terms in the product.

Addition And Subtraction Of Algebraic Expressions Exercise 8.4

Question 1. Multiply the binomial:

1. (2x + 5) and (4x – 3)
2. (y – 8) and (3y – 4)
3. (2.51 – 0.5 m) and (2.51 + 0.5m)
4. (x + 3b) and (x + 5)
5. (2pq + 3q²) and (3pq – 2q²)
6. \(\left(\frac{3}{4} x^2+3 b^2\right) \text { and } 4\left(x^2-\frac{2}{3} b^2\right)\)

Solution:

(i) (2x + 5) xnd (4x – 3)

(2x + 5) x (4x – 3)

= (2x) x (4x – 3) + 5 x (4x – 3) by distributive law

= (2x) x (4x) – (2x) x (3) + (5) x (4x) – (5) x (3)

= 8x² – 6x + 20x – 15

= 8x² + (20x – 6x) – 15 Combining like terms

= 8x² + 14x – 15

2. (y – 8) and (3y – 4) by distributive law

(y – 8) x (3y – 4)

= y x (3y – 4) – 8 x (3y – 4) – (8) x (3y) + 8 x 4

= (y) x (3y) – (y) x (4)

= 3y² – 4y -24y + 32 Combining like terms

3. (2.5 l – 0.5 m) xnd (2.5 l + 0.5 m)

(2.5 l – 0.5 m) x (2.5 l + 0.5 m)

= (2.5/) x (2.5/ + 0.5 m) – (0.5 m) x (2.5/ + 0.5 m) by distributive law

= (2.5 Z) x (2.5 Z) + (2.5 /) x (0.5 m) – (0.5 m) x (2.5 /) – (0.5 m) x (0.5 m)

= 6.25 12 + 1.25 Zm – 1.25 m/- 0.25 m² Combining like terms

= 6.25 l² + (1.25 lm – 1.25 lm) – 0.25 m²

= 6.25 l² – 0.25 m²

4. (x + 3 b) and (x + 5)

(a + 35) x (x + 5)

= a x (c +5) + (3b) x (x + 5) by distributive law

= (a) x (x) + (a) x (5) + (3b) x (x) + (3b) x (5)

= ax + 5a + 3bx + 15b

5.  (2pq + 3q²) xnd. (3pq – 2q²)

(2pq + 3q²) x (3pq – 2q²)

= (2pq) x (3pq – 2q²) + (3q²) x (3pq – 2q²) by distributive law

= (2pq) x (3pq) – (2pq) x (2q²) + (3q²) x (3pq) – (3q²) x (2q²)

= 6p2q2 – 4pq³ + 9pq³ – Qqi

= 6p²q² + (9pq3 – 4pq3) – 6q4 Combining like terms

= 6p²q² + 5pqz – 6q⁴

⇒ \(\left(\frac{3}{4} x^2+3 b^2\right) \text { and } 4\left(x^2-\frac{2}{3} b^2\right)\)

⇒ \(\left(\frac{3}{4} x^2+3 b^2\right) \times 4\left(x^2-\frac{2}{3} b^2\right)=\left(\frac{3}{4} x^2+3 b^2\right) \times\left(4 x^2-\frac{8}{3} b^2\right)\)

⇒ \(\frac{3}{4} x^2 \times\left(4 x^2-\frac{8}{3} b^2\right)+3 b^2 \times\left(4 x^2-\frac{8}{3} b^2\right)\) by distributive law

⇒ \(\left(\frac{3}{4} x^2\right) \times\left(4 x^2\right)-\left(\frac{3}{4} x^2\right) \times\left(\frxc{8}{3} b^2\right)+\left(3 b^2\right) \times\left(4 x^2\right)-\left(3 b^2\right) \times\left(\frac{8}{3} b^2\right)\)

= 3a⁴ – 2a²b² + 12b²a² – 8b⁴

= 3a⁴ + (12a²b² -2a²b²) -8b⁴

= 3a⁴ + 10a²b² – 8b⁴Combining like terms

Question 2. Find the product:

1. (5 – 2x) (3 + x)
2. (x + 7y) (7x-y)
3. (x² + b) (x + b²)
4. (p² – q²) (2p + q)

Solution:

(i) (5 – 2x) (3 + x)

= (5) x (3 + x) – (2x) x (3 + x)

= (5) x (3) + (5) x (x) – (2x) X (3) – (2x) X (X)

= 15 + 5x – 6x – 2x²

= 15 – x – 2x²

2.  (x + 7y) (7x – y)

= (x) x (7x – y) + (7y) x (7x – y)

= (x) x (7x) – (x) x (y) + (7y) x (7x) – (7y) x (y)

= 7x² – xy + 49yx – 7y²

= 7x² + 48xy – 7y²

3.  (a² + b) (a + b²)

= a² x (a+ b²) + b x (a + b²)

= (a²) x (a) + (a²) x (b²) + (b) x (a) + (b) x (b²)

= a³ + a²b² + ab + b³

= a³ + a²b² + ab + b³

4.  (p² – q²) (2p + q)

= p² x (2p + q) – q² x (2p + q)

= (p²) x (2p) + (p²) x (q) – (q²) x (2p)- (q²) x (q)

= 2p³+ p²q – 2q²p – q³

= 2p³ + p²q – 2pq² – q³

Question 3. Simplify :

1. (x² – 5) (x+ 5) + 25
2. (x² + 5) (b³ + 3) + 5
3. (l + s²) (l² – s)
4. (x + b)(c – d) + (x – 6) (c + d)+ 2 (xc + bd)
5. (x + y) (2x + y) + (x + 2y)(x – y)
6.  (x + y) (x² – xy + y²)
7. (1.5x – 4y)(1.5x + 4y + 3)- 4.5x + I2y
8. (x + b + c) (x. + b – c).

Solution:

1.   (x² – 5) (x + 5) + 25

= x²(x + 5) – 5(x + 5) + 25

= (x² x X) + (x² x 5) – (5 x X) – (5 x 5) + 25

= x³ + 5X² – 5x – 25 + 25

= x³ + 5×2 – 5x

2.  (a² + 5) (b³ + 3) + 5

= a²(b³ + 3) + 5(b³ + 3) + 5

= (a² x b³) + (a² x 3) + (5 x b³) +(5a³) + 5

= a²b³ + 3a² + 5b³ + 15 + 5

= a²b³ + 3a² + 5b³ + 20

3.  (t + s²) (t² – s)

= t(t² – s) + s²(t² – s)

= (tx t²) – (t X s) + (s² X t²) – (s² X s)

= t² -ts + s²t² – s³

4.  (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)

= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2(oc + bd)

= ac – ad + be – bd + ac + ad – be – bd + 2ac + 2bd by distributive law

= (ac + ac + 2ac) + (ad- ad) + (be- be) + (2bd – bd – bd)

= 4ac combines the like terms

5.  (x + y) (2x + y) + (x + 2y) (x – y)

= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)

= (X X 2x) + (x X y) + (y X 2x) + (y x y) + (x x x) – (x X y) (2y x x)- (2y x y

By distributive law

– 2×2 + xy + 2xy + y² + x2 – xy + 2xy – 2y²

= (2x² + x²) + (xy + 2xy – xy + 2xy) + (y2 – 2y²)

= 3x² + 4xy – y²

6.  (x + y) (x² – xy + y²)

x(x² – xy + y²)= y(x² – xy + y²)

(x x x²) – (x x xy) + (x x y²) + (y x x²) – (y x xy) + (y x y² ) | by distributive law

x³ – x²y + xy² + x²y – xy² + y3

= x^-3 + x²y – x²y) + (xy² – xy²) + y3   combining the like terms

= x^–3 + y²

7.  (1.5x – 43-) (1.5x + 4y + 3) – 4.5x + 12y

= 1.5x (1.5x + 4y + 3)- 4y (1.5x + 4y + 3) – 4.5x + 12y | by distributive law

= (1.5x x 1.5x) + (1.5x x 4y) + (1.5x X 3) – (4y x 1.5x) – (4y x 4y) – (4y x 3) – 4.5x + 12y | by distributive law

2.25x² + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y

= 2.25x² + (6xy – 6xy) – 16y² + (4.5x – 4.5x) + (12y – 12y) I combining the like terms

= 2.25x² – 16y²

8.  (a + b + c) (a + b – c)

= a(a + b-c) + b(a + b-c) + c(a + 6-c) I by distributive law

= a2 + ab – ac + ab + b² – be + ae + be – c²    I by distributive law

= a²+ (ab + ab) + (ac – ac) + b² + (be – be) – c² combining the like terms

= a² + 2ab + b2 – c²

Addition And Subtraction Of Algebraic Expressions Multiple-Choice Questions And Solutions

Question 1. The expression x + 3 is in

1. one variable
2. two variables
3. no variable
4. none of these.

Solution: 1. one variable

Question 2. The expression 4xy + 7 is in

1. one variable
2. two variables
3. no variable
4. none of these.

Solution: 2. two variables

Question 3. The expression x + y + z is in

1. one variable
2. no variable
3. three variables
4. two variables.

Solution: 3. three variables

Question 4. The value of 5x when x = 5 is

1. 5
2. 10
3. 25
4. -5.

Solution: 3. 25

Question 5. The value of x² – 2x + 1 when x = 1 is

1. l
2. 2
3. -2
4. 0.

Solution: 4. 0

Question 6. The value of x² + y2 when x = 1, y = 2 is

1. l
2. 2
3. 4
4. 5.

Solution: 4. 5

Question 7. The value of x² – 2yx + y2 when x = 1, y = 2 is

1. l
2. -l
3. 2
4. -2.

Solution: 1. 1

Question 8. The value of x² – xy + y2 when x = 0, y = 1 is

1. 0
2. -l
3. l
4. none of these.

Solution: 3. 1

Question 9. The sum of lx, lOx, and 12x is

1. 17x
2. 22x
3. 19x
4. 29x.

Solution: 4. 29x.

Question 10. The sum of 8pq and -17pq is

1. pg
2. 9pqr
3. -9pq
4. -pg.

Solution: 3.-9pq

Question 11. The sum of 5x², -7X², 8x², 11x² and -9x² is

1. 2x²
2. 4x²
3. 6x²
4. 8x²

Solution: 4. 8x².

Question 12. The sum ofx² y²-z² and z²-x² is

1. 0
2. 3x²
3. 3y²
4. 3z²

Solution: 1. 0

Question 13. What do you get when you subtract – 3xy from 5xy?

1. 3xy
2. 5xy
3. 8xy
4. xy.

Solution: 3. 8xy

Question 14. The result of the subtraction of law from 0 is 4

1. 0
2. 7x
3. -7x
4. x

Solution: 3. -7x

Question 15. The result of subtraction of 3x from – 4x is

1. -7x
2. 7x
3. x
4. -x.

Solution: 1. -7x

Question 16. The product of 4mn and 0 is

1. 0
2. 1
3. mn
4. 4mn.

Solution: 1. 0

Question 17. The product of 5x and 2y is

1. xy
2. 2×7
3. 5xy
4. 10xy.

Solution: 4. 10xy.

Question 18. The product of lx and -12x is

1. 84x²
2. -84x²
3. x²
4. -x².

Solution: 2. -84x²

Question 19. The area of a rectangle whose length and breadth are 9 and 4y respectively is

1. 4y³
2. 9y³
3. 36y³
4. 13y³

Solution: 3. 36y³

Question 20. The area of a rectangle with length 2l2m and breadth 3Im? is

1. 6l3m³
2. l3m³
3. 2l3m³
4. 4l3m³

Solution: 1. 6l3m³

Question 21. The volume of a cube of side 2a is

1. 4a³
2. 2a
3. 8a³
4. 8.

Solution: 3. 8a³

Question 22. The volume of a cuboid of dimensions a, 6, c is

1. abc
2. a2b2c2
3. a3b3c3
4. none of these.

Solution: 1. abc

Question 23. The product of x², – x³, – x⁴ is

1. x⁹
2. x⁵
3. x⁷
4. x⁶

Solution: 1. x⁹

Addition And Subtraction Of Algebraic Expressions True-False

Write whether the following statements are True or False:

1. In a polynomial, the exponents of the variables are always non-negative integers:   True

2. a (b + c) = ab + ae is called the distributive property : True

3. a² – b² is the product of (a – b) and (a + b): True

Addition And Subtraction Of Algebraic Expressions Fill in the Blanks

1. The value of, when 92 -72 = 8p, is___________ : 4

2. On subtracting -a²b² from 2ab², we get___________:  2a²b²

3. The difference between the squares of two consecutive natural numbers is their_______: Sum

4. Area of a rectangle with length 3 ab² and breadth 4 ac² is_______:  12a²b²c²

5. Write the product of the monomials -12a, -15a², a²b:  180 a³b³

6. Add a² + b² – c²  b²  + c² – a²  and c²  + a² – b² :  a² + b² + c²

7. Find the value of, if lOOOy = (981)² – (19)²:  962