Type 1: Factorization by Taking Out Common Factor (20 Problems)

1. 12x+18y12x+18y
2. 15a2b−25ab215a2b−25ab2
3. 8p3q+12pq28p3q+12pq2
4. 21mn−35m2n21mn−35m2n
5. 9x2y3−27xy49x2y3−27xy4
6. 4ab2+12a2b4ab2+12a2b
7. 16x4y−24xy316x4y−24xy3
8. 10pq2r+15p2qr10pq2r+15p2qr
9. 18a3b2−27a2b318a3b2−27a2b3
10. 20xy3z−30x2yz220xy3z−30x2yz2
11. 7m2n+14mn27m2n+14mn2
12. 24p2q−36pq224p2q−36pq2
13. 32a3b4+48a4b332a3b4+48a4b3
14. 5xy2z3−10x2yz45xy2z3−10x2yz4
15. 6pqr2+9p2q2r6pqr2+9p2q2r
16. 14ab3−21a2b214ab3−21a2b2
17. 28x2yz+42xy2z228x2yz+42xy2z2
18. 9mn4−18m2n39mn4−18m2n3
19. 12p3qr−18pq2r212p3qr−18pq2r2
20. 25a4b2+50a2b325a4b2+50a2b3

Type 2: Factorization by Grouping Terms (20 Problems)

1. ax+ay+bx+byax+ay+bx+by
2. 2×2+3xy+4x+6y2x2+3xy+4x+6y
3. p2+pq+pr+q2+qr+r2p2+pq+pr+q2+qr+r2
4. 3m2+5mn+2m+10n3m2+5mn+2m+10n
5. a2b+ab2+ac+bca2b+ab2+ac+bc
6. 4x2y+6xy2+2xz+3yz4x2y+6xy2+2xz+3yz
7. 5pq+7pr+5q2+7qr5pq+7pr+5q2+7qr
8. x3+x2y+xz+yzx3+x2y+xz+yz
9. 2a2b+3ab2+4ac+6bc2a2b+3ab2+4ac+6bc
10. mn2+m2n+pn+pmmn2+m2n+pn+pm
11. 7xy+7xz+7y2+7yz7xy+7xz+7y2+7yz
12. ab2+a2b+cb+acab2+a2b+cb+ac
13. 3p2q+5pq2+2pr+10qr3p2q+5pq2+2pr+10qr
14. x2y+xy2+x2z+xyzx2y+xy2+x2z+xyz
15. 4mn+6mp+4n2+6np4mn+6mp+4n2+6np
16. a3+a2b+ac+bca3+a2b+ac+bc
17. 2x2y+3xy2+xz+yz2x2y+3xy2+xz+yz
18. pqr+pqs+pt+qtpqr+pqs+pt+qt
19. 5ab2+7a2b+3ac+21bc5ab2+7a2b+3ac+21bc
20. xy2+x2y+xz2+yz2xy2+x2y+xz2+yz2

Type 3: Factorization Using Standard Identities (20 Problems)

1. x2−9x2−9
2. a2+10a+25a2+10a+25
3. 16b2−2516b2−25
4. p2−14p+49p2−14p+49
5. 9q2+6q+19q2+6q+1
6. 4×2−36y24x2−36y2
7. 25m2+30m+925m2+30m+9
8. 49n2−6449n2−64
9. y2−2y+1y2−2y+1
10. 81z2−10081z2−100
11. a2+8a+16a2+8a+16
12. 36b2c2−436b2c2−4
13. p2−6p+9p2−6p+9
14. 16q2r2+24qr+916q2r2+24qr+9
15. 25×2−125x2−1
16. 4y4−169z24y4−169z2
17. m2+2m+1m2+2m+1
18. 9a2b2−4c29a2b2−4c2
19. 64p2−81q264p2−81q2
20. 100r2+60r+9100r2+60r+9

Type 4: Factorization of Quadratic Expressions (Middle-Term Splitting) (20 Problems)

1. x2+7x+12x2+7x+12
2. y2+8y+15y2+8y+15
3. z2−5z+6z2−5z+6
4. a2+9a+20a2+9a+20
5. b2−7b+12b2−7b+12
6. 2×2+5x+32x2+5x+3
7. 3y2−10y+83y2−10y+8
8. p2+11p+24p2+11p+24
9. q2−13q+36q2−13q+36
10. 4m2+12m+94m2+12m+9
11. 2n2−7n+32n2−7n+3
12. 3r2+14r+153r2+14r+15
13. s2−8s+12s2−8s+12
14. 6t2+7t−36t2+7t−3
15. 2u2+11u+122u2+11u+12
16. 5v2−16v+125v2−16v+12
17. 4w2+4w−34w2+4w−3
18. 3×2−5x−23x2−5x−2
19. 2y2+7y+32y2+7y+3
20. 6z2−13z+66z2−13z+6

Type 5: Division of Algebraic Expressions via Factorization (20 Problems)

1. 12x+186612x+18
2. 15a2−25a5a5a15a2−25a
3. x2−9x−3x−3x2−9
4. y2+7y+10y+5y+5y2+7y+10
5. 10x−255510x−25
6. m2−14m−32m+2m+2m2−14m−32
7. 5p2−25p+20p−1p−15p2−25p+20
8. 10y(6y+21)5(2y+7)5(2y+7)10y(6y+21)
9. 9x2y2(3z−24)27xy(z−8)27xy(z−8)9x2y2(3z−24)
10. 96abc(3a−12)(5b−30)144(a−4)(b−6)144(a−4)(b−6)96abc(3a−12)(5b−30)
11. x2+5xxxx2+5x
12. 4p2−162p−42p−44p2−16
13. q2−49q−7q−7q2−49
14. 2r2+8r+82r+42r+42r2+8r+8
15. 3s2−12s3s3s3s2−12s
16. t2+6t+9t+3t+3t2+6t+9
17. 18u2−24u6u6u18u2−24u
18. v2−25v+5v+5v2−25
19. 7w2+21w7w7w7w2+21w
20. 8×2−324x−84x−88x2−32

Analytical Factorization Questions

1. Prove the identity: Show algebraically that x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx). Why does this work for all real numbers x, y, z?
2. Complete the square creatively: Factorise x4+6×2+9x4+6x2+9 completely. Then explain why adding and subtracting 4×24x2 inside would also work but is unnecessary here.
3. Pattern analysis: Notice that a4+4b4=(a2+2ab+2b2)(a2−2ab+2b2)a4+4b4=(a2+2ab+2b2)(a2−2ab+2b2). Prove this using difference of squares. Why can’t standard middle-term splitting work directly?
4. Substitution insight: For x6+7×3+10x6+7x3+10, let y=x3y=x3. Factorise completely. Now analyze: what happens if the constant term was 1 instead of 10? Does it still factor nicely over integers?
5. Degree analysis: Consider x4+4x4+4. Show it factors as (x2+2x+2)(x2−2x+2)(x2+2x+2)(x2−2x+2). Explain why both quadratic factors are irreducible over the reals.
6. Symmetry breaking: Factorise x4+x2y2+y4x4+x2y2+y4. Compare this with x4+4y4x4+4y4. Why does one need Sophie Germain identity while the other doesn’t?
7. Functional equation: If P(x)=x4−5×2+4P(x)=x4−5x2+4 factors completely into linear factors over reals, find all roots. Now construct a similar quartic that has exactly two real roots using factorization.
8. Coefficient constraint: For which integer values of k does 3×2+kx+23x2+kx+2 factor into two binomials with integer coefficients? Prove your answer using discriminant and factor pairs.
9. Division algorithm test: Divide x3−6×2+11x−6x3−6x2+11x−6 by x−2x−2 using factorization (not polynomial division). Verify the remainder theorem holds. What pattern emerges?
10. Olympiad proof: Prove that if nn is odd, then xn+ynxn+yn has x+yx+y as a factor. Use this to factorise x5+y5x5+y5 completely. Extend: what changes if n is even?

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