## ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 10 Algebraic Expressions and Identities Ex 10.3

Question 1.

Multiply:

(i) (5x – 2) by (3x + 4)

(ii) (ax + b) by (cx + d)

(iii) (4p – 7) by (2 – 3p)

(iv) (2x^{2} + 3) by (3x – 5)

(v) (1.5a – 2.5b) by (1.5a + 2.56)

(vi) \(\left(\frac{3}{7} p^{2}+4 q^{2}\right) \text { by } 7\left(p^{2}-\frac{3}{4} q^{2}\right)\)

Solution:

(i) (5x – 2) by (3x + 4)

= 5x (3x + 4) – 2 (3x + 4)

15x^{2} + 20x – 6x – 8

= 15x^{2}+ 14x – 8

(ii) (ax + b) by (cx + d)

= ax (cx + d) + b (cx + d)

= acx^{2} + adx + bcx + bd

(iii) (4p – 7) by (2 – 3p)

= (4p – 7) (2 -3p)

= 4p(2 – 3p) -7(2 – 3p)

= 8p – 12p^{2} – 14 + 21p

= 29p – 12p^{2} – 14

(iv) (2x^{2} + 3) by (3x – 5)

= (2x^{2} + 3) (3x – 5)

= 2x^{2}(3x – 5) + 3(3x – 5)

= 6x^{3} – 10x^{2} + 9x – 15

(v) (1.5a – 2.5b) by (1.5a + 2.5b)

= (1.5a – 2.5b) (1.5a + 2.5b)

= 1.5a(1.5 + 2.5b) – 2.5b(1.5a + 2.5b)

= 2.25a^{2} + 3.75ab – 3.75a6 – 6.25b^{2}

= 2.25a^{2} – 6.25b^{2}

Question 2.

Multiply:

(i) (x – 2y + 3) by (x + 2y)

(ii) (3 – 5x + 2 × 2) by (4x – 5)

Solution:

(i) (x – 2y + 3) by (x + 2y)

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

= x^{2} + 2xy – 2xy – 4y^{2} + 3x + 6y

= x^{2} – 4y^{2} + 3x + 6y

(ii) (3 – 5x + 2x^{2}) by (4x – 5)

= (4x – 5) (3 – 5x + 2x^{2})

= 4x(3 – 5x + 2x^{2}) – 5(3 – 5x + 2x^{2})

= 12x – 20x^{2} + 8x^{3} – 15 + 25x – 10x^{2}

= 8x^{3} – 30x^{2} + 37x – 15

Question 3.

Multiply:

(i) (3x^{2} – 2x – 1) by (2x^{2} + x – 5)

(ii) (2 – 3y – 5y^{2}) by (2y – 1 + 3y^{2})

Solution:

(i) (3x^{2} – 2x – 1) by (2x^{2} + x – 5)

= (3x^{2} – 2x – 1) (2x^{2} + x – 5)

= 3x^{2}(2x^{2} + x – 5) – 2x(2x^{2} + x – 5) -1(2x^{2} + x – 5)

= 6x^{4} + 3x^{3} – 15x^{2} – 4x^{3} – 2x^{2} + 10x – 2x^{2} – x + 5

= 6x^{4} – x^{3} – 19x^{2} + 9x + 5

(ii) (2 – 3y – 5y^{2}) by (2y- 1 + 3y^{2})

= 2(2y – 1 + 3y^{2} )- 3y (2y – 1 + 3y^{2}) -5y^{2}(2y – 1 + 3y^{2})

= 4y – 2 + 6y^{2} – 6y^{2} + 3y – 9y^{3} – 10y^{3} + 5y^{2} – 15y^{4}

= -15y^{4} – 19y^{3} + 5y^{2} + 7y – 2

Question 4.

Simplify:

(i) (x^{2} + 3) (x – 3) + 9

(ii) (x + 3) (x – 3) (x + 4) (x – 4)

(iii) (x + 5) (x + 6) (x + 7)

(iv) (p + q – 2r) (2p – q + r) – 4qr

(v) (p + q) (r + s) + (p – q)(r – s) – 2(pr + qs)

(vi) (x + y + z) (x – y + z) + (x + y – z) (-x + y + z) – 4zx

Solution:

(i) (x^{2} + 3) (x – 3) + 9

= x^{2} (x – 3) + 3(x – 3) + 9

= x^{2} – 3x^{2} + 3x – 9 + 9

= x^{3} – 3x^{2} + 3x

(ii) (x + 3) (x – 3) (x + 4) (x – 4)

= {(x + 3 (x – 3)} {(x + 4) (x – 4)}

= {x (x – 3) + 3 (x – 3)} {x (x – 4) + 4 (x – 4)}

= (x^{2} – 3x + 3x – 9) {x^{2} – 4x + 4x – 16}

= (x^{2} – 9) (x^{2} – 16)

= x^{2} (x^{2} – 16) – 9 (x^{2} – 16)

= x^{4} – 16x^{2} – 9x^{2} + 144

= x^{4} – 25x^{2} + 144

(iii) (x + 5) (x + 6) (x + 7)

= (x^{2} + 6x + 5x + 30) (x + 7)

= (x^{2} + 11x + 30) (x + 7)

= x(x^{2}+ 11x + 30) + 7(x^{2}+ 11x + 30)

= x^{3} + 11x^{2} + 30x + 7x^{2} + 77x + 210

= x^{3} + 18x^{2} + 107x + 210

(iv) (p + q – 2r)(2p – q + r) – 4qr

= p(2p – q + r) + q(2p – q + r) – 2r(2p – q + r) – 4qr

= 2p^{2} – pq + pr + 2pq – q^{2} + qr – 4pr + 2qr – 2r^{2} – 4qr

= 2p^{2} – q^{2} – 2r^{2} + pq – 3pr – 2qr

(v) (p + q)(r + s) + (p – q) (r – s) – 2(pr + qs)

= pr + ps + qr + qs + pr – ps – qr + qs – 2pr – 2qs = 0

(vi) (x + y + z)(x – y + z) + (x + y – z)(-x + y + z) – 4zx

= x^{2} – xy + xz + xy – y^{2} + yz + xz – yz + z^{2} – x^{2} + xy + xz

– xy + x^{2} + yx + xz – yz – z^{2} – 4zx = 0

Question 5.

If two adjacent sides of a rectangle are 5x^{2} + 25xy + 4y^{2} and 2x^{2} – 2xy + 3y^{2}, find its area.

Solution:

Adjacents sides of a rectangle are

5x^{2} + 25xy + 4y^{2} and

2x^{2} – 2xy + 3y^{2}

∴ Area of rectangle = Product of two sides

= (5x^{2} + 25xy + 4y^{2}) (2x^{2} – 2xy + 3y^{2})

= 10x^{4}– 10x^{3}y+ 15x^{2}y^{2} + 50x^{3}y – 50x^{2}y^{2} + 75xy^{3} + 8x^{2}y^{2} – 8xy^{3} + 12y^{4}

= 10x^{4} + 40x^{3}y – 27x^{2}y^{2} + 67xy^{3} + 12y^{4}