Exercise 3.3: Factorization
1. $ab(x – y) – bc(x – y)$
Solution: $ab(x – y) – bc(x – y)$
$= (x – y)(ab – bc)$
$= b(x – y)(a – c)$
Answer:$b(x – y)(a – c)$
2. $9x^2 + 24x + 16$
Solution: $9x^2 + 24x + 16$
$= (3x)^2 + 2 \cdot (3x) \cdot 4 + (4)^2$
$= (3x + 4)^2$
$= (3x + 4)(3x + 4)$
Answer:$(3x + 4)(3x + 4)$
3. $a^4 – 27a^2 + 1$
Solution: $a^4 – 27a^2 + 1$
$= (a^2)^2 – 2 \cdot a^2 \cdot 1 + (1)^2 – 25a^2$
$= (a^2 – 1)^2 – (5a)^2$
$= (a^2 – 1 + 5a)(a^2 – 1 – 5a)$
$= (a^2 + 5a – 1)(a^2 – 5a – 1)$
Answer:$(a^2 + 5a – 1)(a^2 – 5a – 1)$
4. $x^4 – 6x^2y^2 + y^4$
Solution: $x^4 – 6x^2y^2 + y^4$
$= (x^2)^2 – 2 \cdot x^2 \cdot y^2 + (y^2)^2 – 4x^2y^2$
$= (x^2 – y^2)^2 – (2xy)^2$
$= (x^2 – y^2 + 2xy)(x^2 – y^2 – 2xy)$
$= (x^2 + 2xy – y^2)(x^2 – 2xy – y^2)$
Answer:$(x^2 + 2xy – y^2)(x^2 – 2xy – y^2)$
5. $(a^2 – b^2)(x^2 – y^2) + 4abxy$
Solution: $(a^2 – b^2)(x^2 – y^2) + 4abxy$
$= a^2x^2 – a^2y^2 – b^2x^2 + b^2y^2 + 2abxy + 2abxy$
$= (a^2x^2 + 2abxy + b^2y^2) – (a^2y^2 – 2abxy + b^2x^2)$
$= (ax + by)^2 – (ay – bx)^2$
$= \{(ax + by) + (ay – bx)\}\{(ax + by) – (ay – bx)\}$
$= (ax + by + ay – bx)(ax + by – ay + bx)$
Answer:$(ax – bx + ay + by)(ax + bx – ay + by)$
6. $4a^2 – 12ab + 9b^2 – 4c^2$
Solution: $4a^2 – 12ab + 9b^2 – 4c^2$
$= (2a)^2 – 2 \cdot 2a \cdot 3b + (3b)^2 – (2c)^2$
$= (2a – 3b)^2 – (2c)^2$
$= (2a – 3b + 2c)(2a – 3b – 2c)$
Answer:$(2a – 3b + 2c)(2a – 3b – 2c)$
7. $a^2 + 6a + 8 – y^2 + 2y$
Solution: $a^2 + 6a + 8 – y^2 + 2y$
$= a^2 + 6a + 9 – 1 – y^2 + 2y$
$= (a^2 + 6a + 9) – (y^2 – 2y + 1)$
$= (a + 3)^2 – (y – 1)^2$
$= \{(a + 3) + (y – 1)\}\{(a + 3) – (y – 1)\}$
$= (a + y + 2)(a – y + 4)$
Answer:$(a + y + 2)(a – y + 4)$
8. $16x^2 – 25y^2 – 8xz + 10yz$
Solution: $16x^2 – 25y^2 – 8xz + 10yz$
$= (4x)^2 – (5y)^2 – 2z(4x – 5y)$
$= (4x + 5y)(4x – 5y) – 2z(4x – 5y)$
$= (4x – 5y)(4x + 5y – 2z)$
Answer:$(4x – 5y)(4x + 5y – 2z)$
9. $x^2 + 13x + 36$
Solution: $x^2 + 13x + 36$
$= x^2 + 9x + 4x + 36$
$= x(x + 9) + 4(x + 9)$
$= (x + 9)(x + 4)$
Answer:$(x + 9)(x + 4)$
10. $x^4 + x^2 – 20$
Solution: $x^4 + x^2 – 20$
$= x^4 + 5x^2 – 4x^2 – 20$
$= x^2(x^2 + 5) – 4(x^2 + 5)$
$= (x^2 + 5)(x^2 – 4)$
$= (x^2 + 5)(x + 2)(x – 2)$
Answer:$(x^2 + 5)(x + 2)(x – 2)$
11. $a^2 – 30a + 216$
Solution: $a^2 – 30a + 216$
$= a^2 – 18a – 12a + 216$
$= a(a – 18) – 12(a – 18)$
$= (a – 18)(a – 12)$
Answer:$(a – 18)(a – 12)$
12. $a^8 – a^4 – 2$
Solution: $a^8 – a^4 – 2$
$= a^8 – 2a^4 + a^4 – 2$
$= a^4(a^4 – 2) + 1(a^4 – 2)$
$= (a^4 – 2)(a^4 + 1)$
Answer:$(a^4 + 1)(a^4 – 2)$
13. $x^2 – 37x – 650$
Solution: $x^2 – 37x – 650$
$= x^2 – 50x + 13x – 650$
$= x(x – 50) + 13(x – 50)$
$= (x – 50)(x + 13)$
Answer:$(x – 50)(x + 13)$
14. $9x^2y^2 – 5xy^2 – 14y^2$
Solution: $9x^2y^2 – 5xy^2 – 14y^2$
$= y^2(9x^2 – 5x – 14)$
$= y^2(9x^2 – 14x + 9x – 14)$
$= y^2\{x(9x – 14) + 1(9x – 14)\}$
$= y^2(9x – 14)(x + 1)$
Answer:$y^2(x + 1)(9x – 14)$
15. $4x^4 – 27x^2 – 81$
Solution: $4x^4 – 27x^2 – 81$
$= 4x^4 – 36x^2 + 9x^2 – 81$
$= 4x^2(x^2 – 9) + 9(x^2 – 9)$
$= (x^2 – 9)(4x^2 + 9)$
$= (x + 3)(x – 3)(4x^2 + 9)$
Answer:$(4x^2 + 9)(x + 3)(x – 3)$
16. $ax^2 + (a^2 + 1)x + a$
Solution: $ax^2 + (a^2 + 1)x + a$
$= ax^2 + a^2x + x + a$
$= ax(x + a) + 1(x + a)$
$= (x + a)(ax + 1)$
Answer:$(x + a)(ax + 1)$
17. $3(a^2 + 2a)^2 – 22(a^2 + 2a) + 40$
Solution: $3(a^2 + 2a)^2 – 22(a^2 + 2a) + 40$
Let $a^2 + 2a = x$
Given expression $= 3x^2 – 22x + 40$
$= 3x^2 – 12x – 10x + 40$
$= 3x(x – 4) – 10(x – 4)$
$= (x – 4)(3x – 10)$
$= (a^2 + 2a – 4)\{3(a^2 + 2a) – 10\}$ [Substituting $x = a^2 + 2a$]
$= (a^2 + 2a – 4)(3a^2 + 6a – 10)$
Answer:$(a^2 + 2a – 4)(3a^2 + 6a – 10)$
18. $(a – 1)x^2 + a^2xy + (a + 1)y^2$
Solution: $(a – 1)x^2 + a^2xy + (a + 1)y^2$
Let $a – 1 = p$ and $a + 1 = q$
Multiplying, $pq = (a – 1)(a + 1) = a^2 – 1 \implies a^2 = pq + 1$
Given expression $= px^2 + (pq + 1)xy + qy^2$
$= px^2 + pqxy + xy + qy^2$
$= px(x + qy) + y(x + qy)$
$= (x + qy)(px + y)$
$= \{x + (a + 1)y\}\{(a – 1)x + y\}$ [Substituting values of $p$ and $q$]
$= (x + ay + y)(ax – x + y)$
Answer:$(x + ay + y)(ax – x + y)$
19. $x^3 + 3x^2 + 3x + 2$
Solution: $x^3 + 3x^2 + 3x + 2$
$= x^3 + 3 \cdot x^2 \cdot 1 + 3 \cdot x \cdot 1^2 + 1^3 + 1$
$= (x + 1)^3 + 1^3$
$= \{(x + 1) + 1\}\{(x + 1)^2 – (x + 1) \cdot 1 + 1^2\}$
$= (x + 2)(x^2 + 2x + 1 – x – 1 + 1)$
$= (x + 2)(x^2 + x + 1)$
Answer:$(x + 2)(x^2 + x + 1)$
20. $a^3 – 6a^2 + 12a – 9$
Solution: $a^3 – 6a^2 + 12a – 9$
$= a^3 – 3 \cdot a^2 \cdot 2 + 3 \cdot a \cdot 2^2 – 2^3 – 1$
$= (a – 2)^3 – 1^3$
$= \{(a – 2) – 1\}\{(a – 2)^2 + (a – 2) \cdot 1 + 1^2\}$
$= (a – 3)(a^2 – 4a + 4 + a – 2 + 1)$
$= (a – 3)(a^2 – 3a + 3)$
Answer:$(a – 3)(a^2 – 3a + 3)$
21. $a^3 – 9b^3 + (a + b)^3$
Solution: $a^3 – 9b^3 + (a + b)^3$
$= a^3 – b^3 – 8b^3 + (a + b)^3$
$= (a^3 – b^3) + \{(a + b)^3 – (2b)^3\}$
$= (a – b)(a^2 + ab + b^2) + \{(a + b) – 2b\}\{(a + b)^2 + (a + b)(2b) + (2b)^2\}$
$= (a – b)(a^2 + ab + b^2) + (a – b)(a^2 + 2ab + b^2 + 2ab + 2b^2 + 4b^2)$
$= (a – b)(a^2 + ab + b^2) + (a – b)(a^2 + 4ab + 7b^2)$
$= (a – b)(a^2 + ab + b^2 + a^2 + 4ab + 7b^2)$
$= (a – b)(2a^2 + 5ab + 8b^2)$
Answer:$(a – b)(2a^2 + 5ab + 8b^2)$
22. $8x^3 + 12x^2 + 6x – 63$
Solution: $8x^3 + 12x^2 + 6x – 63$
$= (2x)^3 + 3 \cdot (2x)^2 \cdot 1 + 3 \cdot (2x) \cdot 1^2 + 1^3 – 64$
$= (2x + 1)^3 – (4)^3$
$= \{(2x + 1) – 4\}\{(2x + 1)^2 + (2x + 1) \cdot 4 + 4^2\}$
$= (2x – 3)(4x^2 + 4x + 1 + 8x + 4 + 16)$
$= (2x – 3)(4x^2 + 12x + 21)$
Answer:$(2x – 3)(4x^2 + 12x + 21)$
23. $8a^3 + \frac{b^3}{27}$
Solution: $8a^3 + \frac{b^3}{27}$
$= (2a)^3 + \left(\frac{b}{3}\right)^3$
$= \left(2a + \frac{b}{3}\right)\left\{(2a)^2 – (2a)\left(\frac{b}{3}\right) + \left(\frac{b}{3}\right)^2\right\}$
$= \left(2a + \frac{b}{3}\right)\left(4a^2 – \frac{2ab}{3} + \frac{b^2}{9}\right)$
Answer:$\left(2a + \frac{b}{3}\right)\left(4a^2 – \frac{2ab}{3} + \frac{b^2}{9}\right)$
24. $\frac{a^6}{27} – b^6$
Solution: $\frac{a^6}{27} – b^6$
$= \left(\frac{a^2}{3}\right)^3 – (b^2)^3$
$= \left(\frac{a^2}{3} – b^2\right)\left\{\left(\frac{a^2}{3}\right)^2 + \left(\frac{a^2}{3}\right)(b^2) + (b^2)^2\right\}$
$= \left(\frac{a^2}{3} – b^2\right)\left(\frac{a^4}{9} + \frac{a^2b^2}{3} + b^4\right)$
Answer:$\left(\frac{a^2}{3} – b^2\right)\left(\frac{a^4}{9} + \frac{a^2b^2}{3} + b^4\right)$
25. $4a^2 + \frac{1}{4a^2} – 2 + 4a – \frac{1}{a}$
Solution: $4a^2 + \frac{1}{4a^2} – 2 + 4a – \frac{1}{a}$
$= (2a)^2 – 2 \cdot 2a \cdot \frac{1}{2a} + \left(\frac{1}{2a}\right)^2 + 2\left(2a – \frac{1}{2a}\right)$
$= \left(2a – \frac{1}{2a}\right)^2 + 2\left(2a – \frac{1}{2a}\right)$
$= \left(2a – \frac{1}{2a}\right)\left(2a – \frac{1}{2a} + 2\right)$
Answer:$\left(2a – \frac{1}{2a}\right)\left(2a – \frac{1}{2a} + 2\right)$
26. $(3a + 1)^3 – (2a – 3)^3$
Solution: $(3a + 1)^3 – (2a – 3)^3$
$= \{(3a + 1) – (2a – 3)\}\{(3a + 1)^2 + (3a + 1)(2a – 3) + (2a – 3)^2\}$
$= (3a + 1 – 2a + 3)\{(9a^2 + 6a + 1) + (6a^2 – 9a + 2a – 3) + (4a^2 – 12a + 9)\}$
$= (a + 4)(9a^2 + 6a + 1 + 6a^2 – 7a – 3 + 4a^2 – 12a + 9)$
$= (a + 4)(19a^2 – 13a + 7)$
Answer:$(a + 4)(19a^2 – 13a + 7)$
27. $(x + 2)(x + 3)(x + 4)(x + 5) – 48$
Solution: $(x + 2)(x + 3)(x + 4)(x + 5) – 48$
$= \{(x + 2)(x + 5)\}\{(x + 3)(x + 4)\} – 48$
$= (x^2 + 7x + 10)(x^2 + 7x + 12) – 48$
$= (a + 10)(a + 12) – 48$ [Let $x^2 + 7x = a$]
$= a^2 + 22a + 120 – 48$
$= a^2 + 22a + 72$
$= a^2 + 18a + 4a + 72$
$= a(a + 18) + 4(a + 18)$
$= (a + 18)(a + 4)$
$= (x^2 + 7x + 18)(x^2 + 7x + 4)$ [Substituting $a = x^2 + 7x$]
Answer:$(x^2 + 7x + 18)(x^2 + 7x + 4)$
28. $(x – 1)(x – 3)(x – 5)(x – 7) – 65$
Solution: $(x – 1)(x – 3)(x – 5)(x – 7) – 65$
$= \{(x – 1)(x – 7)\}\{(x – 3)(x – 5)\} – 65$
$= (x^2 – 8x + 7)(x^2 – 8x + 15) – 65$
$= (a + 7)(a + 15) – 65$ [Let $x^2 – 8x = a$]
$= a^2 + 22a + 105 – 65$
$= a^2 + 22a + 40$
$= a^2 + 20a + 2a + 40$
$= a(a + 20) + 2(a + 20)$
$= (a + 20)(a + 2)$
$= (x^2 – 8x + 20)(x^2 – 8x + 2)$ [Substituting $a = x^2 – 8x$]
Answer:$(x^2 – 8x + 20)(x^2 – 8x + 2)$
29. $2b^2c^2 + 2c^2a^2 + 2a^2b^2 – a^4 – b^4 – c^4$
Solution: $2b^2c^2 + 2c^2a^2 + 2a^2b^2 – a^4 – b^4 – c^4$
$= 4a^2b^2 – (a^4 + b^4 + c^4 + 2a^2b^2 – 2b^2c^2 – 2c^2a^2)$
$= (2ab)^2 – (a^2 + b^2 – c^2)^2$
$= (2ab + a^2 + b^2 – c^2)(2ab – a^2 – b^2 + c^2)$
$= \{(a^2 + 2ab + b^2) – c^2\}\{c^2 – (a^2 – 2ab + b^2)\}$
$= \{(a + b)^2 – c^2\}\{c^2 – (a – b)^2\}$
$= (a + b + c)(a + b – c)(c + a – b)(c – a + b)$
$= (a + b + c)(a + b – c)(a – b + c)(b + c – a)$
Answer:$(a + b + c)(a + b – c)(a – b + c)(b + c – a)$
30. $14(x + z)^2 – 29(x + z)(x + 1) – 15(x + 1)^2$
Solution: $14(x + z)^2 – 29(x + z)(x + 1) – 15(x + 1)^2$
Let $x + z = a$ and $x + 1 = b$
Given expression $= 14a^2 – 29ab – 15b^2$
$= 14a^2 – 35ab + 6ab – 15b^2$
$= 7a(2a – 5b) + 3b(2a – 5b)$
$= (2a – 5b)(7a + 3b)$
$= \{2(x + z) – 5(x + 1)\}\{7(x + z) + 3(x + 1)\}$ [Substituting values of $a$ and $b$]
$= (2x + 2z – 5x – 5)(7x + 7z + 3x + 3)$
$= (2z – 3x – 5)(10x + 7z + 3)$
Answer:$(2z – 3x – 5)(10x + 7z + 3)$
31. Show that, $(x + 1)(x + 2)(3x – 1)(3x – 4) = (3x^2 + 2x – 1)(3x^2 + 2x – 8)$
Solution: L.H.S. $= (x + 1)(x + 2)(3x – 1)(3x – 4)$
$= \{(x + 1)(3x – 1)\}\{(x + 2)(3x – 4)\}$
$= (3x^2 – x + 3x – 1)(3x^2 – 4x + 6x – 8)$
$= (3x^2 + 2x – 1)(3x^2 + 2x – 8) = \text{R.H.S.}$ (Shown)