Q1. Factorise:
(i) $6x^3 - 8x^2$
(ii) $35a^3b^2c + 42ab^2c^2$
(iii) $36x^2y^2 - 30x^3y^3 + 48x^3y^2$
(iv) $8(2a + b)^3 - 12(2a + b)^2$
(v) $9a(x - 2y)^4 - 12a(x - 2y)^3$
Show Answer
(i) $6x^3 - 8x^2$
$\Rightarrow 2x^2(3x - 4)$ Ans
(ii) $35a^3b^2c + 42ab^2c^2$
$\Rightarrow 7ab^2c(5a^2 + 6c)$ Ans
(iii) $36x^2y^2 - 30x^3y^3 + 48x^3y^2$
$\Rightarrow 6x^2y^2(6 - 5xy + 8x)$ Ans
(iv) $8(2a + 3b)^3 - 12(2a + 3b)^2$
Taking $2a + 3b = x$, we have:
$\Rightarrow 8x^3 - 12x^2$
$\Rightarrow 4x^2(2x - 3)$
$\Rightarrow 4(2a + 3b)^2(2(2a + 3b) - 3)$
$\Rightarrow 4(2a + 3b)^2(4a + 6b - 3)$ Ans
(v) $9a(x - 2y)^4 - 12a(x - 2y)^3$
Taking $x - 2y = x$, we have:
$\Rightarrow 9ax^4 - 12ax^3$
$\Rightarrow 3ax^3(3x - 4)$
$\Rightarrow 3a(x - 2y)^3(3(x - 2y) - 4)$
$\Rightarrow 3a(x - 2y)^3(3x - 6y - 4)$ Ans
$\Rightarrow 2x^2(3x - 4)$ Ans
(ii) $35a^3b^2c + 42ab^2c^2$
$\Rightarrow 7ab^2c(5a^2 + 6c)$ Ans
(iii) $36x^2y^2 - 30x^3y^3 + 48x^3y^2$
$\Rightarrow 6x^2y^2(6 - 5xy + 8x)$ Ans
(iv) $8(2a + 3b)^3 - 12(2a + 3b)^2$
Taking $2a + 3b = x$, we have:
$\Rightarrow 8x^3 - 12x^2$
$\Rightarrow 4x^2(2x - 3)$
$\Rightarrow 4(2a + 3b)^2(2(2a + 3b) - 3)$
$\Rightarrow 4(2a + 3b)^2(4a + 6b - 3)$ Ans
(v) $9a(x - 2y)^4 - 12a(x - 2y)^3$
Taking $x - 2y = x$, we have:
$\Rightarrow 9ax^4 - 12ax^3$
$\Rightarrow 3ax^3(3x - 4)$
$\Rightarrow 3a(x - 2y)^3(3(x - 2y) - 4)$
$\Rightarrow 3a(x - 2y)^3(3x - 6y - 4)$ Ans
Q2. Factorise:
(i) $a^2 - ab - 3a + 3b$
(ii) $x^2y + xy^2 + 5x - 5y$
(iii) $a^2 - ab(1 - b) - b^3$
(iv) $xy^2 + (x - 1)y - 1$
(v) $(ax + by)^2 + (bx - ay)^2$
(vi) $ab(x^2 + y^2) - xy(a^2 + b^2)$
(vii) $m - 1 - (m - 1)^2 + am - a$
Show Answer
(i) $a^2 - ab - 3a + 3b$
$\Rightarrow (a^2 - ab) - (3a + 3b)$
$\Rightarrow a(a - b) - 3(a + b)$
$\Rightarrow (a - b)(a - 3)$ Ans
(ii) $x^2y - xy^2 + 5x - 5y$
$\Rightarrow (x^2y - xy^2) + (5x - 5y)$
$\Rightarrow xy(x - y) + 5(x - y)$
$\Rightarrow (x - y)(xy + 5)$ Ans
(iii) $a^2 - ab(1 - b) - b^3$
$\Rightarrow a^2 - ab + ab^2 - b^3$
$\Rightarrow (a^2 - ab) + (ab^2 - b^3)$
$\Rightarrow a(a - b) + b^2(a - b)$
$\Rightarrow (a - b)(a + b^2)$ Ans
(iv) $xy^2 + (x - 1)y - 1$
$\Rightarrow xy^2 + xy - y - 1$
$\Rightarrow (xy^2 + xy) - (y - 1)$
$\Rightarrow xy(y + 1) - 1(y + 1)$
$\Rightarrow (y + 1)(xy - 1)$ Ans
(v) $(ax + by)^2 + (bx - ay)^2$
$\Rightarrow (a^2x^2 + b^2y^2) + (b^2x^2 - a^2y^2)$
In $(a^2x^2 + b^2y^2)$, taking $ax = a \text{ and } by = b$, we have:
$\because a^2 + b^2 = a^2 + 2ab + b^2$
$\therefore (ax)^2 + 2axby + (by)^2$
Similarly, in $(b^2x^2 - a^2y^2)$, taking $bx = a \text{ and } ay = b$, we have:
$\because a^2 - b^2 = a^2 - 2ab + b^2$
$\therefore (bx)^2 - 2bxay + (ay)^2$
Now we have:
$\Rightarrow (ax)^2 + 2axby + (by)^2 + (bx)^2 - 2bxay + (ay)^2$
$\Rightarrow [(ax)^2 + (by)^2] + [(bx)^2 + (ay)^2]$
Upon rearranging we have:
$\Rightarrow a^2x^2 + + b^2x^2 + a^2y^2 + b^2y^2$
$\Rightarrow x^2(a^2 + + b^2) + y^2(a^2 + b^2)$
$\Rightarrow (a^2 + b^2)(x^2 + y^2)$ Ans
(vi) $ab(x^2 + y^2) - xy(a^2 + b^2)$
$\Rightarrow (abx^2 + aby^2) - (a^2xy - b^2xy)$
$\Rightarrow abx^2 + aby^2 - a^2xy - b^2xy$
Upon rearranging we have:
$\Rightarrow abx^2 - a^2xy - b^2xy + aby^2$
$\Rightarrow ax(bx - ay) - by(bx - ay)$
$\Rightarrow (bx - ay)(ax - by)$ Ans
(vii) $m - 1 - (m - 1)^2 + am - a$
$\Rightarrow m - 1 - (m - 1)^2 + a(m - 1)$
Taking $m - 1 = x$, we have:
$\Rightarrow x - x^2 + ax$
$\Rightarrow x(1 - x + a)$
$\therefore (m - 1)(1 - (m - 1) + a)$
$\Rightarrow (m - 1)(1 - m + 1 + a)$
$\Rightarrow (m - 1)(2 - m + a)$ Ans
$\Rightarrow (a^2 - ab) - (3a + 3b)$
$\Rightarrow a(a - b) - 3(a + b)$
$\Rightarrow (a - b)(a - 3)$ Ans
(ii) $x^2y - xy^2 + 5x - 5y$
$\Rightarrow (x^2y - xy^2) + (5x - 5y)$
$\Rightarrow xy(x - y) + 5(x - y)$
$\Rightarrow (x - y)(xy + 5)$ Ans
(iii) $a^2 - ab(1 - b) - b^3$
$\Rightarrow a^2 - ab + ab^2 - b^3$
$\Rightarrow (a^2 - ab) + (ab^2 - b^3)$
$\Rightarrow a(a - b) + b^2(a - b)$
$\Rightarrow (a - b)(a + b^2)$ Ans
(iv) $xy^2 + (x - 1)y - 1$
$\Rightarrow xy^2 + xy - y - 1$
$\Rightarrow (xy^2 + xy) - (y - 1)$
$\Rightarrow xy(y + 1) - 1(y + 1)$
$\Rightarrow (y + 1)(xy - 1)$ Ans
(v) $(ax + by)^2 + (bx - ay)^2$
$\Rightarrow (a^2x^2 + b^2y^2) + (b^2x^2 - a^2y^2)$
In $(a^2x^2 + b^2y^2)$, taking $ax = a \text{ and } by = b$, we have:
$\because a^2 + b^2 = a^2 + 2ab + b^2$
$\therefore (ax)^2 + 2axby + (by)^2$
Similarly, in $(b^2x^2 - a^2y^2)$, taking $bx = a \text{ and } ay = b$, we have:
$\because a^2 - b^2 = a^2 - 2ab + b^2$
$\therefore (bx)^2 - 2bxay + (ay)^2$
Now we have:
$\Rightarrow (ax)^2 + 2axby + (by)^2 + (bx)^2 - 2bxay + (ay)^2$
$\Rightarrow [(ax)^2 + (by)^2] + [(bx)^2 + (ay)^2]$
Upon rearranging we have:
$\Rightarrow a^2x^2 + + b^2x^2 + a^2y^2 + b^2y^2$
$\Rightarrow x^2(a^2 + + b^2) + y^2(a^2 + b^2)$
$\Rightarrow (a^2 + b^2)(x^2 + y^2)$ Ans
(vi) $ab(x^2 + y^2) - xy(a^2 + b^2)$
$\Rightarrow (abx^2 + aby^2) - (a^2xy - b^2xy)$
$\Rightarrow abx^2 + aby^2 - a^2xy - b^2xy$
Upon rearranging we have:
$\Rightarrow abx^2 - a^2xy - b^2xy + aby^2$
$\Rightarrow ax(bx - ay) - by(bx - ay)$
$\Rightarrow (bx - ay)(ax - by)$ Ans
(vii) $m - 1 - (m - 1)^2 + am - a$
$\Rightarrow m - 1 - (m - 1)^2 + a(m - 1)$
Taking $m - 1 = x$, we have:
$\Rightarrow x - x^2 + ax$
$\Rightarrow x(1 - x + a)$
$\therefore (m - 1)(1 - (m - 1) + a)$
$\Rightarrow (m - 1)(1 - m + 1 + a)$
$\Rightarrow (m - 1)(2 - m + a)$ Ans
Q3. Factorise:
(i) $a^2 - (b - c)^2$
(ii) $25(2x - y)^2 - 16(x - 2y)^2$
(iii) $16(5x + 4)^2 - 9(3x - 2)^2$
(iv) $9x^2 - \dfrac{1}{16}$
(v) $25(x - 2y)^2 - 4$
Show Answer
(i) $a^2 - (b - c)^2$
$(a - (b - c))^2$
$\because a^2 - b^2 = (a + b)(a - b)$
$\Rightarrow (a + (b - c))(a - (b - c))$
$\Rightarrow (a + b - c)(a - b + c)$ Ans
(ii) $25(2x - y)^2 - 16(x - 2y)^2$
$\Rightarrow (5(2x - y))^2 - (4(x - 2y))^2$
$\Rightarrow (10x - 5y)^2 - (4x - 8y)^2$
$\because a^2 - b^2 = a^2 - 2ab + b^2$
$\therefore ((10x)^2 - 2(10x)(5y) + (5y)^2) - ((4x)^2 - 2(4x)(8y) + (8y)^2)$
$\Rightarrow (100x^2 - 100xy + 25y^2) - (16x^2 - 64xy + 64y^2)$
$\Rightarrow 100x^2 - 100xy + 25y^2 - 16x^2 + 64xy - 64y^2$
$\Rightarrow (100x^2 - 16x^2) + (64xy - 100xy) + (25y^2 - 64y^2)$
$\Rightarrow 84x^2 - 36xy - 39y^2$
$\Rightarrow 84x^2 - 78xy + 42xy - 39y^2$
$\Rightarrow (84x^2 - 78xy) + (42xy - 39y^2)$
$\Rightarrow 6x(14x - 13y) + 3y(14x - 13y)$
$\Rightarrow (14x - 13y)(6x + 3y)$
$\Rightarrow (14x - 13y)3(2x + y)$
$\Rightarrow 3(14x - 13y)(2x + y)$ Ans
(iii) $16(5x + 4)^2 - 9(3x - 2)^2$
$\Rightarrow (4(5x + 4))^2 - (3(3x - 2))^2$
$\Rightarrow (20x + 16)^2 - (9x - 6)^2$
$\because (a + b)^2 = a^2 + 2ab + b^2$
And
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore ((20x)^2 + 2(20x)(16) + (16)^2) - ((9x)^2 - 2(9x)(6) + (6)^2)$
$\Rightarrow (400x^2 + 640x + 256) - (81x^2 - 108x + 36)$
$\Rightarrow (400x^2 - 81x^2) + (640x + 108x) + (256 - 36)$
$\Rightarrow 319x^2 + 748x + 220$
$\because 319 \times 220 = 70180, 110 + 638 = 748 \text{ and } 110 \times 638 = 70180$
$\therefore 319x^2 + 110x + 638x + 220$
$\Rightarrow (319x^2 + 110x) + (638x + 220)$
$\Rightarrow 11x(29x + 10) + 22(29x + 10)$
$\Rightarrow (29x + 10)(11x + 22)$
$\Rightarrow (29x + 10)11(x + 2)$
$\Rightarrow 11(29x + 10)(x + 2)$ Ans
(iv) $9x^2 - \dfrac{1}{16}$
$\Rightarrow (3x)^2 - (\dfrac{1}{4})^2$
$\Rightarrow (3x - \dfrac{1}{4})^2$
$\Rightarrow (3x + \dfrac{1}{4})(3x - \dfrac{1}{4})$ Ans
(v) $25(x - 2y)^2 - 4$
$\Rightarrow (5(x - 2y))^2 - (2)^2$
$\Rightarrow (5x - 10y)^2 - (2)^2$
$\Rightarrow (5x - 10y - 2)^2$
$\Rightarrow (5x - 10y + 2)(5x - 10y - 2)$ Ans
$(a - (b - c))^2$
$\because a^2 - b^2 = (a + b)(a - b)$
$\Rightarrow (a + (b - c))(a - (b - c))$
$\Rightarrow (a + b - c)(a - b + c)$ Ans
(ii) $25(2x - y)^2 - 16(x - 2y)^2$
$\Rightarrow (5(2x - y))^2 - (4(x - 2y))^2$
$\Rightarrow (10x - 5y)^2 - (4x - 8y)^2$
$\because a^2 - b^2 = a^2 - 2ab + b^2$
$\therefore ((10x)^2 - 2(10x)(5y) + (5y)^2) - ((4x)^2 - 2(4x)(8y) + (8y)^2)$
$\Rightarrow (100x^2 - 100xy + 25y^2) - (16x^2 - 64xy + 64y^2)$
$\Rightarrow 100x^2 - 100xy + 25y^2 - 16x^2 + 64xy - 64y^2$
$\Rightarrow (100x^2 - 16x^2) + (64xy - 100xy) + (25y^2 - 64y^2)$
$\Rightarrow 84x^2 - 36xy - 39y^2$
$\Rightarrow 84x^2 - 78xy + 42xy - 39y^2$
$\Rightarrow (84x^2 - 78xy) + (42xy - 39y^2)$
$\Rightarrow 6x(14x - 13y) + 3y(14x - 13y)$
$\Rightarrow (14x - 13y)(6x + 3y)$
$\Rightarrow (14x - 13y)3(2x + y)$
$\Rightarrow 3(14x - 13y)(2x + y)$ Ans
(iii) $16(5x + 4)^2 - 9(3x - 2)^2$
$\Rightarrow (4(5x + 4))^2 - (3(3x - 2))^2$
$\Rightarrow (20x + 16)^2 - (9x - 6)^2$
$\because (a + b)^2 = a^2 + 2ab + b^2$
And
$\because (a - b)^2 = a^2 - 2ab + b^2$
$\therefore ((20x)^2 + 2(20x)(16) + (16)^2) - ((9x)^2 - 2(9x)(6) + (6)^2)$
$\Rightarrow (400x^2 + 640x + 256) - (81x^2 - 108x + 36)$
$\Rightarrow (400x^2 - 81x^2) + (640x + 108x) + (256 - 36)$
$\Rightarrow 319x^2 + 748x + 220$
$\because 319 \times 220 = 70180, 110 + 638 = 748 \text{ and } 110 \times 638 = 70180$
$\therefore 319x^2 + 110x + 638x + 220$
$\Rightarrow (319x^2 + 110x) + (638x + 220)$
$\Rightarrow 11x(29x + 10) + 22(29x + 10)$
$\Rightarrow (29x + 10)(11x + 22)$
$\Rightarrow (29x + 10)11(x + 2)$
$\Rightarrow 11(29x + 10)(x + 2)$ Ans
(iv) $9x^2 - \dfrac{1}{16}$
$\Rightarrow (3x)^2 - (\dfrac{1}{4})^2$
$\Rightarrow (3x - \dfrac{1}{4})^2$
$\Rightarrow (3x + \dfrac{1}{4})(3x - \dfrac{1}{4})$ Ans
(v) $25(x - 2y)^2 - 4$
$\Rightarrow (5(x - 2y))^2 - (2)^2$
$\Rightarrow (5x - 10y)^2 - (2)^2$
$\Rightarrow (5x - 10y - 2)^2$
$\Rightarrow (5x - 10y + 2)(5x - 10y - 2)$ Ans
Q4. Factorise:
(i) $a^2 - 23a + 42$
(ii) $a^2 - 23a - 108$
(iii) $1 - 18x - 63x^2$
(iv) $5x^2 + 4xy - 12y^2$
(v) $x(3x + 14) + 8$
(vi) $5 - 4x(1 + 3x)$
(vii) $x^2y^2 - 3xy - 40$
(viii) $(3x - 2y)^2 - 5(3x - 2y) - 24$
(ix) $(a + b)^2 - (a + b) - 35$
Show Answer
(i) $a^2 - 23a + 42$
$\Rightarrow a^2 - 21a - 2a + 42$
$\Rightarrow (a^2 - 21a) - (2a + 42)$
$\Rightarrow a(a - 21) - 2(a - 21)$
$\Rightarrow (a - 21)(a - 2)$ Ans
(ii) $a^2 - 23a - 108$
$\Rightarrow a^2 - 27a + 4a + 108$
$\Rightarrow (a^2 - 27a) + (4a + 108)$
$\Rightarrow a(a - 27) + 4(a - 27)$
$\Rightarrow (a - 27)(a + 4)$ Ans
(iii) $1 - 18x - 63x^2$
$\Rightarrow 1 - 21x + 3x - 63x^2$
$\Rightarrow (1 - 21x) + (3x - 63x^2)$
$\Rightarrow 1(1 - 21x) + 3x(1 - 21x)$
$\Rightarrow (1 - 21x)(1 + 3x)$ Ans
(iv) $5x^2 + 4xy - 12y^2$
$\Rightarrow 5x^2 - 10xy + 6xy - 12y^2$
$\Rightarrow (5x^2 - 10xy) + (6xy - 12y^2)$
$\Rightarrow 5x(x - 2y) + 6y(x - 2y)$
$\Rightarrow (x - 2y)(5x + 6y)$ Ans
(v) $x(3x + 14) + 8$
$\Rightarrow 3x^2 + 14x + 8$
$\Rightarrow 3x^2 + 12x + 2x + 8$
$\Rightarrow (3x^2 + 12x) + (2x + 8)$
$\Rightarrow 3x(x + 4) + 2(x + 4)$
$\Rightarrow (x + 4)(3x + 2)$ Ans
(vi) $5 - 4x(1 + 3x)$
$\Rightarrow 5 - 4x - 12x^2$
$\Rightarrow 5 - 10x + 6x - 12x^2$
$\Rightarrow (5 - 10x) + (6x - 12x^2)$
$\Rightarrow 5(1 - 2x) + 6x(1 - 2x)$
$\Rightarrow (1 - 2x)(5 + 6x)$ Ans
(vii) $x^2y^2 - 3xy - 40$
$\Rightarrow x^2y^2 - 8xy + 5xy - 40$
$\Rightarrow (x^2y^2 - 8xy) + (5xy - 40)$
$\Rightarrow xy(xy - 8) + 5(xy - 8)$
$\Rightarrow (xy - 8)(xy + 5)$ Ans
(viii) $(3x - 2y)^2 - 5(3x - 2y) - 24$
By taking $(3x - 2y) = x$, we have:
$\Rightarrow x^2 - 5x - 24$
$\Rightarrow x^2 - 8x + 3x - 24$
$\Rightarrow (x^2 - 8x) + (3x - 24)$
$\Rightarrow x(x - 8) + 3(x - 8)$
$\Rightarrow (x - 8)(x + 3)$
$\therefore (3x - 2y - 8)(3x - 2y + 3)$ Ans
(ix) $12(a + b)^2 - (a + b) - 35$
By taking $(a + b) = x$, we have:
$\Rightarrow 12x^2 - x - 35$
$\because 12 \times 35 = 420 \text{ and } 21 \times 20 = 420$
$\Rightarrow 12x^2 - 21x + 20x - 35$
$\Rightarrow 3x(4x - 7) + 5(4x - 7)$
$\Rightarrow (4x - 7)(3x + 5)$
$\therefore (4(a + b) - 7)(3(a + b) + 5)$
$\Rightarrow (4a + 4b - 7)(3a + 3b + 5)$ Ans
$\Rightarrow a^2 - 21a - 2a + 42$
$\Rightarrow (a^2 - 21a) - (2a + 42)$
$\Rightarrow a(a - 21) - 2(a - 21)$
$\Rightarrow (a - 21)(a - 2)$ Ans
(ii) $a^2 - 23a - 108$
$\Rightarrow a^2 - 27a + 4a + 108$
$\Rightarrow (a^2 - 27a) + (4a + 108)$
$\Rightarrow a(a - 27) + 4(a - 27)$
$\Rightarrow (a - 27)(a + 4)$ Ans
(iii) $1 - 18x - 63x^2$
$\Rightarrow 1 - 21x + 3x - 63x^2$
$\Rightarrow (1 - 21x) + (3x - 63x^2)$
$\Rightarrow 1(1 - 21x) + 3x(1 - 21x)$
$\Rightarrow (1 - 21x)(1 + 3x)$ Ans
(iv) $5x^2 + 4xy - 12y^2$
$\Rightarrow 5x^2 - 10xy + 6xy - 12y^2$
$\Rightarrow (5x^2 - 10xy) + (6xy - 12y^2)$
$\Rightarrow 5x(x - 2y) + 6y(x - 2y)$
$\Rightarrow (x - 2y)(5x + 6y)$ Ans
(v) $x(3x + 14) + 8$
$\Rightarrow 3x^2 + 14x + 8$
$\Rightarrow 3x^2 + 12x + 2x + 8$
$\Rightarrow (3x^2 + 12x) + (2x + 8)$
$\Rightarrow 3x(x + 4) + 2(x + 4)$
$\Rightarrow (x + 4)(3x + 2)$ Ans
(vi) $5 - 4x(1 + 3x)$
$\Rightarrow 5 - 4x - 12x^2$
$\Rightarrow 5 - 10x + 6x - 12x^2$
$\Rightarrow (5 - 10x) + (6x - 12x^2)$
$\Rightarrow 5(1 - 2x) + 6x(1 - 2x)$
$\Rightarrow (1 - 2x)(5 + 6x)$ Ans
(vii) $x^2y^2 - 3xy - 40$
$\Rightarrow x^2y^2 - 8xy + 5xy - 40$
$\Rightarrow (x^2y^2 - 8xy) + (5xy - 40)$
$\Rightarrow xy(xy - 8) + 5(xy - 8)$
$\Rightarrow (xy - 8)(xy + 5)$ Ans
(viii) $(3x - 2y)^2 - 5(3x - 2y) - 24$
By taking $(3x - 2y) = x$, we have:
$\Rightarrow x^2 - 5x - 24$
$\Rightarrow x^2 - 8x + 3x - 24$
$\Rightarrow (x^2 - 8x) + (3x - 24)$
$\Rightarrow x(x - 8) + 3(x - 8)$
$\Rightarrow (x - 8)(x + 3)$
$\therefore (3x - 2y - 8)(3x - 2y + 3)$ Ans
(ix) $12(a + b)^2 - (a + b) - 35$
By taking $(a + b) = x$, we have:
$\Rightarrow 12x^2 - x - 35$
$\because 12 \times 35 = 420 \text{ and } 21 \times 20 = 420$
$\Rightarrow 12x^2 - 21x + 20x - 35$
$\Rightarrow 3x(4x - 7) + 5(4x - 7)$
$\Rightarrow (4x - 7)(3x + 5)$
$\therefore (4(a + b) - 7)(3(a + b) + 5)$
$\Rightarrow (4a + 4b - 7)(3a + 3b + 5)$ Ans
Q5. Factorise:
(i) $15(5x - 4)^2 - 10(5x - 4)$
(ii) $3a^2x - bx + 3a^2 - b$
(iii) $b(c - d)^2 + a(d - c) + 3(c - d)$
(iv) $ax^2 + b^2y - ab^2 - x^2y$
(v) $1 - 3x - 3y - 4(x + y)^2$
Show Answer
(i) $15(5x - 4)^2 - 10(5x - 4)$
By taking $(5x - 4) = x$, we have:
$\Rightarrow 15x^2 - 10x$
$\Rightarrow 5x(3x - 2)$
$\therefore 5(5x - 4)(3(5x - 4) - 2)$
$\Rightarrow 5(5x - 4)(15x - 12 - 2)$
$\Rightarrow 5(5x - 4)(15x - 14)$ Ans
(ii) $3a^2x - bx + 3a^2 - b$
$\Rightarrow (3a^2x - bx) + (3a^2 - b)$
$\Rightarrow x(3a^2 - b) + 1(3a^2 - b)$
$\Rightarrow (3a^2 - b)(x + 1)$ Ans
(iii) $b(c - d)^2 + a(d - c) + 3(c - d)$
In $a(d - c)$, by taking $a(-c + d) == -a(c - d)$
We have:
$\Rightarrow b(c - d)^2 - a(c - d) + 3(c - d)$
Now by taking $(c - d) = x$, we have:
$\Rightarrow bx^2 - ax + 3x$
$\Rightarrow x(bx - a + 3)$
$\Rightarrow (c - d)(b(c - d) - a + 3)$
$\Rightarrow (c - d)(bc - bd - a + 3)$ Ans
(iv) $ax^2 + b^2y - ab^2 - x^2y$
Upon rearranging the terms, we have:
$\Rightarrow ax^2 - x^2y - ab^2 + b^2y$
$\Rightarrow (ax^2 - x^2y) - (ab^2 + b^2y)$
$\Rightarrow x^2(a - y) - b^2(a - y)$
$\Rightarrow (a - y)(x^2 - b^2)$
$\Rightarrow (a - y)(x + b)(x - b)$ Ans
(v) $1 - 3x - 3y - 4(x + y)^2$
$\Rightarrow 1 - 3(x + y) - 4(x + y)^2$
By taking $(x + y) = x$, we have:
$\Rightarrow 1 - 3x - 4x^2$
$\Rightarrow 1 - 4x + x - 4x^2$
$\Rightarrow 1(1 - 4x) + x(1 - 4x)$
$\Rightarrow (1 - 4x)(1 + x)$
$\Rightarrow (1 - 4(x + y))(1 + (x + y))$
$\Rightarrow (1 - 4x - 4y)(1 + x + y)$ Ans
By taking $(5x - 4) = x$, we have:
$\Rightarrow 15x^2 - 10x$
$\Rightarrow 5x(3x - 2)$
$\therefore 5(5x - 4)(3(5x - 4) - 2)$
$\Rightarrow 5(5x - 4)(15x - 12 - 2)$
$\Rightarrow 5(5x - 4)(15x - 14)$ Ans
(ii) $3a^2x - bx + 3a^2 - b$
$\Rightarrow (3a^2x - bx) + (3a^2 - b)$
$\Rightarrow x(3a^2 - b) + 1(3a^2 - b)$
$\Rightarrow (3a^2 - b)(x + 1)$ Ans
(iii) $b(c - d)^2 + a(d - c) + 3(c - d)$
In $a(d - c)$, by taking $a(-c + d) == -a(c - d)$
We have:
$\Rightarrow b(c - d)^2 - a(c - d) + 3(c - d)$
Now by taking $(c - d) = x$, we have:
$\Rightarrow bx^2 - ax + 3x$
$\Rightarrow x(bx - a + 3)$
$\Rightarrow (c - d)(b(c - d) - a + 3)$
$\Rightarrow (c - d)(bc - bd - a + 3)$ Ans
(iv) $ax^2 + b^2y - ab^2 - x^2y$
Upon rearranging the terms, we have:
$\Rightarrow ax^2 - x^2y - ab^2 + b^2y$
$\Rightarrow (ax^2 - x^2y) - (ab^2 + b^2y)$
$\Rightarrow x^2(a - y) - b^2(a - y)$
$\Rightarrow (a - y)(x^2 - b^2)$
$\Rightarrow (a - y)(x + b)(x - b)$ Ans
(v) $1 - 3x - 3y - 4(x + y)^2$
$\Rightarrow 1 - 3(x + y) - 4(x + y)^2$
By taking $(x + y) = x$, we have:
$\Rightarrow 1 - 3x - 4x^2$
$\Rightarrow 1 - 4x + x - 4x^2$
$\Rightarrow 1(1 - 4x) + x(1 - 4x)$
$\Rightarrow (1 - 4x)(1 + x)$
$\Rightarrow (1 - 4(x + y))(1 + (x + y))$
$\Rightarrow (1 - 4x - 4y)(1 + x + y)$ Ans
Q6. Factorise:
(i) $2a^3 - 50a$
(ii) $54a^2b^2 - 6$
(iii) $64a^2b - 144b^3$
(iv) $(2x - y)^2 - (2x - y)^2$
(v) $x^2 - 2xy + y^2 - z^2$
(vi) $x^2 - y^2 - 2yz - z^2$
(vii) $7a^5 - 567a$
(viii) $5x^2 - \dfrac{20x^4}{9}$
Show Answer
(i) $2a^3 - 50a$
$\Rightarrow 2a(a^2 - 25)$
$\Rightarrow 2a(a^2 - 5^2)$
$\Rightarrow 2a(a - 5)^2$
$\Rightarrow 2a(a + 5)(a - 5)$ Ans
(ii) $54a^2b^2 - 6$
$\Rightarrow 6(9a^2b^2 - 1)$
$\Rightarrow 6((3ab)^2 - (1)^2)$
$\Rightarrow 6(3ab - 1)^2$
$\Rightarrow 6(3ab + 1)(3ab - 1)$ Ans
(iii) $64a^2b - 144b^3$
$\Rightarrow b(64a^2 - 144b^2)$
$\Rightarrow b((8a)^2 - (12b)^2)$
$\Rightarrow b(8a - 12b)^2$
$\Rightarrow b(4(2a - 3b))^2$
$\therefore 16b(2a + 3b)(2a - 3b)$
(iv) $(2x - y)^3 - (2x - y)$
By taking $(2x - y) = x$, we have:
$\Rightarrow x^3 - x$
$\Rightarrow x(x^2 - 1)$
$\Rightarrow x(x^2 - (1)^2)$
$\Rightarrow x(x - 1)^2$
$\Rightarrow x(x + 1)(x - 1)$
$\Rightarrow (2x - y)((2x - y) + 1)((2x - y) - 1)$
$\Rightarrow (2x - y)(2x - y + 1)(2x - y - 1)$ Ans
(v) $x^2 - 2xy + y^2 - z^2$
We have: $(x^2 - 2xy + y^2) - z^2$
Where $(x^2 - 2xy + y^2)$ is perfect square:
$\therefore (x^2 - 2xy + y^2) = (x^2 - y^2)$
$\Rightarrow (x - y)^2 - z^2$
$\Rightarrow (x - y + z)(x - y - z)$ Ans
(vi) $x^2 - y^2 - 2yz - z^2$
We have: $x^2 - (y^2 - 2yz - z^2)$
Where $(y^2 - 2yz - z^2)$ is perfect square:
$\therefore (y^2 - 2yz - z^2) = (y^2 - z^2)$
$\Rightarrow x^2 - (y - z)^2$
$\Rightarrow (x + y + z)(x - y - z)$ Ans
(vii) $7a^5 - 567a$
$\Rightarrow 7a(a^4 - 81)$
$\Rightarrow 7a((a^2)^2 - (9)^2)$
$\Rightarrow 7a(a^2 - 9)^2$
$\Rightarrow 7a(a^2 + 9)(a^2 - 9)$
$\Rightarrow 7a(a^2 + 9)(a^2 - 3^2)$
$\Rightarrow 7a(a^2 + 9)(a - 3)^2$
$\Rightarrow 7a(a^2 + 9)(a + 3)(a - 3)$ Ans
(viii) $5x^2 - \dfrac{20x^4}{9}$
$\Rightarrow 5x^2(1 - \dfrac{4x^2}{9})$
$\Rightarrow 5x^2((1)^2 - \dfrac{(2x)^2}{(3)^2})$
$\Rightarrow 5x^2(1 - \dfrac{2x}{3})^2$
$\Rightarrow 5x^2(1 + \dfrac{2x}{3})(1 - \dfrac{2x}{3})$ Ans
$\Rightarrow 2a(a^2 - 25)$
$\Rightarrow 2a(a^2 - 5^2)$
$\Rightarrow 2a(a - 5)^2$
$\Rightarrow 2a(a + 5)(a - 5)$ Ans
(ii) $54a^2b^2 - 6$
$\Rightarrow 6(9a^2b^2 - 1)$
$\Rightarrow 6((3ab)^2 - (1)^2)$
$\Rightarrow 6(3ab - 1)^2$
$\Rightarrow 6(3ab + 1)(3ab - 1)$ Ans
(iii) $64a^2b - 144b^3$
$\Rightarrow b(64a^2 - 144b^2)$
$\Rightarrow b((8a)^2 - (12b)^2)$
$\Rightarrow b(8a - 12b)^2$
$\Rightarrow b(4(2a - 3b))^2$
$\therefore 16b(2a + 3b)(2a - 3b)$
(iv) $(2x - y)^3 - (2x - y)$
By taking $(2x - y) = x$, we have:
$\Rightarrow x^3 - x$
$\Rightarrow x(x^2 - 1)$
$\Rightarrow x(x^2 - (1)^2)$
$\Rightarrow x(x - 1)^2$
$\Rightarrow x(x + 1)(x - 1)$
$\Rightarrow (2x - y)((2x - y) + 1)((2x - y) - 1)$
$\Rightarrow (2x - y)(2x - y + 1)(2x - y - 1)$ Ans
(v) $x^2 - 2xy + y^2 - z^2$
We have: $(x^2 - 2xy + y^2) - z^2$
Where $(x^2 - 2xy + y^2)$ is perfect square:
$\therefore (x^2 - 2xy + y^2) = (x^2 - y^2)$
$\Rightarrow (x - y)^2 - z^2$
$\Rightarrow (x - y + z)(x - y - z)$ Ans
(vi) $x^2 - y^2 - 2yz - z^2$
We have: $x^2 - (y^2 - 2yz - z^2)$
Where $(y^2 - 2yz - z^2)$ is perfect square:
$\therefore (y^2 - 2yz - z^2) = (y^2 - z^2)$
$\Rightarrow x^2 - (y - z)^2$
$\Rightarrow (x + y + z)(x - y - z)$ Ans
(vii) $7a^5 - 567a$
$\Rightarrow 7a(a^4 - 81)$
$\Rightarrow 7a((a^2)^2 - (9)^2)$
$\Rightarrow 7a(a^2 - 9)^2$
$\Rightarrow 7a(a^2 + 9)(a^2 - 9)$
$\Rightarrow 7a(a^2 + 9)(a^2 - 3^2)$
$\Rightarrow 7a(a^2 + 9)(a - 3)^2$
$\Rightarrow 7a(a^2 + 9)(a + 3)(a - 3)$ Ans
(viii) $5x^2 - \dfrac{20x^4}{9}$
$\Rightarrow 5x^2(1 - \dfrac{4x^2}{9})$
$\Rightarrow 5x^2((1)^2 - \dfrac{(2x)^2}{(3)^2})$
$\Rightarrow 5x^2(1 - \dfrac{2x}{3})^2$
$\Rightarrow 5x^2(1 + \dfrac{2x}{3})(1 - \dfrac{2x}{3})$ Ans
Q7. Factorise $xy^2 - xz^2$, Hence, find the value of:
(i) $9 \times 8^2 - 9 \times 2^2$
(ii) $40 \times 5.5^2 - 40 \times 4.5^2$
Show Answer
$\Rightarrow xy^2 - xz^2 = x(y^2 - z^2)$
$\Rightarrow x(y^2 - z^2) = x(y - z)^2$
$\therefore x(y - z)^2 = x(y + z)(y - z)$ Ans
(i) $9 \times 8^2- 9 \times 2^2$
Taking $9 = x, 8^2 = y^2 \text{ and } 2^2 = z$
Now, applying the formula $x(y + z)(y - z)$
$\therefore 9(8 + 2)(8 - 2) = 9(10)(6)$
$\Rightarrow 9 \times 60 = 540$ Ans
(ii) $40 \times 5.5^2 - 40 \times 4.5^2$
Taking $40 = x, 5.5^2 = y^2 \text{ and } 4.5^2 = z$
Now, applying the formula $x(y + z)(y - z)$
$\therefore 40(5.5 + 4.5)(5.5 - 4.5) = 40(10)(1)$
$\Rightarrow 40 \times 10 = 400$ Ans
$\Rightarrow x(y^2 - z^2) = x(y - z)^2$
$\therefore x(y - z)^2 = x(y + z)(y - z)$ Ans
(i) $9 \times 8^2- 9 \times 2^2$
Taking $9 = x, 8^2 = y^2 \text{ and } 2^2 = z$
Now, applying the formula $x(y + z)(y - z)$
$\therefore 9(8 + 2)(8 - 2) = 9(10)(6)$
$\Rightarrow 9 \times 60 = 540$ Ans
(ii) $40 \times 5.5^2 - 40 \times 4.5^2$
Taking $40 = x, 5.5^2 = y^2 \text{ and } 4.5^2 = z$
Now, applying the formula $x(y + z)(y - z)$
$\therefore 40(5.5 + 4.5)(5.5 - 4.5) = 40(10)(1)$
$\Rightarrow 40 \times 10 = 400$ Ans
Q8. Factorise:
(i) $(a - 3b)^3 - 36b^2$
(ii) $25(a - 5b)^2 - 4(a - 3b)^2$
(iii) $a^2 - 0.36b^2$
(iv) $a^4 - 625$
(v) $x^4 - 5x^2 - 36$
(vi) $15(2x - y)^2 - 16(2x - y) - 15$
Show Answer
(i) $(a - 3b)^3 - 36b^2$
$\Rightarrow (a - 3b)^2 - (6b)^2$
$\Rightarrow (a - 3b + 6b)(a - 3b - 6b)$
$\Rightarrow (a + 3b)(a - 9b)$ Ans
(ii) $25(a - 5b)^2 - 4(a - 3b)^2$
$\Rightarrow (5(a - 5b))^2 - (2(a - 3b))^2$
$\Rightarrow (5a - 25b)^2 - (2a - 6b)^2$
$\Rightarrow ((5a)^2 - 2 \times 5 \times 25b + (25b)^2) - ((2a)^2 - 2 \times 2a \times 6b + (6b)^2)$
$\Rightarrow (25a^2 - 250ab + 625b^2) - (4a^2 - 24ab - 36b^2)$
$\Rightarrow (25a^2 - 4a^2) - (250ab + 24ab) + (625b^2 - 36b^2)$
$\Rightarrow 21a^2 - 226ab + 589b^2$
$\because 21 \times 589 = 12369, 133 \times 93 = 12369 \text{ and } 133 + 93 = 226$
$\therefore 21a^2 - 93ab - 133ab + 589b^2$
$\Rightarrow (21a^2 - 93ab) - (133ab + 589b^2)$
$\Rightarrow 3a(7a - 31b) - 19b(7a - 31b)$
$\Rightarrow (7a - 31b)(3a - 19b)$ Ans
(iii) $a^2 - 0.36b^2$
$\Rightarrow a^2 - (0.6)^2$
$\Rightarrow (a - 0.6)^2$
$\Rightarrow (a + 0.6)(a - 0.6)$ Ans
(iv) $a^4 - 625$
$\Rightarrow (a^2)^2 - (25)^2$
$\Rightarrow (a^2 - 25)^2$
$\Rightarrow (a^2 + 25)(a^2 - 25)$
$\Rightarrow (a^2 + 25)(a^2 - (5)^2)$
$\Rightarrow (a^2 + 25)(a - 5)^2$
$\Rightarrow (a^2 + 25)(a + 5)(a - 5)$ Ans
(v) $x^4 - 5x^2 - 36$
$\Rightarrow x^4 + 4x^2 - 9x^2 - 36$
$\Rightarrow (x^4 + 4x^2) - (9x^2 - 36)$
$\Rightarrow x^2(x^2 + 4) - 9(x^2 + 4)$
$\Rightarrow (x^2 + 4)(x^2 - 9)$
$\Rightarrow (x^2 + 4)(x^2 - 3^2)$
$\Rightarrow (x^2 + 4)(x - 3)^2$
$\Rightarrow (x^2 + 4)(x + 3)(x - 3)$ Ans
(vi) $15(2x - y)^2 - 16(2x - y) - 15$
Taking $2x - y = x$, we have:
$\Rightarrow 15x^2 - 16x - 15$
$\because 15 \times 15 = 225, 25 \times 9 = 225 \text{ and } 25 - - = 16$
$\therefore 15x^2 - 25x + 9x - 15$
$\Rightarrow (15x^2 - 25x) + (9x - 15)$
$\Rightarrow 5x(3x - 5) + 3(3x - 5)$
$\Rightarrow (3x - 5)(5x + 3)$
$\Rightarrow (3(2x - y) - 5)(5(2x - y) + 3)$
$\Rightarrow (6x - 3y - 5)(10x - 5y + 3)$ Ans
$\Rightarrow (a - 3b)^2 - (6b)^2$
$\Rightarrow (a - 3b + 6b)(a - 3b - 6b)$
$\Rightarrow (a + 3b)(a - 9b)$ Ans
(ii) $25(a - 5b)^2 - 4(a - 3b)^2$
$\Rightarrow (5(a - 5b))^2 - (2(a - 3b))^2$
$\Rightarrow (5a - 25b)^2 - (2a - 6b)^2$
$\Rightarrow ((5a)^2 - 2 \times 5 \times 25b + (25b)^2) - ((2a)^2 - 2 \times 2a \times 6b + (6b)^2)$
$\Rightarrow (25a^2 - 250ab + 625b^2) - (4a^2 - 24ab - 36b^2)$
$\Rightarrow (25a^2 - 4a^2) - (250ab + 24ab) + (625b^2 - 36b^2)$
$\Rightarrow 21a^2 - 226ab + 589b^2$
$\because 21 \times 589 = 12369, 133 \times 93 = 12369 \text{ and } 133 + 93 = 226$
$\therefore 21a^2 - 93ab - 133ab + 589b^2$
$\Rightarrow (21a^2 - 93ab) - (133ab + 589b^2)$
$\Rightarrow 3a(7a - 31b) - 19b(7a - 31b)$
$\Rightarrow (7a - 31b)(3a - 19b)$ Ans
(iii) $a^2 - 0.36b^2$
$\Rightarrow a^2 - (0.6)^2$
$\Rightarrow (a - 0.6)^2$
$\Rightarrow (a + 0.6)(a - 0.6)$ Ans
(iv) $a^4 - 625$
$\Rightarrow (a^2)^2 - (25)^2$
$\Rightarrow (a^2 - 25)^2$
$\Rightarrow (a^2 + 25)(a^2 - 25)$
$\Rightarrow (a^2 + 25)(a^2 - (5)^2)$
$\Rightarrow (a^2 + 25)(a - 5)^2$
$\Rightarrow (a^2 + 25)(a + 5)(a - 5)$ Ans
(v) $x^4 - 5x^2 - 36$
$\Rightarrow x^4 + 4x^2 - 9x^2 - 36$
$\Rightarrow (x^4 + 4x^2) - (9x^2 - 36)$
$\Rightarrow x^2(x^2 + 4) - 9(x^2 + 4)$
$\Rightarrow (x^2 + 4)(x^2 - 9)$
$\Rightarrow (x^2 + 4)(x^2 - 3^2)$
$\Rightarrow (x^2 + 4)(x - 3)^2$
$\Rightarrow (x^2 + 4)(x + 3)(x - 3)$ Ans
(vi) $15(2x - y)^2 - 16(2x - y) - 15$
Taking $2x - y = x$, we have:
$\Rightarrow 15x^2 - 16x - 15$
$\because 15 \times 15 = 225, 25 \times 9 = 225 \text{ and } 25 - - = 16$
$\therefore 15x^2 - 25x + 9x - 15$
$\Rightarrow (15x^2 - 25x) + (9x - 15)$
$\Rightarrow 5x(3x - 5) + 3(3x - 5)$
$\Rightarrow (3x - 5)(5x + 3)$
$\Rightarrow (3(2x - y) - 5)(5(2x - y) + 3)$
$\Rightarrow (6x - 3y - 5)(10x - 5y + 3)$ Ans
Q9. Factorise $a^2b - b^3$. Using this result. find the value of $101^2 \times 100 - 100^3$
Show Answer
$\Rightarrow a^2b - b^3 = b(a^2 - b^2)$
$\Rightarrow b(a^2 - b^2) = b(a - b)^2$
$\Rightarrow b(a - b)^2 = b(a + b)(a - b)$
Given: $101^2 \times 100 - 100^3$
Taking $a^2 = 101^2, b = 100 and b^3 = 100^3$
And applying the formula $b(a + b)(a - b)$, we have:
$\Rightarrow 100(101 + 100)(101 - 100)$
$\Rightarrow 100(201)(1)$
$\Rightarrow 100 \times 201 = 20100$ Ans
$\Rightarrow b(a^2 - b^2) = b(a - b)^2$
$\Rightarrow b(a - b)^2 = b(a + b)(a - b)$
Given: $101^2 \times 100 - 100^3$
Taking $a^2 = 101^2, b = 100 and b^3 = 100^3$
And applying the formula $b(a + b)(a - b)$, we have:
$\Rightarrow 100(101 + 100)(101 - 100)$
$\Rightarrow 100(201)(1)$
$\Rightarrow 100 \times 201 = 20100$ Ans
Q10. Evaluating (using factors): $301^2 \times 300 - 300^3$
Show Answer
Taking $a = 301^2, b = 300$
And applying the formula $b(a + b)(a - b)$, we have:
$\Rightarrow 300(301 + 300)(301 - 300)$
$\Rightarrow 300(601)(1)$
$\Rightarrow 300 \times 601 = 1,80,300$ Ans
And applying the formula $b(a + b)(a - b)$, we have:
$\Rightarrow 300(301 + 300)(301 - 300)$
$\Rightarrow 300(601)(1)$
$\Rightarrow 300 \times 601 = 1,80,300$ Ans
Q11. Factorise:
(i) $20a^2 - 10a - 150$
(ii) $9x^2 - (x^2 - 4)^2$
(iii) $9(x + y)^2 - x^2$
(iv) $4a^2 - b^2 + 2a + b$
(v) $9x^2 - 4(y + 2x)^2$
Show Answer
(i) $20a^2 - 10a - 150$
$\Rightarrow 20a^2 - 60a + 50a - 150$
$\Rightarrow (20a^2 - 60a) + (50a - 150)$
$\Rightarrow 20a(a - 3) + 50(a - 3)$
$\Rightarrow (a - 3)(20a + 50)$
$\Rightarrow (a - 3)10(2a + 5)$
$\Rightarrow 10(a - 3)(2a + 5)$ Ans
(ii) $9x^2 - (x^2 - 4)^2$
$\Rightarrow (3x)^2 - (x^2 - 4)^2$
$\Rightarrow (3x + \overline{x^2 - 4})(3x - \overline{x^2 - 4})$
$\Rightarrow (3x + x^2 - 4)(3x - x^2 + 4)$
Upon rearranging the equation we have:
$\Rightarrow (x^2 + 3x - 4)(4 + 3x - x^2)$
$\Rightarrow (x^2 + 4x - x - 4)(4 + 4x - x - x^2)$
$\Rightarrow [(x^2 + 4x) - (x - 4)][(4 + 4x) - (x - x^2)]$
$\Rightarrow [x(x + 4) - 1(x + 4)][4(1 + x) - x(1 + x)]$
$\Rightarrow (x + 4)(x - 1)(1 + x)(4 - x)$ Ans
(iii) $9(x + y)^2 - x^2$
$\Rightarrow (3(x + y))^2 - x^2$
$\Rightarrow (3x + 3y)^2 - x^2$
$\Rightarrow (3x + 3y + x)(3x + 3y - x)$
$\Rightarrow ((3x + x) + 3y)((3x - x) + 3y)$
$\Rightarrow (4x + 3y)(2x + 3y)$ Ans
(iv) $4a^2 - b^2 + 2a + b$
$\Rightarrow (2a)^2 - b^2 + (2a + b)$
$\Rightarrow (2a + b)(2a - b) + 1(2a + b)$
$\Rightarrow (2a + b)(2a - b + 1)$ Ans
(v) $9x^2 - 4(y + 2x)^2$
$\Rightarrow (3x)^2 - (2(y + 2x))^2$
$\Rightarrow (3x)^2 - (2y + 4x))^2$
$\Rightarrow (3x + \overline{2y + 4x})(3x - \overline{2y + 4x})$
$\Rightarrow (3x + 2y + 4x)(3x - 2y - 4x)$
$\Rightarrow ((3x + 4x) + 2y)((3x - 4x) - 2y)$
$\Rightarrow (7x + 2y)(- x - 2y)$
$\Rightarrow -(7x + 2y)(x - 2y)$ Ans
$\Rightarrow 20a^2 - 60a + 50a - 150$
$\Rightarrow (20a^2 - 60a) + (50a - 150)$
$\Rightarrow 20a(a - 3) + 50(a - 3)$
$\Rightarrow (a - 3)(20a + 50)$
$\Rightarrow (a - 3)10(2a + 5)$
$\Rightarrow 10(a - 3)(2a + 5)$ Ans
(ii) $9x^2 - (x^2 - 4)^2$
$\Rightarrow (3x)^2 - (x^2 - 4)^2$
$\Rightarrow (3x + \overline{x^2 - 4})(3x - \overline{x^2 - 4})$
$\Rightarrow (3x + x^2 - 4)(3x - x^2 + 4)$
Upon rearranging the equation we have:
$\Rightarrow (x^2 + 3x - 4)(4 + 3x - x^2)$
$\Rightarrow (x^2 + 4x - x - 4)(4 + 4x - x - x^2)$
$\Rightarrow [(x^2 + 4x) - (x - 4)][(4 + 4x) - (x - x^2)]$
$\Rightarrow [x(x + 4) - 1(x + 4)][4(1 + x) - x(1 + x)]$
$\Rightarrow (x + 4)(x - 1)(1 + x)(4 - x)$ Ans
(iii) $9(x + y)^2 - x^2$
$\Rightarrow (3(x + y))^2 - x^2$
$\Rightarrow (3x + 3y)^2 - x^2$
$\Rightarrow (3x + 3y + x)(3x + 3y - x)$
$\Rightarrow ((3x + x) + 3y)((3x - x) + 3y)$
$\Rightarrow (4x + 3y)(2x + 3y)$ Ans
(iv) $4a^2 - b^2 + 2a + b$
$\Rightarrow (2a)^2 - b^2 + (2a + b)$
$\Rightarrow (2a + b)(2a - b) + 1(2a + b)$
$\Rightarrow (2a + b)(2a - b + 1)$ Ans
(v) $9x^2 - 4(y + 2x)^2$
$\Rightarrow (3x)^2 - (2(y + 2x))^2$
$\Rightarrow (3x)^2 - (2y + 4x))^2$
$\Rightarrow (3x + \overline{2y + 4x})(3x - \overline{2y + 4x})$
$\Rightarrow (3x + 2y + 4x)(3x - 2y - 4x)$
$\Rightarrow ((3x + 4x) + 2y)((3x - 4x) - 2y)$
$\Rightarrow (7x + 2y)(- x - 2y)$
$\Rightarrow -(7x + 2y)(x - 2y)$ Ans
Q12. Factorise:
(i) $2x^3 + 5x^2y - 12xy^2$
(ii) $16y^3 - 4y$
(iii) $ab - ac - b^2 + bc$
(iv) $5x^2 - 4(x^2 + 1)$
(v) $ab(x^2 + y^2) - xy(a^2 + b^2)^2$
(vi) $9 - 4(x - 1)^2$
Show Answer
(i) $2x^3 + 5x^2y - 12xy^2$
$\Rightarrow 2x^3 + 8x^2y - 3x^2y - 12xy^2$
$\Rightarrow 2x^2(x + 4y) - 3xy(x + 4y)$
$\Rightarrow (x + 4y)(2x^2 - 3xy)$
$\Rightarrow (x + 4y)x(2x - 3y)$
$\Rightarrow x(x + 4y)(2x - 3y)$ Ans
(ii) $16y^3 - 4y$
$\Rightarrow 4y(4y^2 - 1)$
$\Rightarrow 4y((2y)^2 - (1)^2)$
$\Rightarrow 4y(2y + 1)(2y - 1)$
(iii) $ab - ac - b^2 + bc$
$\Rightarrow (ab - ac) - (b^2 + bc)$
$\Rightarrow a(b - c) - b(b - c)$
$\Rightarrow (b - c)(a - b)$ Ans
(iv) $5x^2 - 4(x^2 + 1)$
$\Rightarrow 5x^2 - 4x^2 - 4$
$\Rightarrow x^2 - 4$
$\Rightarrow x^2 - 2^2$
$\Rightarrow (x + 2)(x - 2)$ Ans
(v) $ab(x^2 + y^2) - xy(a^2 + b^2)$
$\Rightarrow abx^2 + aby^2 - a^2xy - b^2xy$
Upon rearranging the equation, we have:
$\Rightarrow abx^2 - a^2xy - b^2xy + aby^2$
$\Rightarrow ax(bx - ay) - by(bx - ay)$
$\Rightarrow (bx - ay)(ax - by)$ Ans
(vi) $9 - 4(x - 1)^2$
$\Rightarrow (3)^2 - (2(x - 1))^2$
$\Rightarrow (3)^2 - (2x + 2)^2$
$\Rightarrow (3 + 2x - 2)(3 - 2x + 2)$
$\Rightarrow (1 + 2x)(5 - 2x)$ Ans
$\Rightarrow 2x^3 + 8x^2y - 3x^2y - 12xy^2$
$\Rightarrow 2x^2(x + 4y) - 3xy(x + 4y)$
$\Rightarrow (x + 4y)(2x^2 - 3xy)$
$\Rightarrow (x + 4y)x(2x - 3y)$
$\Rightarrow x(x + 4y)(2x - 3y)$ Ans
(ii) $16y^3 - 4y$
$\Rightarrow 4y(4y^2 - 1)$
$\Rightarrow 4y((2y)^2 - (1)^2)$
$\Rightarrow 4y(2y + 1)(2y - 1)$
(iii) $ab - ac - b^2 + bc$
$\Rightarrow (ab - ac) - (b^2 + bc)$
$\Rightarrow a(b - c) - b(b - c)$
$\Rightarrow (b - c)(a - b)$ Ans
(iv) $5x^2 - 4(x^2 + 1)$
$\Rightarrow 5x^2 - 4x^2 - 4$
$\Rightarrow x^2 - 4$
$\Rightarrow x^2 - 2^2$
$\Rightarrow (x + 2)(x - 2)$ Ans
(v) $ab(x^2 + y^2) - xy(a^2 + b^2)$
$\Rightarrow abx^2 + aby^2 - a^2xy - b^2xy$
Upon rearranging the equation, we have:
$\Rightarrow abx^2 - a^2xy - b^2xy + aby^2$
$\Rightarrow ax(bx - ay) - by(bx - ay)$
$\Rightarrow (bx - ay)(ax - by)$ Ans
(vi) $9 - 4(x - 1)^2$
$\Rightarrow (3)^2 - (2(x - 1))^2$
$\Rightarrow (3)^2 - (2x + 2)^2$
$\Rightarrow (3 + 2x - 2)(3 - 2x + 2)$
$\Rightarrow (1 + 2x)(5 - 2x)$ Ans
Q13. Factorise:
(i) $c^2 - (c - 3d)^2$
(ii) $2(a + b)^3 - 8a - 8b$
(iii) $25(x - y)^2 - 4(x + 4)^2$
(iv) $a - b - 4a^2 + 4b^2$
(v) $x(x + z) - y(y + z)$
Show Answer
(i) $c^2 - (c - 3d)^2$
$\Rightarrow (c)^2 - (c - 3d)^2$
$\Rightarrow (c + \overline{c - 3d})(c - \overline{c - 3d})$
$\Rightarrow (c + c - 3d)(c - c + 3d)$
$\Rightarrow (2c - 3d)(3d)$
$\Rightarrow 3d(2c - 3d)$ Ans
(ii) $2(a + b)^3 - 8a - 8b$
$\Rightarrow 2(a + b)^3 - 8(a + b)$
$\Rightarrow 2(a + b)(a + b)^2 - (2)^3(a + b)$
$\Rightarrow 2(a + b)(a + b)^2 - (2)^2$
$\Rightarrow 2(a + b)(a + b + 2)(a + b - 2)$ Ans
(iii) $25(x - y)^2 - 4(x + 4)^2$
$\Rightarrow (5(x - y))^2 - (2(x + 4))^2$
$\Rightarrow (5x - 5y)^2 - (2x + 8)^2$
$\Rightarrow (5x - 5y + \overline{2x + 8})(5x - 5y - \overline{2x + 8})$
$\Rightarrow (5x - 5y + 2x + 8)(5x - 5y - 2x - 8)$
$\Rightarrow (7x - 5y + 8)(3x - 5y - 8)$ Ans
(iv) $a - b - 4a^2 + 4b^2$
$\Rightarrow (a - b) - 4(a^2 + b^2)$
$\Rightarrow (a - b) - 4(a + b)^2$
$\Rightarrow (a - b) - [4(a + b)(a - b)]$
$\Rightarrow (a - b) - [(a - b)(4a + 4b)]$
$\Rightarrow (a - b) [1 - (4a + 4b)]$
$\Rightarrow (a - b) [1 - 4a + 4b]$ Ans
(v) $x(x + z) - y(y + z)$
$\Rightarrow x^2 + xz - y^2 - yz$
Upon rearranging the equation, we have:
$\Rightarrow (x^2 - y^2) + (xz - yz)$
$\Rightarrow (x - y)^2 + z(x - y)$
$\Rightarrow (x + y)(x - y) + z(x - y)$
$\Rightarrow (x - y)(x + y + z)$ Ans
$\Rightarrow (c)^2 - (c - 3d)^2$
$\Rightarrow (c + \overline{c - 3d})(c - \overline{c - 3d})$
$\Rightarrow (c + c - 3d)(c - c + 3d)$
$\Rightarrow (2c - 3d)(3d)$
$\Rightarrow 3d(2c - 3d)$ Ans
(ii) $2(a + b)^3 - 8a - 8b$
$\Rightarrow 2(a + b)^3 - 8(a + b)$
$\Rightarrow 2(a + b)(a + b)^2 - (2)^3(a + b)$
$\Rightarrow 2(a + b)(a + b)^2 - (2)^2$
$\Rightarrow 2(a + b)(a + b + 2)(a + b - 2)$ Ans
(iii) $25(x - y)^2 - 4(x + 4)^2$
$\Rightarrow (5(x - y))^2 - (2(x + 4))^2$
$\Rightarrow (5x - 5y)^2 - (2x + 8)^2$
$\Rightarrow (5x - 5y + \overline{2x + 8})(5x - 5y - \overline{2x + 8})$
$\Rightarrow (5x - 5y + 2x + 8)(5x - 5y - 2x - 8)$
$\Rightarrow (7x - 5y + 8)(3x - 5y - 8)$ Ans
(iv) $a - b - 4a^2 + 4b^2$
$\Rightarrow (a - b) - 4(a^2 + b^2)$
$\Rightarrow (a - b) - 4(a + b)^2$
$\Rightarrow (a - b) - [4(a + b)(a - b)]$
$\Rightarrow (a - b) - [(a - b)(4a + 4b)]$
$\Rightarrow (a - b) [1 - (4a + 4b)]$
$\Rightarrow (a - b) [1 - 4a + 4b]$ Ans
(v) $x(x + z) - y(y + z)$
$\Rightarrow x^2 + xz - y^2 - yz$
Upon rearranging the equation, we have:
$\Rightarrow (x^2 - y^2) + (xz - yz)$
$\Rightarrow (x - y)^2 + z(x - y)$
$\Rightarrow (x + y)(x - y) + z(x - y)$
$\Rightarrow (x - y)(x + y + z)$ Ans
Q14. Factorise:
(i) $9x^2 - 4a^2 + 4ay - y^2$
(ii) $(x + y)^2 - 5(x + y) - 6$
(iii) $a^2 + 2ab + b^2 - c^2$
(iv) $x^2 - y^2 + 2yz - z^2$
(v) $9 - x^2 + 2xy - y^2$
Show Answer
(i) $9x^2 - 4a^2 + 4ay - y^2$
$\because (4a^2 + 4ay - y^2) = (2a - y)^2$
$\therefore (3x)^2 - (2a - y)^2$
$\Rightarrow (3x + \overline{2a - y})(3x - \overline{2a - y})$
$\Rightarrow (3x + 2a - y)(3x - 2a + y)$ Ans
(ii) $(x + y)^2 - 5(x + y) - 6$
Taking $(x + y) = x$, we have:
$\therefore x^2 - 5x - 6$
$\Rightarrow x^2 - 6x + x - 6$
$\Rightarrow (x^2 - 6x) + (x - 6)$
$\Rightarrow x(x - 6) + 1(x - 6)$
$\Rightarrow (x - 6)(x + 1)$
$\therefore (x + y - 6)(x + y + 1)$ Ans
(iii) $a^2 + 2ab + b^2 - c^2$
$\because a^2 + 2ab + b^2 = (a + b)^2$
$\therefore (a + b)^2 - c^2$
$\Rightarrow (a + b + c)(a + b - c)$ Ans
(iv) $x^2 - y^2 + 2yz - z^2$
$\because y^2 + 2yz - z^2 = (y - z)^2$
$\therefore x^2 - (y - z)^2$
$\Rightarrow (x + \overline{y - z})(x - \overline{y - z})$
$\Rightarrow (x + y - z)(x - y + z)$ Ans
(v) $9 - x^2 + 2xy - y^2$
$\because x^2 + 2xy - y^2 = (x - y)^2$
$\therefore 3^2 - (x - y)^2$
$\Rightarrow (3 + \overline{x - y})(3 - \overline{x - y})$
$\Rightarrow (3 + x - y)(3 - x + y)$ Ans
$\because (4a^2 + 4ay - y^2) = (2a - y)^2$
$\therefore (3x)^2 - (2a - y)^2$
$\Rightarrow (3x + \overline{2a - y})(3x - \overline{2a - y})$
$\Rightarrow (3x + 2a - y)(3x - 2a + y)$ Ans
(ii) $(x + y)^2 - 5(x + y) - 6$
Taking $(x + y) = x$, we have:
$\therefore x^2 - 5x - 6$
$\Rightarrow x^2 - 6x + x - 6$
$\Rightarrow (x^2 - 6x) + (x - 6)$
$\Rightarrow x(x - 6) + 1(x - 6)$
$\Rightarrow (x - 6)(x + 1)$
$\therefore (x + y - 6)(x + y + 1)$ Ans
(iii) $a^2 + 2ab + b^2 - c^2$
$\because a^2 + 2ab + b^2 = (a + b)^2$
$\therefore (a + b)^2 - c^2$
$\Rightarrow (a + b + c)(a + b - c)$ Ans
(iv) $x^2 - y^2 + 2yz - z^2$
$\because y^2 + 2yz - z^2 = (y - z)^2$
$\therefore x^2 - (y - z)^2$
$\Rightarrow (x + \overline{y - z})(x - \overline{y - z})$
$\Rightarrow (x + y - z)(x - y + z)$ Ans
(v) $9 - x^2 + 2xy - y^2$
$\because x^2 + 2xy - y^2 = (x - y)^2$
$\therefore 3^2 - (x - y)^2$
$\Rightarrow (3 + \overline{x - y})(3 - \overline{x - y})$
$\Rightarrow (3 + x - y)(3 - x + y)$ Ans
Q15. Factorise:
(i) $(2a - 3)^2 - 2(2a - 3)(a - 1) + (a - 1)^2$
(ii) $a^2 - 2ab + b^2 - a + b$
(iii) $(a^2 - 3a)(a^2 - 3a + 7) + 10$
(iv) $(a^2 - a)(4a^2 - 4a - 5) - 6$
(v) $5 - (3a^2 - 2a)(6 - 3a^2 + 2a)$
Show Answer
(i) $(2a - 3)^2 - 2(2a - 3)(a - 1) + (a - 1)^2$
Taking $(2a - 3) = x$ and $(a - 1) = y$, we have:
$\Rightarrow x^2 - 2xy + y^2 = (x - y)^2$
$\Rightarrow (2a - 3 - \overline{a - 1})^2$
$\Rightarrow (2a - 3 - a + 1)^2$
$\Rightarrow (a - 2)^2$ Ans
(ii) $a^2 - 2ab + b^2 - a + b$
$\because a^2 - 2ab + b^2 = (a - b)^2$
$\therefore (a - b)^2 - (a + b)$
$\Rightarrow (a - b)(a - b) - 1(a - b)$
$\Rightarrow (a - b)(a - b - 1)$ Ans
(iii) $(a^2 - 3a)(a^2 - 3a + 7) + 10$
Taking $a^2 - 3a = x$, we have:
$\therefore x(x + 7) + 10$
$\Rightarrow x^2 + 7x + 10$
$\Rightarrow x^2 + 5x + 2x + 10$
$\Rightarrow (x^2 + 5x) + (2x + 10)$
$\Rightarrow x(x + 5) + 2(x + 5)$
$\Rightarrow (a^2 - 3a + 5)(a^2 - 3a + 2)$
$\Rightarrow (a^2 - 3a + 5)(a^2 - 2a - a + 2)$
$\Rightarrow (a^2 - 3a + 5)((a^2 - 2a) - (a + 2))$
$\Rightarrow (a^2 - 3a + 5)(a(a - 2) + 1(a - 2))$
$\Rightarrow (a^2 - 3a + 5)(a - 2)(a + 1)$ Ans
(iv) $(a^2 - a)(4a^2 - 4a - 5) - 6$
$\Rightarrow (a^2 - a)[4(a^2 - a) - 5] - 6$
Taking $a^2 - a = x$, we have:
$\Rightarrow x(4x - 5) - 6$
$\Rightarrow 4x^2 - 5x - 6$
$\Rightarrow 4x^2 - 8x + 3x - 6$
$\Rightarrow (4x^2 - 8x) + (3x - 6)$
$\Rightarrow 4x(x - 2) + 3(x - 2)$
$\Rightarrow (x - 2)(4x + 3)$
$\therefore (a^2 - a - 2)(4(a^2 - a) + 3)$
$\Rightarrow (a^2 - a - 2)(4a^2 - 4a + 3)$
$\Rightarrow [a^2 - 2a + a - 2](4a^2 - 4a + 3)$
$\Rightarrow [(a^2 - 2a) + (a - 2)](4a^2 - 4a + 3)$
$\Rightarrow [a(a - 2) + 1(a - 2)](4a^2 - 4a + 3)$
$\Rightarrow (a - 2)(a + 1)(4a^2 - 4a + 3)$ Ans
(v) $5 - (3a^2 - 2a)(6 - 3a^2 + 2a)$
$\Rightarrow 5 - (3a^2 - 2a)(6 - \overline{3a^2 + 2a})$
$\Rightarrow 5 - (3a^2 - 2a)(6 - 3a^2 - 2a)$
Taking $3a^2 - 2a = x$, we have:
$\Rightarrow 5 - x(6 - x)$
$\Rightarrow 5 - 6x + x^2$
$\Rightarrow 5 - 5x - x + x^2$
$\Rightarrow (5 - 5x) - (x + x^2)$
$\Rightarrow 5(1 - x) - x(1 - x)$
$\Rightarrow (1 - x)(5 - x)$
Now substituting $x = 3a^2 - 2a$:
$\Rightarrow (1 - 3a^2 - 2a)(5 - 3a^2 - 2a)$
Upon rearranging the polynomials, we have:
$\Rightarrow [1 - a + 3a - 3a^2][5 + 5a - 3a - 3a^2]$
$\Rightarrow [(1 - a) + (3a - 3a^2)][(5 + 5a) - (3a - 3a^2)]$
$\Rightarrow [1(1 - a) + 3a(1 - a)][5(1 + a) - 3a(1 + a)]$
$\therefore (1 - a)(1 + 3a)(1 + a)(5 - 3a)$ Ans
Taking $(2a - 3) = x$ and $(a - 1) = y$, we have:
$\Rightarrow x^2 - 2xy + y^2 = (x - y)^2$
$\Rightarrow (2a - 3 - \overline{a - 1})^2$
$\Rightarrow (2a - 3 - a + 1)^2$
$\Rightarrow (a - 2)^2$ Ans
(ii) $a^2 - 2ab + b^2 - a + b$
$\because a^2 - 2ab + b^2 = (a - b)^2$
$\therefore (a - b)^2 - (a + b)$
$\Rightarrow (a - b)(a - b) - 1(a - b)$
$\Rightarrow (a - b)(a - b - 1)$ Ans
(iii) $(a^2 - 3a)(a^2 - 3a + 7) + 10$
Taking $a^2 - 3a = x$, we have:
$\therefore x(x + 7) + 10$
$\Rightarrow x^2 + 7x + 10$
$\Rightarrow x^2 + 5x + 2x + 10$
$\Rightarrow (x^2 + 5x) + (2x + 10)$
$\Rightarrow x(x + 5) + 2(x + 5)$
$\Rightarrow (a^2 - 3a + 5)(a^2 - 3a + 2)$
$\Rightarrow (a^2 - 3a + 5)(a^2 - 2a - a + 2)$
$\Rightarrow (a^2 - 3a + 5)((a^2 - 2a) - (a + 2))$
$\Rightarrow (a^2 - 3a + 5)(a(a - 2) + 1(a - 2))$
$\Rightarrow (a^2 - 3a + 5)(a - 2)(a + 1)$ Ans
(iv) $(a^2 - a)(4a^2 - 4a - 5) - 6$
$\Rightarrow (a^2 - a)[4(a^2 - a) - 5] - 6$
Taking $a^2 - a = x$, we have:
$\Rightarrow x(4x - 5) - 6$
$\Rightarrow 4x^2 - 5x - 6$
$\Rightarrow 4x^2 - 8x + 3x - 6$
$\Rightarrow (4x^2 - 8x) + (3x - 6)$
$\Rightarrow 4x(x - 2) + 3(x - 2)$
$\Rightarrow (x - 2)(4x + 3)$
$\therefore (a^2 - a - 2)(4(a^2 - a) + 3)$
$\Rightarrow (a^2 - a - 2)(4a^2 - 4a + 3)$
$\Rightarrow [a^2 - 2a + a - 2](4a^2 - 4a + 3)$
$\Rightarrow [(a^2 - 2a) + (a - 2)](4a^2 - 4a + 3)$
$\Rightarrow [a(a - 2) + 1(a - 2)](4a^2 - 4a + 3)$
$\Rightarrow (a - 2)(a + 1)(4a^2 - 4a + 3)$ Ans
(v) $5 - (3a^2 - 2a)(6 - 3a^2 + 2a)$
$\Rightarrow 5 - (3a^2 - 2a)(6 - \overline{3a^2 + 2a})$
$\Rightarrow 5 - (3a^2 - 2a)(6 - 3a^2 - 2a)$
Taking $3a^2 - 2a = x$, we have:
$\Rightarrow 5 - x(6 - x)$
$\Rightarrow 5 - 6x + x^2$
$\Rightarrow 5 - 5x - x + x^2$
$\Rightarrow (5 - 5x) - (x + x^2)$
$\Rightarrow 5(1 - x) - x(1 - x)$
$\Rightarrow (1 - x)(5 - x)$
Now substituting $x = 3a^2 - 2a$:
$\Rightarrow (1 - 3a^2 - 2a)(5 - 3a^2 - 2a)$
Upon rearranging the polynomials, we have:
$\Rightarrow [1 - a + 3a - 3a^2][5 + 5a - 3a - 3a^2]$
$\Rightarrow [(1 - a) + (3a - 3a^2)][(5 + 5a) - (3a - 3a^2)]$
$\Rightarrow [1(1 - a) + 3a(1 - a)][5(1 + a) - 3a(1 + a)]$
$\therefore (1 - a)(1 + 3a)(1 + a)(5 - 3a)$ Ans