Q1. Find whether the trinomial is a perfect square or not:
(i) $x^2 + 14x + 49$
(ii) $a^2 - 10a + 25$
(iii) $4x^2 + 4x + 1$
(iv) $9b^2 + 12b + 16$
(v) $16x^2 - 16xy + y^2$
(vi) $x^2 - 4x + 16$
Show Answer
(i) $x^2 + 14x + 49$
$\Rightarrow x^2 + 14x + (7)^2$
$\Rightarrow x^2 + (2 \times x \times 7) + (7)^2$
By taking $x = a \text{ and } 7 = b$ we have:
$\Rightarrow a^2 + 2ab + b^2 = (a + b)^2$
$\Rightarrow a^2 + 14x + 49$ Hence Perfect Square
(ii) $a^2 - 10a + 25$
$\Rightarrow a^2 - 10a + (5)^2$
$\Rightarrow a^2 - (2 \times a \times 5) + (5)^2$
By taking $a^2 = a \text{ and } 5 = b$ we have:
$\Rightarrow a^2 - 2ab + b^2 = (a - b)^2$
$\Rightarrow a^2 - 10a + 25$ Hence Perfect Square
(iii) $4x^2 + 4x + 1$
$\Rightarrow (2x)^2 + 4x + (1)^2$
$\Rightarrow (2x)^2 + (2 \times 2x \times 1) + (1)^2$
By taking $2x = a \text{ and } 1 = b$ we have:
$\Rightarrow a^2 + 2ab + b^2 = (a + b)^2$
$\Rightarrow 4x^2 + 4x + 1$ Hence Perfect Square
(iv) $9b^2 + 12b + 16$
$\Rightarrow (3b)^2 + 12b + (4)^2$
$\Rightarrow (3b)^2 + (3b \times 4) + (4)^2$
By taking $3b = a \text{ and } 4 = b$ we have:
$\Rightarrow a^2 + ab + b^2$
$\Rightarrow 4x^2 + 4x + 1$ Hence not a Perfect Square
(v) $16x^2 - 16xy + y^2$
$\Rightarrow (4x)^2 - 16xy + (y)^2$
$\Rightarrow (4x)^2 - (2 \times 4x \times y) + (4)^2$
By taking $4x = a \text{ and } y = b$ we have:
$\Rightarrow a^2 - 8ab + b^2$
$\Rightarrow 16x^2 - 16xy + y^2$ Hence not a Perfect Square
(vi) $x^2 - 4x + 16$
$\Rightarrow (x)^2 - 4x + (4)^2$
$\Rightarrow (4x)^2 - (2 \times x \times 4) + (4)^2$
By taking $x = a \text{ and } 4 = b$ we have:
$\Rightarrow a^2 - 8ab + b^2$
$\Rightarrow x^2 - 4x + 16$ Hence not a Perfect Square
$\Rightarrow x^2 + 14x + (7)^2$
$\Rightarrow x^2 + (2 \times x \times 7) + (7)^2$
By taking $x = a \text{ and } 7 = b$ we have:
$\Rightarrow a^2 + 2ab + b^2 = (a + b)^2$
$\Rightarrow a^2 + 14x + 49$ Hence Perfect Square
(ii) $a^2 - 10a + 25$
$\Rightarrow a^2 - 10a + (5)^2$
$\Rightarrow a^2 - (2 \times a \times 5) + (5)^2$
By taking $a^2 = a \text{ and } 5 = b$ we have:
$\Rightarrow a^2 - 2ab + b^2 = (a - b)^2$
$\Rightarrow a^2 - 10a + 25$ Hence Perfect Square
(iii) $4x^2 + 4x + 1$
$\Rightarrow (2x)^2 + 4x + (1)^2$
$\Rightarrow (2x)^2 + (2 \times 2x \times 1) + (1)^2$
By taking $2x = a \text{ and } 1 = b$ we have:
$\Rightarrow a^2 + 2ab + b^2 = (a + b)^2$
$\Rightarrow 4x^2 + 4x + 1$ Hence Perfect Square
(iv) $9b^2 + 12b + 16$
$\Rightarrow (3b)^2 + 12b + (4)^2$
$\Rightarrow (3b)^2 + (3b \times 4) + (4)^2$
By taking $3b = a \text{ and } 4 = b$ we have:
$\Rightarrow a^2 + ab + b^2$
$\Rightarrow 4x^2 + 4x + 1$ Hence not a Perfect Square
(v) $16x^2 - 16xy + y^2$
$\Rightarrow (4x)^2 - 16xy + (y)^2$
$\Rightarrow (4x)^2 - (2 \times 4x \times y) + (4)^2$
By taking $4x = a \text{ and } y = b$ we have:
$\Rightarrow a^2 - 8ab + b^2$
$\Rightarrow 16x^2 - 16xy + y^2$ Hence not a Perfect Square
(vi) $x^2 - 4x + 16$
$\Rightarrow (x)^2 - 4x + (4)^2$
$\Rightarrow (4x)^2 - (2 \times x \times 4) + (4)^2$
By taking $x = a \text{ and } 4 = b$ we have:
$\Rightarrow a^2 - 8ab + b^2$
$\Rightarrow x^2 - 4x + 16$ Hence not a Perfect Square
Q2. Factorise completely:
(i) $2 - 8x^2$
(ii) $8x^2y - 18y^3$
(iii) $ax^2 - ay^2$
(iv) $25x^3 - x$
(v) $a^4 - b^4$
Show Answer
(i) $2 - 8x^2$
$\Rightarrow 2(1 - 4x^2)$
$\Rightarrow 2((1)^2 - (2x)^2)$
$\Rightarrow 2(1 - 2x)^2$
$\Rightarrow 2(1 + 2x)(1 - 2x)$ Ans
(ii) $8x^2y - 18y^3$
$\Rightarrow 2y(4x^2 - 9y^2)$
$\Rightarrow 2((2x)^2 - (3y)^2)$
$\Rightarrow 2(2x - 3y)^2$
$\Rightarrow 2(2x + 3y)(2x - 3y)$ Ans
(iii) $ax^2 - ay^2$
$\Rightarrow a(x^2 - y^2)$
$\Rightarrow a(x - y)^2$
$\Rightarrow a(x + y)(x - y)$ Ans
(iv) $25x^3 - x$
$\Rightarrow x(25x^2 - 1)$
$\Rightarrow x((5x)^2 - (1)^2)$
$\Rightarrow x(5x - 1)^2$
$\Rightarrow x(5x + 1)(5x - 1)$ Ans
(v) $a^4 - b^4$
$\Rightarrow ((a^2)^2 - (b^2)^2)$
$\Rightarrow (a^2 - b^2)^2$
$\Rightarrow (a^2 + b^2)(a^2 - b^2)$
$\Rightarrow (a^2 + b^2)(a - b)^2$
$\Rightarrow (a^2 + b^2)(a + b)(a - b)$ Ans
$\Rightarrow 2(1 - 4x^2)$
$\Rightarrow 2((1)^2 - (2x)^2)$
$\Rightarrow 2(1 - 2x)^2$
$\Rightarrow 2(1 + 2x)(1 - 2x)$ Ans
(ii) $8x^2y - 18y^3$
$\Rightarrow 2y(4x^2 - 9y^2)$
$\Rightarrow 2((2x)^2 - (3y)^2)$
$\Rightarrow 2(2x - 3y)^2$
$\Rightarrow 2(2x + 3y)(2x - 3y)$ Ans
(iii) $ax^2 - ay^2$
$\Rightarrow a(x^2 - y^2)$
$\Rightarrow a(x - y)^2$
$\Rightarrow a(x + y)(x - y)$ Ans
(iv) $25x^3 - x$
$\Rightarrow x(25x^2 - 1)$
$\Rightarrow x((5x)^2 - (1)^2)$
$\Rightarrow x(5x - 1)^2$
$\Rightarrow x(5x + 1)(5x - 1)$ Ans
(v) $a^4 - b^4$
$\Rightarrow ((a^2)^2 - (b^2)^2)$
$\Rightarrow (a^2 - b^2)^2$
$\Rightarrow (a^2 + b^2)(a^2 - b^2)$
$\Rightarrow (a^2 + b^2)(a - b)^2$
$\Rightarrow (a^2 + b^2)(a + b)(a - b)$ Ans
Q3. Factorise completely:
(i) $16x^4 - 81y^4$
(ii) $625 - x^4$
(iii) $x^2 - y^2 - 3x - 3y$
(iv) $x^2 - y^2 - 2x + 2y$
(v) $3x^2 + 15x - 72$
Show Answer
(i) $16x^4 - 81y^4$
$\Rightarrow [(4x^2)^2 - (9y^2)^2]$
$\Rightarrow (4x^2 - 9y^2)^2$
$\Rightarrow (4x^2 + 9y^2)(4x^2 - 9y^2)$
$\Rightarrow (4x^2 + 9y^2)((2x)^2 - (3y)^2)^2$
$\Rightarrow (4x^2 + 9y^2)(2x - 3y)^2$
$\Rightarrow (4x^2 + 9y^2)(2x + 3y)(2x - 3y)$ Ans
(ii) $625 - x^4$
$\Rightarrow [(25^2)^2 - (x^2)^2]$
$\Rightarrow (25^2 - x^2)^2$
$\Rightarrow (25^2 + x^2)(25^2 - x^2)$
$\Rightarrow (25^2 + x^2)((5)^2 - (x)^2)$
$\Rightarrow (25x^2 + x^2)(5 - x)^2$
$\Rightarrow (25x^2 + x^2)(5 + x)(5 - x)$ Ans
(iii) $x^2 - y^2 - 3x - 3y$
$\Rightarrow (x^2 - y^2) - (3x - 3y)$
$\Rightarrow (x - y)^2 - (3x - 3y)$
$\Rightarrow (x + y)(x - y) - 3(x - y)$
$\Rightarrow (x + y)(x - y - 3)$ Ans
(iv) $x^2 - y^2 - 2x + 2y$
$\Rightarrow (x - y)^2 - (2x + 2y)$
$\Rightarrow (x + y)(x - y) - 2(x + y)$
$\Rightarrow (x - y)(x + y - 2)$ Ans
(v) $3x^2 + 15x - 72$
$\Rightarrow 3[x^2 + 5x - 24]$
$\Rightarrow 3[x^2 + 8x - 3x - 24]$
$\Rightarrow 3[x(x + 8) - 3(x + 8)]$
$\Rightarrow 3(x + 8)(x - 3)$ Ans
$\Rightarrow [(4x^2)^2 - (9y^2)^2]$
$\Rightarrow (4x^2 - 9y^2)^2$
$\Rightarrow (4x^2 + 9y^2)(4x^2 - 9y^2)$
$\Rightarrow (4x^2 + 9y^2)((2x)^2 - (3y)^2)^2$
$\Rightarrow (4x^2 + 9y^2)(2x - 3y)^2$
$\Rightarrow (4x^2 + 9y^2)(2x + 3y)(2x - 3y)$ Ans
(ii) $625 - x^4$
$\Rightarrow [(25^2)^2 - (x^2)^2]$
$\Rightarrow (25^2 - x^2)^2$
$\Rightarrow (25^2 + x^2)(25^2 - x^2)$
$\Rightarrow (25^2 + x^2)((5)^2 - (x)^2)$
$\Rightarrow (25x^2 + x^2)(5 - x)^2$
$\Rightarrow (25x^2 + x^2)(5 + x)(5 - x)$ Ans
(iii) $x^2 - y^2 - 3x - 3y$
$\Rightarrow (x^2 - y^2) - (3x - 3y)$
$\Rightarrow (x - y)^2 - (3x - 3y)$
$\Rightarrow (x + y)(x - y) - 3(x - y)$
$\Rightarrow (x + y)(x - y - 3)$ Ans
(iv) $x^2 - y^2 - 2x + 2y$
$\Rightarrow (x - y)^2 - (2x + 2y)$
$\Rightarrow (x + y)(x - y) - 2(x + y)$
$\Rightarrow (x - y)(x + y - 2)$ Ans
(v) $3x^2 + 15x - 72$
$\Rightarrow 3[x^2 + 5x - 24]$
$\Rightarrow 3[x^2 + 8x - 3x - 24]$
$\Rightarrow 3[x(x + 8) - 3(x + 8)]$
$\Rightarrow 3(x + 8)(x - 3)$ Ans
Q4. Factorise completely:
(i) $2a^2 - 8a - 64$
(ii) $5b^2 + 45b + 90$
(iii) $3x^2y - 11xy + 6y$
(iv) $5ap^2 + 11ap + 2a$
(v) $a^2 + 2ab + b^2 - c^2$
Show Answer
(i) $2a^2 - 8a - 64$
$\Rightarrow 2[a^2 - 4a - 32]$
$\Rightarrow 2[a^2 - 8a + 4a - 32]$
$\Rightarrow 2[(a^2 - 8a) + (4a - 32)]$
$\Rightarrow 2[a(a - 8) + 4(a - 8)]$
$\Rightarrow 2[(a - 8)(a + 4)]$
$\therefore 2(a - 8)(a + 4)$ Ans
(ii) $5b^2 + 45b + 90$
$\Rightarrow 5[b^2 + 9b + 18]$
$\Rightarrow 5[b^2 + 6b + 3b + 18]$
$\Rightarrow 5[(b^2 + 6b) + (3b + 18)]$
$\Rightarrow 5[b(b + 6) + 3(b + 6)]$
$\Rightarrow 5[(b + 6)(b + 3)]$
$\therefore 5(b + 6)(b + 3)$ Ans
(iii) $3x^2y + 11xy + 6y$
$\Rightarrow 3x^2y + 9xy + 2xy + 6y$
$\Rightarrow (3x^2y + 9xy) + (2xy + 6y)$
$\Rightarrow 3xy(x + 3) + 2y(x + 3)$
$\Rightarrow (x + 3)(3xy + 2y)$
$\Rightarrow (x + 3)y(3x + 2)$
$\therefore y(x + 3)(3x + 2)$ Ans
(iv) $5ap^2 + 11ap + 2a$
$\Rightarrow 5ap^2 + 10ap + ap + 2a$
$\Rightarrow (5ap^2 + 10ap) + (ap + 2a)$
$\Rightarrow 5ap(p + 2) + a(p + 2)$
$\Rightarrow (p + 2)(5ap + a)$
$\Rightarrow (p + 2)a(5p + 1)$
$\therefore a(p + 2)(5p + 1)$ Ans
(v) $a^2 + 2ab + b^2 - c^2$
In the given expression, $a^2 + 2ab + b^2$ is a perfect square:
$\because a^2 + 2ab + b^2 = (a + b)^2$
$\therefore (a + b)^2 - c^2$
$\Rightarrow (a + b + c)(a + b - c)$ Ans
$\Rightarrow 2[a^2 - 4a - 32]$
$\Rightarrow 2[a^2 - 8a + 4a - 32]$
$\Rightarrow 2[(a^2 - 8a) + (4a - 32)]$
$\Rightarrow 2[a(a - 8) + 4(a - 8)]$
$\Rightarrow 2[(a - 8)(a + 4)]$
$\therefore 2(a - 8)(a + 4)$ Ans
(ii) $5b^2 + 45b + 90$
$\Rightarrow 5[b^2 + 9b + 18]$
$\Rightarrow 5[b^2 + 6b + 3b + 18]$
$\Rightarrow 5[(b^2 + 6b) + (3b + 18)]$
$\Rightarrow 5[b(b + 6) + 3(b + 6)]$
$\Rightarrow 5[(b + 6)(b + 3)]$
$\therefore 5(b + 6)(b + 3)$ Ans
(iii) $3x^2y + 11xy + 6y$
$\Rightarrow 3x^2y + 9xy + 2xy + 6y$
$\Rightarrow (3x^2y + 9xy) + (2xy + 6y)$
$\Rightarrow 3xy(x + 3) + 2y(x + 3)$
$\Rightarrow (x + 3)(3xy + 2y)$
$\Rightarrow (x + 3)y(3x + 2)$
$\therefore y(x + 3)(3x + 2)$ Ans
(iv) $5ap^2 + 11ap + 2a$
$\Rightarrow 5ap^2 + 10ap + ap + 2a$
$\Rightarrow (5ap^2 + 10ap) + (ap + 2a)$
$\Rightarrow 5ap(p + 2) + a(p + 2)$
$\Rightarrow (p + 2)(5ap + a)$
$\Rightarrow (p + 2)a(5p + 1)$
$\therefore a(p + 2)(5p + 1)$ Ans
(v) $a^2 + 2ab + b^2 - c^2$
In the given expression, $a^2 + 2ab + b^2$ is a perfect square:
$\because a^2 + 2ab + b^2 = (a + b)^2$
$\therefore (a + b)^2 - c^2$
$\Rightarrow (a + b + c)(a + b - c)$ Ans
Q5. Factorise completely:
(i) $x^2 + 6xy + 9y^2 + x + 3y$
(ii) $4a^2 - 12ab + 9b^2 + 4a - 6b$
(iii) $2a^2b^2 - 98b^4$
(iv) $a^2 - 16b^2 - 2a - 8b$
Show Answer
(i) $x^2 + 6xy + 9y^2 + x + 3y$
In the given expression, $x^2 + 6xy + 9y^2$ is a perfect square:
$\Rightarrow x^2 + 6xy + (3y)^2$
$\Rightarrow x^2 + 2 \times x \times 3y + (3y)^2$
By taking, $x = a \text{ and } 3y = b$, we have:
$\because a^2 + 2ab + b^2 = (a + b)^2 = (x + 3y)^2$
$\therefore (x + 3y)^2 + (x + 3y)$
$\Rightarrow (x + 3y)(x + 3y) + 1(x + 3y)$
$\Rightarrow (x + 3y)(x + 3y + 1)$ Ans
(ii) $4a^2 - 12ab + 9b^2 + 4a - 6b$
First factorising $4a^2 - 12ab + 9b^2:$
$\Rightarrow 4a^2 - 6ab - 6ab + 9b^2$
$\Rightarrow (4a^2 - 6ab) - (6ab + 9b^2)$
$\Rightarrow 2a(2a - 3b) - 3b(2a - 3b)$
$\Rightarrow (2a - 3b)(2a - 3b) = (2a - 3b)^2$
Now, $(2a - 3b)^2 + (4a - 6b)$
$\Rightarrow (2a + 3b)(2a - 3b) + 2(2a - 3b)$
$\Rightarrow (2a - 3b)(2a + 3b + 2)$ Ans
(iii) $2a^2b^2 - 98b^4$
$\Rightarrow 2(a^2b^2 - 49b^4)$
$\Rightarrow 2((ab)^2 - (7b^2)^2$
$\Rightarrow 2(ab - 7b^2)^2$
$\Rightarrow 2(ab + 7b^2)(ab - 7b^2)$
$\Rightarrow 2[b(a + 7b)b(a - 7b)]$
$\therefore 2b^2(a + 7b)(a - 7b)$ Ans
(iv) $a^2 - 16b^2 - 2a - 8b$
$\Rightarrow (a^2 - (4b)^2) - (2a - 8b)$
$\Rightarrow (a - 4b)^2 - (2a - 8b)$
$\Rightarrow (a + 4b)(a - 4b) - 2(a + 4b)$
$\Rightarrow (a + 4b)(a - 4b - 2)$ Ans
In the given expression, $x^2 + 6xy + 9y^2$ is a perfect square:
$\Rightarrow x^2 + 6xy + (3y)^2$
$\Rightarrow x^2 + 2 \times x \times 3y + (3y)^2$
By taking, $x = a \text{ and } 3y = b$, we have:
$\because a^2 + 2ab + b^2 = (a + b)^2 = (x + 3y)^2$
$\therefore (x + 3y)^2 + (x + 3y)$
$\Rightarrow (x + 3y)(x + 3y) + 1(x + 3y)$
$\Rightarrow (x + 3y)(x + 3y + 1)$ Ans
(ii) $4a^2 - 12ab + 9b^2 + 4a - 6b$
First factorising $4a^2 - 12ab + 9b^2:$
$\Rightarrow 4a^2 - 6ab - 6ab + 9b^2$
$\Rightarrow (4a^2 - 6ab) - (6ab + 9b^2)$
$\Rightarrow 2a(2a - 3b) - 3b(2a - 3b)$
$\Rightarrow (2a - 3b)(2a - 3b) = (2a - 3b)^2$
Now, $(2a - 3b)^2 + (4a - 6b)$
$\Rightarrow (2a + 3b)(2a - 3b) + 2(2a - 3b)$
$\Rightarrow (2a - 3b)(2a + 3b + 2)$ Ans
(iii) $2a^2b^2 - 98b^4$
$\Rightarrow 2(a^2b^2 - 49b^4)$
$\Rightarrow 2((ab)^2 - (7b^2)^2$
$\Rightarrow 2(ab - 7b^2)^2$
$\Rightarrow 2(ab + 7b^2)(ab - 7b^2)$
$\Rightarrow 2[b(a + 7b)b(a - 7b)]$
$\therefore 2b^2(a + 7b)(a - 7b)$ Ans
(iv) $a^2 - 16b^2 - 2a - 8b$
$\Rightarrow (a^2 - (4b)^2) - (2a - 8b)$
$\Rightarrow (a - 4b)^2 - (2a - 8b)$
$\Rightarrow (a + 4b)(a - 4b) - 2(a + 4b)$
$\Rightarrow (a + 4b)(a - 4b - 2)$ Ans