ICSE

ML Aggarwal Class 8 Factorisation Check Your Progress

Question 1.
Find the HCF of the given polynomials:
(i) 14pq, 28p²q²
(ii) 8abc, 24ab², 12a²b
Solution:
(i) 14pq, 28p²q² (HCF of 14, 28 = 14)
HCF of 1 4pq, 28p²q² = 14pq
(ii) 8abc, 24ab², 12a²b
HCF of 8, 24, 12 = 4
HCF of 8abc, 24ab², 12a²b = 4ab

Question 2.
Factorise the following:
(i) 10x² – 18x³ + 14x⁴
(ii) 5x²y + 10xyz + 15xy²
(iii) p²x² + c²x² – ac² – ap²
(iv) 15(x + y)² – 5x – 5y
(v) (ax + by)² + (ay – bx)²
(vi) ax + by + cx + bx + cy + ay
(vii) 49x² – 70xy + 25y²
(viii) 4a² + 12ab + 9b²
(ix) 49p² – 36q²
(x) 100x³ – 25xy²
(xi) x² – 2xy + y² – z²
(xii) x⁸ – y⁸
(xiii) 12x³ – 14x² – 10x
(xiv) p² – 10p + 21
(xv) 2x² – x – 6
(xvi) 6x² – 5xy – 6y²
(xvii) x² + 2xy – 99y²
Solution:
(i) 10x² – 18x³ + 14x⁴
HCF of 10, 18, 14 = 2
∴ 10x² – 18x^{3/sup> + 14 x 4
= 2x2 (5 – 9x + 7x2)}

(ii) 5x²y + 10xyz + 15xy²
HCF of 5, 10, 15 = 5
∴ 5x²y + 10xyz + 15xy²
= 5xy(x + 2z + 3y)

(iii) p²x² + c²x² – ac² – ap
= p²x² – ap² + c²x² – ac²
p²(x² – a) + c²(x² – a)
= (x² – a) (p² + c²)

(iv) 15(x + y)² – 5x – 5y
= 15(x + y)² – 5(x + y)
= 5(x + y) [3(x + y) – 1]
= 5(x + y) (3x + 3y – 1)

(v) (ax + by)² + (ay – bx)²
a²x² + b²y² + 2abxy + a²y² + b²x² – 2abxy
= a²x² + a²y² + b²x² + b²y²
= a²(x² + y²) + b²(x² + y²)
= (x² + y²) (a² + b²)

(vi) ax + by + cx + bx + cy + ay
= ax + bx + cx + ay + by + cy (grouping)
= x(a + b + c) + y(a + b + c)
= (a + b + c) (x + y)

(vii) 49x² – 70xy + 25y²
= (7x)² – 2 × 7x × 5y + (5y)²
{∵ (a – b)² = a² – 2ab + b²}
= (7x – 5y)²

(viii) 4a² + 12ab + 9b²
= (2a)2² + 2 × 2a × 3b + (3b)²
{∵ (a + b)² = a² + 2ab + b²}
= (2a + 3b)²

(ix) 49p² – 36q²
= (7p)² – (6q)²
= (7p + 6q) (7p – 6q)
{∵ a² – b² = (a + b) (a – b)}

(x) 100x³ – 25xy²
= 25x(x² – y²) = 25x{(x)² – (y)²}
= 25x(x + y) (x – y)

(xi) x² – 2xy + y² – z²
= (x – y)² – (z)²
{∵ a² -2ab + b² = (a – b)²)
a² – b² = (a + b)(a – b)}
= (x – y + z)(x – y – z)

(xii) x⁸ – y⁸
= (x⁴)² – (y⁴)²
{∵ a² – b² = (a + b)(a- b)}
= (x⁴ + y⁴)(x⁴ – y⁴)
= (x⁴ + y⁴) {(x²)² – (y²)¹}
= (x⁴ + y⁴) (x² + y²) (x² – y²)
= (x⁴ + y⁴ (x² + y²) (x + y) (x – y)

(xiii) 12x³ – 14x² – 10x
= 2x(6x² – 7x – 5)
{∵ 6 × (-5) = -30
∴ -30 = -10 × 3
-7 = -10 + 3}
= 2x{6x² + 3x – 10x – 5}
= 2x{3x(2x + 1) – 5(2x + 1)}
= 2x(2x + 1) (3x – 5)

(xiv) p² – 10p + 21
= p² – 3p – 7p + 21
{∵ 21 =-3 × (-7)
-10 = -3 – 7}
= p(p – 3) – 7(p – 3)
= (p – 3)(p – 7)

(xv) 2x² – x – 6
= 2x² – 4x + 3x – 6
{ ∵ -6 × 2 = -12
∴ -12 = -4 × 3
-1 = -4 + 3}
= 2x(x – 2) + 3(x – 2)
= (x – 2) (2x + 3)

(xvi) 6x² – 5xy – 6y²
= 6x² – 9xy + 4xy – 6y²
{∵ 6 × (-6) = -36
∴ – 36 = -9 × 4
– 5 = -9 + 4}
= 3x(2x – 3y) + 2y(2x – 3y)
= (2x – 3y) (3x + 2y)

(xvii) x² + 2xy – 99y²
= x² + 11xy – 9xy – 99y²
{∵ -99 = -11 × 9
-2 = -11 + 9 }
= x(x + 11y) – 9y(x + 11y)
= (x + 11y) (x – 9y)

Question 3.
Divide as directed:
(i) 15(y + 3)(y² – 16) ÷ 5(y² – y – 12)
(ii) (3x³ – 6x² – 24x) ÷ (x – 4) (x + 2)
(iii) (x⁴ – 81) ÷ (x³ + 3x² + 9x + 27)
Solution:
(i) 15(y + 3)(y² – 16) ÷ 5(y² – y – 12)
y² – 16 = (y)² – (4)²
= (y + 4)(y – 4)
y² – y – 12 = y² – 4y + 3y – 12
= y(y – 4) + 3(y – 4)
= (y – 4)(y + 3)
Now, \(\frac{15(y+3)\left(y^{2}-16\right)}{5\left(y^{2}-y-12\right)}\)
\(\frac{15 \times(y+3)(y+4)(y-4)}{5(y-4)(y+3)}\)
= 3(y + 4)

(ii) (3x³ – 6x² – 24x) ÷ (x – 4) (x + 2)
3x³ – 6x² – 24x = 3x(x² – 2x – 8)
= 3x{x² – 4x + 2x – 8}
= 3x{x(x – 4) + 2(x – 4)}
= 3x(x – 4) (x + 2)
∴ \(\frac{3 x^{3}-6 x^{2}-24 x}{(x-4)(x+2)}=\frac{3 x(x-4)(x+2)}{(x-4)(x+2)}=3 x\)

(iii) (x⁴ – 81) ÷ (x³ + 3x² + 9x + 27)
x⁴ – 81 = (x²)² – (9)² = (x² + 9) (x² – 9)
= (x² + 9) {(x)² – (3)²}
= (x² + 9) (x + 3) (x – 3)
x³ + 3x² + 9x + 27
= (x)² + (x + 3) + 9 (x + 3)
= (x² + 9) (x + 3)
Now, \(\frac{x^{4}-81}{x^{3}+3 x^{2}+9 x+27}\)
= \(\frac{\left(x^{2}+9\right)(x+3)(x-3)}{\left(x^{2}+9\right)(x+3)}\)
= x = -3

ML Aggarwal Class 8 Solutions for ICSE Maths

ICSE Class 8 Maths Factorisation

Mental Maths
Question 1.
Fill in the blanks:
(i) When an algebraic expression can be written as the product of two or more expressions then each of these expressions is called ……….. of the given expression.
(ii) The process of finding two or more expressions whose product is the given expression is called ………..
(iii) HCF of two or more monomials = (HCF of their ……….. coefficients) × (HCF of their literal coefficients)
(iv) HCB of literal coefficients = product of each common literal raised to the ……….. power.
(v) To factorise the trinomial of the form x² + px + q, we need to find two integers a and b such that a + b = ……….. and ab = ………..
(vi) To factorise the trinomial of the form ax² + bx + c, where a, b and c are integers, we split b into two parts such that ……….. of these parts is b and their is ……….. ac.
Solution:
(i) When an algebraic expression can be written as the product of two or
more expressions then each of these expressions is called factor of the given expression.
(ii) The process of finding two or more expressions
whose product is the given expression is called factorization.
(iii) HCF of two or more monomials
= (HCF of their numerical coefficients) × (HCF of their literal coefficients)
(iv) HCF of literal coefficients
= product of each common literal raised to the lowest power.
(v) To factorise the trinomial of form x² + px + q,
we need to find two integers a and b such that a + b= p and ab = q.
(vi) To factorise the trinomial of the form ax² + bx + c,
where a, b and c are integers, we split b into two parts such that
algebraic sum of these parts is b and their product is ac.

Question 2.
State whether the following statements are true (T) or false (F):
(i) Factorisation is the reverse process of multiplication.
(ii) HCF of two or more polynomials (with integral coefficients) is the smallest common factor of the given polynomials.
(iii) HCF of 6x²y² and 8xy³ is 2xy².
(iv) Factorisation by grouping is possible only if the given polynomial contains an even number of terms.
(v) To factorise the trinomial of the form ax² + bx + c where, a, b, c are integers we want to find two integers A and B such that
A + B = ac and AB = b
(vi) Factors of 4x² – 12x + 9 are (2x – 3) (2x – 3).
Solution:
(i) Factorisation is the reverse process of multiplication. True
(ii) HCF of two or more polynomials (with integral coefficients) is the
smallest common factor of the given polynomials. False
(iii) HCF of 6x²y² and 8xy² is 2xy². True
(iv) Factorisation by grouping is possible only
if the given polynomial contains an even number of terms. True
(v) To factorise the trinomial of the form ax² + bx + c
where, a, b, c are integers we want to find two integers A and B such that
A + B = ac and AB = b False
Correct :
A + B should be equal to ft and AB = ac
(vi) Factors of
4x² – 12x + 9 are (2x – 3) (2x – 3). True

Multiple Choice Questions
Choose the correct answer from the given four options (3 to 14):
Question 3.
H.C.F. of 6abc, 24ab², 12a²b is
(a) 6ab
(b) 6ab²
(c) 6a²b
(d) 6abc
Solution:
H.C.F. of babe, 24ab², 12a²b
= H.C.F. of 6, 24, 12 × H.C.F. of abc, ab², a²b
= 6 × a × b = 6ab (a)

Question 4.
Factors of 12a²b + 15ab² are
(a) 3a(4ab + 5b2)
(b) 3ab(4a + 5b)
(c) 3b(4a² + 5ab)
(d) none of these
Solution:
12a²b + 15 ab² = 3ab(4a + 5b) (b)

Question 5.
Factors of 6xy – 4y + 6 – 9x are
(a) (3y – 2) (2x – 3)
(b) (3x – 2) (2y – 3)
(c) (2y – 3) (2 – 3x)
(d) none of these
Solution:
6xy – 4y + 6 – 9x
= 6xy – 9x – 4y + 6
= 3x(2y – 3) -2(2y – 3)
= (2y – 3) (3x – 2)

Question 6.
Factors of 49p³q – 36pq are
(a) p(7p + 6q) (7p – 6q)
(b) q(7p – 6) (7p + 6)
(c) pq(7p + 6) (7p – 6)
(d) none of these
Solution:
49p²q – 36pq
= pq(49p² – 36)
=pq[(7p)² – (6)²]
= pq(7p + 6) (7p – 6)

Question 7.
Factors of y(y – z) + 9(z – y) are
(a) (y – z) (y + 9)
(b) (z – y) (y + 9)
(c) (y – z) (y – 9)
(d) none of these
Solution:
y(y – z) + 9(z – y)
= y(y – z) – 9(y – z)
= (y – z) (y – 9) (c)

Question 8.
Factors of (lm + l) + m + 1 are
(a) (lm + l )(m + l)
(b) (lm + m)(l + 1)
(c) l(m + 1)
(d) (l + 1)(m + 1)
Solution:
Factors of lm + l + m + 1 are
l(m + 1) + l (m + 1) = (m + 1)(l + 1) (d)

Question 9.
Factors of z² – 4z – 12 are
(a) (z + 6)(z – 2)
(b) (z – 6)(z + 2)
(c) (z – 6)(z – 2)
(d) (z + 6)(z + 2)
Solution:
Factors of z² – 4z – 12
⇒ z² – 6z + 2z – 12
= z(z – 6) + 2(z – b)
= (z – 6)(z + 2) (b)

Question 10.
Factors of 63a² – 112b² are
(a) 63 (a – 2b)(a + 2b)
(b) 7(3a + 2b)(3a – 2b)
(c) 7(3a + 4b)(3a – 4b)
(d) none of these
Solution:
Factors of 63a² – 112b² are
= 7(9a² – 16b²)
= 7[(3a)² – (4b)²]
= 7(3a + 4b)(3a – 4b) (c)

Question 11.
Factors of p⁴ – 81 are
(a) (p² – 9)(p² + 9)
(b) (p + 3)² (p – 3)²
(c) (p + 3) (p – 3) (p² + 9)
(d) none of these
Solution:
p⁴ – 81 = (p²)² – (9)²
= (p² + 9)(p² – 9)
= (p² + 9){(p)² – (3)²}
= (p² + 9) (p + 3) (p – 3) (c)

Question 12.
Factors of 3x + 7x – 6 are
(a) (3x – 2)(x + 3)
(b) (3x + 2) (x – 3)
(c) (3x – 2)(x – 3)
(d) (3x + 2) (x + 3)
Solution:
3x² + 7x – 6
= 3x² + 9x – 2x – 6
= 3x(x + 3) -2(x + 3)
= (3x – 2) (x + 3) (a)

Question 13.
Factors of 16x² + 40x + 25 are
(a) (4x + 5)(4x + 5)
(b) (4x + 5)(4x – 5)
(c) (4x + 5)(4x + 8)
(d) none of these
Solution:
16x² + 40x + 25
= (4x)² + 2 × 4x × 5 + (5)²
= (4x + 5)²
= (4x + 5)(4x + 5) (a)

Question 14.
Factors of x² – 4xy + 4y² are
(a) (x – 2y)(x + 2y)
(b) (x-2y)(x-2y)
(c) (x + 2y)(x + 2y)
(d) none of these
Solution:
x² – 4xy + 4y²
= (x)² – 2 × x × 2y + (2y)² = (x – 2y)²
= (x- 2y)(x – 2y) (b)

Higher Order Thinking Skills (Hots)
Factorise the following
Question 1.
x² + \(\left(a+\frac{1}{a}\right)\)x + 1
Solution:
x² + \(\left(a+\frac{1}{a}\right)\)x + 1
= x² + ax + \(\frac{x}{a}\) + 1
= x(x + a) + \(\frac{1}{a}\)(x + a)
= (x + a)\(\left(x+\frac{1}{a}\right)\)

Question 2.
36a⁴ – 97a²b² + 36b⁴
Solution:
= 36a⁴ – 97a²b² + 36b⁴
= 36a⁴ – 72a²b² + 36b⁴ – 25a²b²
= (6a²)² – 2 × 6a² × 6b² + (6b²)² – (5ab)²
= (6a² – 6b²)² – (5ab)²
= (6a² – 6b² + 5ab)(6a² – 6b² – 5ab)
= (6a² + 5ab – 6b²)(6a² – 5ab – 6b²)
= [6a² + 9ab – 4ab – 6b²] [6a² – 9ab + 4ab – 6b²]
= [3a(2a + 3b) – 2b(2a + 3b)] [3a(2a – 3b) + 2b(2a – 3b)]
= (2a + 3b)(3a – 2b)(2a – 3b)(3a + 2b)

Question 3.
2x² – \(\sqrt{3}\)x – 3
Solution:
2x² – \(\sqrt{3}\)x – 3
= 2x² – \(2 \sqrt{3}\)x + \(\sqrt{3}\)x – 3
{∵ 2 × (-3) = -6
∴ -6 = \(-2 \sqrt{3} \times \sqrt{3}\)
–\(\sqrt{3}\) = -2\(\sqrt{3}\) + \(\sqrt{3}\)}
= 2x(x – \(\sqrt{3}\) ) + \(\sqrt{3}\) (x – \(\sqrt{3}\) )
= (x – \(\sqrt{3}\) )(2x + \(\sqrt{3}\))

Question 4.
y(y² – 2y) + 2(2y – y²) – 2 + y
Solution:
y(y² – 2y) + 2(2y – y²) – 2 + y
= y³ – 2y² + 4y – 2y² -2 + y
= y³ – 4y² + 5y – 2
= y³ – 2y² + y – 2y² + 4y – 2
= y(y² -2y + 1) – 2(y² -2y + 1)
= (y² – 2y + 1)(y – 2)
= [(y)² – 2 × y × 1 + (1)²] (y – 2)
= (y – 1)²(y – 2)

ML Aggarwal Class 8 Solutions for ICSE Maths

Ex 10.3 Class 8 ML Aggarwal

Question 1.
Multiply:
(i) (5x – 2) by (3x + 4)
(ii) (ax + b) by (cx + d)
(iii) (4p – 7) by (2 – 3p)
(iv) (2x² + 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)
15x² + 20x – 6x – 8
= 15x²+ 14x – 8

(ii) (ax + b) by (cx + d)
= ax (cx + d) + b (cx + d)
= acx² + adx + bcx + bd

(iii) (4p – 7) by (2 – 3p)
= (4p – 7) (2 -3p)
= 4p(2 – 3p) -7(2 – 3p)
= 8p – 12p² – 14 + 21p
= 29p – 12p² – 14

(iv) (2x² + 3) by (3x – 5)
= (2x² + 3) (3x – 5)
= 2x²(3x – 5) + 3(3x – 5)
= 6x³ – 10x² + 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)
= 2.25a² + 3.75ab – 3.75a6 – 6.25b²
= 2.25a² – 6.25b²

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)
= x² + 2xy – 2xy – 4y² + 3x + 6y
= x² – 4y² + 3x + 6y

(ii) (3 – 5x + 2x²) by (4x – 5)
= (4x – 5) (3 – 5x + 2x²)
= 4x(3 – 5x + 2x²) – 5(3 – 5x + 2x²)
= 12x – 20x² + 8x³ – 15 + 25x – 10x²
= 8x³ – 30x² + 37x – 15

Question 3.
Multiply:
(i) (3x² – 2x – 1) by (2x² + x – 5)
(ii) (2 – 3y – 5y²) by (2y – 1 + 3y²)
Solution:
(i) (3x² – 2x – 1) by (2x² + x – 5)
= (3x² – 2x – 1) (2x² + x – 5)
= 3x²(2x² + x – 5) – 2x(2x² + x – 5) -1(2x² + x – 5)
= 6x⁴ + 3x³ – 15x² – 4x³ – 2x² + 10x – 2x² – x + 5
= 6x⁴ – x³ – 19x² + 9x + 5

(ii) (2 – 3y – 5y²) by (2y- 1 + 3y²)
= 2(2y – 1 + 3y² )- 3y (2y – 1 + 3y²) -5y²(2y – 1 + 3y²)
= 4y – 2 + 6y² – 6y² + 3y – 9y³ – 10y³ + 5y² – 15y⁴
= -15y⁴ – 19y³ + 5y² + 7y – 2

Question 4.
Simplify:
(i) (x² + 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) (x² + 3) (x – 3) + 9
= x² (x – 3) + 3(x – 3) + 9
= x² – 3x² + 3x – 9 + 9
= x³ – 3x² + 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)}
= (x² – 3x + 3x – 9) {x² – 4x + 4x – 16}
= (x² – 9) (x² – 16)
= x² (x² – 16) – 9 (x² – 16)
= x⁴ – 16x² – 9x² + 144
= x⁴ – 25x² + 144

(iii) (x + 5) (x + 6) (x + 7)
= (x² + 6x + 5x + 30) (x + 7)
= (x² + 11x + 30) (x + 7)
= x(x²+ 11x + 30) + 7(x²+ 11x + 30)
= x³ + 11x² + 30x + 7x² + 77x + 210
= x³ + 18x² + 107x + 210

(iv) (p + q – 2r)(2p – q + r) – 4qr
= p(2p – q + r) + q(2p – q + r) – 2r(2p – q + r) – 4qr
= 2p² – pq + pr + 2pq – q² + qr – 4pr + 2qr – 2r² – 4qr
= 2p² – q² – 2r² + 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
= x² – xy + xz + xy – y² + yz + xz – yz + z² – x² + xy + xz
– xy + x² + yx + xz – yz – z² – 4zx = 0

Question 5.
If two adjacent sides of a rectangle are 5x² + 25xy + 4y² and 2x² – 2xy + 3y², find its area.
Solution:
Adjacents sides of a rectangle are
5x² + 25xy + 4y² and
2x² – 2xy + 3y²
∴ Area of rectangle = Product of two sides
= (5x² + 25xy + 4y²) (2x² – 2xy + 3y²)
= 10x⁴– 10x³y+ 15x²y² + 50x³y – 50x²y² + 75xy³ + 8x²y² – 8xy³ + 12y⁴
= 10x⁴ + 40x³y – 27x²y² + 67xy³ + 12y⁴

ML Aggarwal Class 8 Solutions for ICSE Maths

Ex 10.4 Class 8 ML Aggarwal

Question 1.
Divide:
(i) – 39pq²r⁵ by – 24p³q³r
(ii) –\(\frac{3}{4}\)a²b³ by \(\frac{6}{7}\)a³b²
Solution:

Question 2.
Divide:
(i) 9x⁴ – 8x³ – 12x + 3 by 3x
(ii) 14p²q³ – 32p³q² + 15pq² – 22p + 18q by – 2p²q.
Solution:

Question 3.
Divide:
(i) 6x² + 13x + 5 by 2x + 1
(ii) 1 + y³ by 1 + y
(iii) 5 + x – 2x² by x + 1
(iv) x³ – 6x² + 12x – 8 by x – 2
Solution:

∴ Quotient = 3x + 5
and remainder = 0

∴ Quotient = y² – y + 1
and remainder = 0

(iii) Arranging the terms of dividend in descending order
of powers of x and then dividing, we get

∴ Quotient = – 2x + 3 and remainder = 2

(iv) x³ – 6x² + 12x – 8 by x – 2

Quotient = x² – 4x + 4 and remainder = 0

Question 4.
Divide:
(i) 6x³ + x² – 26x – 25 by 3x – 7
(ii) m³ – 6m² + 7 by m – 1
Solution:
(i)

∴ Quotient = 2x² + 5x + 3 and remainder = – 4

∴ Quotient = m² – 5m – 5 and remainder = 2.

Question 5.
Divide:
(i) a³ + 2a² + 2a + 1 by + a² + a + 1
(ii) 12x³ – 17x² + 26x – 18 by 3x² – 2x + 5
Solution:

∴ Quotient = a + 1 and remainder = 0.

(ii) 12x³ – 17x² + 26x – 18 by 3x² – 2x + 5

∴ Quotient = 4x – 3 and remainder = -3

Question 6.
If the area of a rectangle is 8x² – 45y² + 18xy and one of its sides is 4x + 15y, find the length of adjacent side.
Solution:
Area of rectangle = 8x² – 45y² + 18xy
One side = 4x + 15y
∴ Second (adjacent) side

ML Aggarwal Class 8 Solutions for ICSE Maths

ICSE Class 8 Maths Algebraic Expressions and Identities

Mental Maths
Question 1.
Fill in the blanks:
(i) A symbol which has a fixed value is called a ……….
(ii) A symbol which can be given various numerical values is called ……….. or ………
(iii) The various parts of an algebraic expression separated by + or – sign are called ………..
(iv) An algebraic expression having ………… terms is called a binomial.
(v) Each of the quantity (constant or literal) multiplied together to form a product is called a ………… of the product.
(vi) The terms having same literal coefficients are called a ………… otherwise they are called …………
(vii) Degree of the polynomial is the greatest sum of the powers of ………… in each term.
(viii)An identity is an equality which is true for ………… of variables in it.
(ix) (a + b)² = …………
(x) (x + a) (x + b) = x² + ………… + ab.
(xi) Dividend = ………… + remainder.
Solution:
(i) A symbol which has a fixed value is called a constant.
(ii) A symbol which can be given various numerical values
is called variable or literal.
(iii) The various parts of an algebraic expression
separated by + or – sign are called terms.
(iv) An algebraic expression having two terms is called a binomial.
(v) Each of the quantity (constant or literal) multiplied together
to form a product is called a factor of the product.
(vi) The terms having same literal coefficients are called
alike terms otherwise they are called unlike terms.
(vii) Degree of the polynomial is the greatest sum of the
powers of variables in each term.
(viii)An identity is an equality which is true
for all values of variables in it.
(ix) (a + b)² = a² + 2ab + b².
(x) (x + a) (x + b) = x² + (a + b)x + ab.
(xi) Dividend = divisor × quotient + remainder.

Question 2.
State whether the following statements are true (T) or false (F):
(i) An algebraic expression having only one term is called a monomial.
(ii) A symbol which has fixed value is called a literal.
(iii) The term of an algebraic expression having no literal factor is called its constant term,
(iv) In any term of an algebraic expression the constant part is called literal coefficient of the term.
(v) An algebraic expression is called polynomial if the powers of the variables involved in it in each term are non-negative integers.
(vi) 5x²y²z, 3x²zy², – \(\frac{4}{5}\) zx²y² are unlike terms.
(vii) 5x + \(\frac{2}{x}\) + 7 is a polynomial of degree 1.
(viii) 3x²y + 7x + 8y + 9 is a polynomial of degree 3.
(ix) Numerical coefficient of -7x³y is 7.
(x) An equation is true for all values of variables in it.
(xi) (a – b)² + 2ab = a² + b².
Solution:
(i) An algebraic expression having only one term is called a monomial. True
(ii) A symbol which has fixed value is called a literal. False
Correct:
It is called constant.
(iii) The term of an algebraic expression having
no literal factor is called its constant term.
True
(iv) In any term of an algebraic expression, the constant part
is called literal coefficient of the term. False
Correct:
It is called the constant coefficient.
(v) An algebraic expression is called polynomial if the powers
of the variables involved in it in each term are non-negative integers. True
(vi) 5x²y²z, 3x²zy², – \(\frac{4}{5}\) zx²y² are unlike terms.
False
Correct:
They are like terms.
(vii) 5x + \(\frac{2}{x}\) + 7 is a polynomial of degree 1. False
Correct:
∵ \(\frac{2}{x}\) has negative power of y i.e. x^-1.
(viii)3x²y + 7x + 8y + 9 is a polynomial of degree 3. True
(ix) Numerical coefficient of -7x³y is 7. False Correct:
It is -7.
(x) An equation is true for all values of variables in it. False
Correct:
It is true for some specific values of variables.
(xi) (a – b)² + 2ab = a² + b². True

Multiple Choice Questions
Choose the correct answer from the given four options (3 to 18):
Question 3.
The literal coefficient of -9xyz² is
(a) -9
(b) xy
(c) xyz²
(d) -9xy
Solution:
Literal co-efficient of -9xyz² is -9 (a)

Question 4.
Which of the following algebraic expression is
(a) 3x² – 2x + 7
(b) \(\frac{5 x^{3}}{2 x}\) + 3x² + 8
(c) 3x + \(\frac{2}{x}\) + 7
(d) \(\sqrt{2}\)x² + \(\sqrt{3}\)x + \(\sqrt{6}\)
Solution:
3x² – 2x + 7 is not an algebraic expression. (c)

Question 5.
Which of the following algebraic expressions is not a monomial?
(a) 3x × y × z
(b) -5pq
(c) 8m² × n ÷ 31
(d) 7x ÷ y – z
Solution:
7x ÷ y – z = \(\frac{7 x}{y}\) – z
It is not monomial.
(As it has two terms) (d)

Question 6.
Degree of the polynomial 7x²yz² + 6x³y²z² – 5x + 8y is
(a) 4
(b) 5
(c) 6
(d) 7
Solution:
Degree of polynomial
7x²yz² + 6x³y²z² -5x + 8y is 3 + 2 + 2 = 7 (d)

Question 7.
a(b -c) + b(c – a) + c(a – b) is equal to
(a) ab + bc + ca
(b) 0
(c) 2(ab + bc + ca)
(d) none of these
Solution:
a(b -c) + b(c – a) + c(a – b)
= ab – ac + bc – ab + ac – bc = 0 (b)

Question 8.
\(\frac{7}{5}\)xy – \(\frac{2}{3}\)xy + \(\frac{8}{9}\) xy is equal to

Solution:
\(\frac{7}{5}\)xy – \(\frac{2}{3}\)xy + \(\frac{8}{9}\) xy
= \(\frac{63 x y-30 x y+40 x y}{45}=\frac{73}{45} x y\) (a)

Question 9.
(3p²qr³) × (-4p³q²r²) x (7pq³r) is equal to
(a) 84p⁶q⁶r⁶
(b) -84p⁶q⁶r⁶
(c) 84p⁶q⁵r⁶
(d) -84p⁶q⁵r⁶
Solution:
(3p²qr³) × (-4p³q²r²) × (7pq³r)
= 3(-4) × 7p²⁺³⁺¹ × q¹⁺²⁺³ × r³⁺²⁺¹
= -84p⁶q⁶r⁶ (b)

Question 10.
3m × (2m² – 5mn + 4n²) is equal to
(a) 6m³ + 15m²n – 12mn²
(b) 6m³ – 15m²n + 12mn²
(c) 6m³ – 15m²n – 12mn²
(d) 6m³ + 15m²n + 12mn²
Solution:
3m × (2m² – 5mn + 4n²)
= 6m³ – 15m²n + 12mn² (b)

Question 11.
Volume of a rectangular box whose adjacent edges are 3x²y, 4y²z and 5z²x respectively is
(a) 60xyz
(b) 60x²y²z²
(c) 60x³y³z³
(d) none of these
Solution:
Volume = 3x²y × 4y²z × 5z²x = 60x³y³²z³ (c)

Question 12.
(x – 1) (x + 2) is equal to
(a) 2x + 3
(b) x² + 2x + 2
(c) x² + 3x + 2
(d) x² + 2x + 3
Solution:
(x + 1) (x + 2) = x² + (1 + 2)x + 1 × 2
= x² + 3x + 2 (c)

Question 13.
If x + \(\frac{1}{x}\) = 2, then x² + \(\frac{1}{x^{2}}\) is equal to
(a) 4
(b) 2
(c) 0
(d) none of these
Solution:
x + \(\frac{1}{x}\) = 2
Squaring both sides,

Question 14.
If x² + y² = 9 and xy = 8, then x + y is equal to
(a) 25
(b) 5
(c) -5
(d) ±5
Solution:
x² + y² = 9, xy = 8
(x + y)² = x² + y² + 2xy
= (9)² + 2 × 8
= 9 + 16 = 25 = ±(5) (d)

Question 15.
(102)² – (98)² is equal to
(a) 200
(b) 400
(c) 600
(d) 800
Solution:
(102)² – (98)²
= (102 + 98) × (102 – 98)
= 200 × 4 = 800 (d)

Question 16.
-50x³y²z² divided by -5xyz is equal to
(a) 10xyz
(b) 10x²yz
(c) -10xyz
(d) -10x²yz
Solution:
\(\frac{-50 x^{3} y^{2} z^{2}}{-5 x y z}\) = 10x²yz (b)

Question 17.
96 × 104 is equal to
(a) 9984
(b) 9974
(c) 9964
(d) none of these
Solution:
96 × 104 = (100 – 4) (100 + 4)
= (100)² – (4)²
= 10000 – 16 = 9984 (a)

Question 18.
If the area of a rectangle is 24(x²yz + xy²z + xyz²) and its length is 8xyz, then its breadth is
(a) 3(x + y + z)
(b) 3xyz
(c) 3(x + y – z)
(d) none of these
Solution:
In a rectangle
Area = 24(x²yz + xy²z + xyz²)
Length = 8xyz
Breadth = \(\frac{24 x y z(x+y+z)}{8 x y z}\) = 3(x + y + z) (a)

Higher Order Thinking Skills (Hots)
Question 1.
Using the identity (a + b) = a² + 2ab + b², derive the formula for (a + b + c)². Hence, find the value of (2x – 3y + 4z)².
Solution:
(a + b)² = a² + 2ab + b²
(a + b + c)² = (a + b)² + c² + 2 (a + b)c
= a² + b² + 2ab + c² + 2ca + 2bc
= a² + b² + c² + 2ab + 2bc + 2ca
(2x – 3y + 4z)² = (2x)² + (3y)² + (4z)² + 2 × 2x × (-3y)
+ 2(-3y) (4z) + 2(4z) (2x)
= 4x² + 9y² + 16z² – 12xy – 24yz + 16zx

Question 2.
Using the product of algebraic expressions, find the formulas for (a + b)² and (a – b).
Solution:
(a + b)² = (a + b)² (a + b)
= (a² + 2ab + b²) (a + b)
= a(a² + 2ab + b²) + b(a² + 2ab + b²)
= a³ + 2a²b + ab² + a²b + 2ab² + b³
= a³² + b³ + 3a²b + 3ab²
= a² + b2² + 3ab(a + b)

(a – b)² = (a – b) (a – b)²
= (a – b) (a² – 2ab + b²)
= a(a² – 2ab + b²) – b(a² – 2ab + b²)
= a³ – 2a²b + ab² – a²b + 2ab² – b³
= a³ – b³ – 3a²b + 3ab²
= a³ – b³ – 3ab(a – b)

ML Aggarwal Class 8 Solutions for ICSE Maths