Chapter 6 — Factorisation of Algebraic Expressions

Class 8 (Maharashtra Board) — Notes, Q&A & Textbook Solutions

Formatting reminder: Questions are red, answers are green. Math is typeset with MathJax for perfect rendering.
Quick theory recap — what to remember
  • Common factor: Factor out greatest common factor: \(ax+ay = a(x+y)\).
  • Difference of squares: \(a^2-b^2=(a-b)(a+b)\).
  • Quadratic trinomial: \(ax^2+bx+c\). For \(x^2+(a+b)x+ab=(x+a)(x+b)\). Use splitting the middle term for general quadratics.
  • Sum/difference of cubes:
    \(a^3+b^3=(a+b)(a^2-ab+b^2)\),
    \(a^3-b^3=(a-b)(a^2+ab+b^2)\).
  • Factorisation steps: Look for GCF, then special patterns, then split middle term if needed; check by expanding back.
Part A — 20 important 1-mark Q&A

Q1. Factorise \(4xy + 8xy^2\).

Ans: \(4xy(1+2y)\).

Q2. Factorise \(p^2 - 9q^2\).

Ans: \((p-3q)(p+3q)\).

Q3. Factorise \(x^2 + 5x + 6\).

Ans: \((x+2)(x+3).\)

Q4. Factorise \(x^2 - 10x + 21\).

Ans: \((x-7)(x-3).\)

Q5. Factorise \(2x^2 -9x +9\).

Ans: \((x-3)(2x-3).\)

Q6. Factorise \(2x^2 +5x -18\).

Ans: \((2x+9)(x-2).\)

Q7. Factorise \(x^3+27y^3.\)

Ans: \((x+3y)(x^2-3xy+9y^2).\)

Q8. Factorise \(x^3-8y^3.\)

Ans: \((x-2y)(x^2+2xy+4y^2).\)

Q9. Factorise \(a^2-b^2.\)

Ans: \((a-b)(a+b).\)

Q10. Factorise \(y^2+24y+144.\)

Ans: \((y+12)^2.\)

Q11. Factorise \(5y^2+5y-10.\)

Ans: \(5(y^2+y-2)=5(y+2)(y-1).\)

Q12. Factorise \(p^2-2p-35.\)

Ans: \((p-7)(p+5).\)

Q13. Factorise \(3x^2+14x+15.\)

Ans: \((3x+5)(x+3).\)

Q14. Factorise \(2x^2+x-45.\)

Ans: \((2x-9)(x+5).\)

Q15. Factorise \(20x^2-26x+8.\)

Ans: \(2(5x-4)(2x-1).\)

Q16. Factorise \(44x^2-x-3.\)

Ans: \((11x-3)(4x+1).\)

Q17. Factorise \(x^3+64y^3.\)

Ans: \((x+4y)(x^2-4xy+16y^2).\)

Q18. Factorise \(125p^3+q^3.\)

Ans: \((5p+q)(25p^2-5pq+q^2).\)

Q19. Factorise \(y^3-27.\)

Ans: \((y-3)(y^2+3y+9).\)

Q20. What is the factorisation of \(a^3-b^3\)?

Ans: \((a-b)(a^2+ab+b^2).\)

Part B — 20 important 2-mark Q&A

Q1. Factorise \(x^2+9x+18\).

Ans: \((x+3)(x+6)\) because \(3+6=9,\;3\cdot6=18.\)

Q2. Factorise \(x^2-10x+9\).

Ans: \((x-1)(x-9)\).

Q3. Factorise \(y^2+24y+144\).

Ans: \((y+12)^2.\)

Q4. Factorise \(5y^2+5y-10\).

Ans: \(5(y^2+y-2)=5(y+2)(y-1).\)

Q5. Factorise \(p^2-2p-35\).

Ans: \((p-7)(p+5).\)

Q6. Factorise \(p^2-7p-44\).

Ans: \((p-11)(p+4)\) since \(-11+4=-7,\; -11\cdot4=-44.\)

Q7. Factorise \(m^2-23m+120\).

Ans: \((m-8)(m-15).\)

Q8. Factorise \(m^2-25m+100\).

Ans: \((m-5)(m-20).\)

Q9. Factorise \(3x^2+14x+15\).

Ans: \((3x+5)(x+3).\)

Q10. Factorise \(2x^2+x-45\) by splitting the middle term.

Ans: Split \(x\) as \(-9x+10x\): \(2x^2-9x+10x-45\) = \(x(2x-9)+5(2x-9)=(2x-9)(x+5).\)

Q11. Factorise \(20x^2-26x+8\).

Ans: \(2(10x^2-13x+4)=2(5x-4)(2x-1).\)

Q12. Factorise \(44x^2-x-3\).

Ans: \((11x-3)(4x+1).\)

Q13. Factorise \(2y^2-4y-30\).

Ans: \(2(y^2-2y-15)=2(y-5)(y+3).\)

Q14. Show that \(x^3+27y^3=(x+3y)(x^2-3xy+9y^2)\) by expansion check.

Ans: Expand RHS: \(x^3-3x^2y+9xy^2+3x^2y-9xy^2+27y^3=x^3+27y^3.\)

Q15. Factorise \(8p^3+125q^3.\)

Ans: \((2p+5q)(4p^2-10pq+25q^2).\)

Q16. Factorise \(250p^3+432q^3.\)

Ans: \(2[(5p)^3+(6q)^3]=2(5p+6q)(25p^2-30pq+36q^2).\)

Q17. Factorise \(x^3-8y^3.\)

Ans: \((x-2y)(x^2+2xy+4y^2).\)

Q18. Factorise \(27p^3-125q^3.\)

Ans: \((3p-5q)(9p^2+15pq+25q^2).\)

Q19. Simplify \( (p+q)^3 + (p-q)^3\).

Ans: Expanding and adding gives \(2p^3 + 6pq^2.\)

Q20. Simplify \( (2x+3y)^3 - (2x-3y)^3\).

Ans: Gives \(72x^2y + 54y^3.\)

Part C — 20 important 3-mark Q&A

Q1. Factorise completely: \(2x^2 - 9x + 9\) (show steps).

Ans: Product \(2\cdot9=18\). Find two numbers with product 18 and sum \(-9\): \(-6\) and \(-3\).
Write \(2x^2-6x-3x+9\) = \(2x(x-3)-3(x-3)\) = \((x-3)(2x-3).\)

Q2. Factorise \(2x^2 +5x -18\) fully (steps).

Ans: \(2\cdot(-18)=-36\). Find numbers sum 5 and product \(-36\): 9 and -4.
Write \(2x^2+9x-4x-18 = x(2x+9)-2(2x+9)=(2x+9)(x-2).\)

Q3. Factorise \(x^2-10x+21\) by grouping.

Ans: Find -7 and -3: \(x^2 -7x -3x +21 = x(x-7)-3(x-7)=(x-7)(x-3).\)

Q4. Factorise \(2y^2 -4y -30\) showing factor 2 first.

Ans: \(2(y^2-2y-15)=2[y^2-5y+3y-15]=2[y(y-5)+3(y-5)]=2(y-5)(y+3).\)

Q5. Prove identity \(a^3+b^3=(a+b)(a^2-ab+b^2)\) by expanding RHS.

Ans: Expand RHS: \(a^3-ab^2+a b^2? \) Carry out full expansion: \((a+b)(a^2-ab+b^2)=a^3 - a^2 b + a b^2 + a^2 b - ab^2 + b^3 = a^3 + b^3.\)

Q6. Factorise \(x^3+64y^3\) and check by expansion.

Ans: \((x+4y)(x^2-4xy+16y^2)\). Expand to verify it equals \(x^3+64y^3.\)

Q7. Factorise \(125k^3 + 27m^3\).

Ans: \((5k+3m)(25k^2 -15km +9m^2).\)

Q8. Factorise \(24a^3 + 81b^3\) completely.

Ans: \(3(8a^3+27b^3)=3(2a+3b)(4a^2 -6ab +9b^2).\)

Q9. Factorise \(27m^3 - 216n^3\).

Ans: \(27(m^3-8n^3)=27(m-2n)(m^2+2mn+4n^2).\)

Q10. Factorise \(343a^3 - 512b^3\).

Ans: \((7a-8b)(49a^2+56ab+64b^2).\)

Q11. Simplify \( (x+y)^3 - (x-y)^3\) (show algebraic steps).

Ans: Expand both and subtract:
\((x+y)^3-(x-y)^3 = 6x^2y + 2y^3.\)

Q12. Simplify \( (3a+5b)^3 - (3a-5b)^3\).

Ans: Put \(A=3a,B=5b\) then difference = \(6A^2B + 2B^3\).
So result = \(6(3a)^2(5b) + 2(5b)^3 = 270a^2b + 250 b^3.\)

Q13. Prove \( (a+b)^3 - a^3 - b^3 = 3ab(a+b)\).

Ans: Expand \( (a+b)^3 = a^3 +3a^2b +3ab^2 + b^3\). Subtract \(a^3+b^3\) gives \(3ab(a+b).\)

Q14. Find \(p^3-(p+1)^3\).

Ans: \(p^3 - (p^3+3p^2+3p+1) = -3p^2 -3p -1.\)

Q15. Simplify \( (3xy-2ab)^3 - (3xy+2ab)^3\).

Ans: Using formula for difference of cubes of \((u-v)^3-(u+v)^3 = -6u^2v -2v^3\) with \(u=3xy,\;v=2ab\),
result = \(-108x^2y^2 ab -16a^3 b^3.\)

Q16. Show \(x^3 - 8 = (x-2)(x^2 +2x +4)\) by multiplication.

Ans: Multiply RHS: \(x^3 - 8\) — expansion verifies identity (difference of cubes).

Q17. Show how to factor \(16a^3 - 128 b^3\) completely.

Ans: Factor 16: \(16(a^3-8b^3)=16(a-2b)(a^2+2ab+4b^2).\)

Q18. Why is splitting the middle term useful? Give one example.

Ans: Splitting transforms quadratic into four-term expression so grouping can be used to factor. Example: \(2x^2+5x-18 = 2x^2+9x-4x-18 = (2x+9)(x-2).\)

Q19. Explain how to test your factorization.

Ans: Multiply the factors (expand) — if you get the original expression, factorisation is correct.

Q20. Factor and simplify \(x^3+27- (x+3)^3\).

Ans: Note \((x+3)^3 = x^3 +9x^2 +27x +27\). So expression \(x^3+27 - (x^3+9x^2+27x+27) = -9x^2 -27x = -9x(x+3).\)

Part D — Worked solutions of textbook Practice Sets

Practice Set 6.1 — Quadratic factorisation (answers)

Below are the exact factorisations for the items listed in your textbook extract:

  1. \(x^2 + 9x + 18 = (x+3)(x+6)\).
  2. \(x^2 - 10x + 9 = (x-1)(x-9)\).
  3. \(y^2 + 24y + 144 = (y+12)^2\).
  4. \(5y^2 + 5y - 10 = 5(y^2+y-2) = 5(y+2)(y-1)\).
  5. \(p^2 - 2p -35 = (p-7)(p+5)\).
  6. \(p^2 - 7p -44 = (p-11)(p+4)\).
  7. \(m^2 -23m + 120 = (m-8)(m-15)\).
  8. \(m^2 -25m +100 = (m-5)(m-20)\).
  9. \(3x^2 +14x +15 = (3x+5)(x+3)\).
  10. \(2x^2 + x -45 = (2x-9)(x+5)\).
  11. \(20x^2 -26x +8 = 2(5x-4)(2x-1)\).
  12. \(44x^2 - x - 3 = (11x-3)(4x+1)\).

Practice Set 6.2 — Sum of cubes (worked answers)

Factorisations using \(a^3+b^3=(a+b)(a^2-ab+b^2)\):

  1. \(x^3 +64y^3 = (x+4y)(x^2 - 4xy +16y^2).\)
  2. \(125p^3 + q^3 = (5p+q)(25p^2 -5pq + q^2).\)
  3. \(125k^3 +27m^3 = (5k+3m)(25k^2 -15km +9m^2).\)
  4. \(2l^3 + 432m^3 = 2\bigl(l^3 +216m^3\bigr) = 2( l+6m)(l^2 -6lm +36m^2).\)
  5. \(24a^3 +81b^3 = 3\bigl(8a^3+27b^3\bigr)=3(2a+3b)(4a^2 -6ab +9b^2).\)
A few items in the Practice Set 6.2 as pasted were unreadable / garbled (OCR artefacts). I answered the clearly readable items above. If you paste or upload the exact lines for items (6)–(8) I will factorise them precisely in the same page.

Practice Set 6.3 — Difference / sum of cubes & simplifications

Using \(a^3-b^3=(a-b)(a^2+ab+b^2)\) and related simplifications:

  1. \(y^3 - 27 = (y-3)(y^2+3y+9).\)
  2. \(x^3 - 64y^3 = (x-4y)(x^2 +4xy +16y^2).\)
  3. \(27m^3 - 216n^3 = 27(m^3 -8n^3) = 27(m-2n)(m^2 +2mn +4n^2).\)
  4. \(125y^3 -1 = (5y-1)(25y^2 +5y +1).\)
  5. \(8p^3 - 27q^3 = (2p-3q)(4p^2 +6pq +9q^2).\)
  6. \(343a^3 -512b^3 = (7a-8b)(49a^2 +56ab +64b^2).\)
  7. \(64x^3 -729y^3 = (4x-9y)(16x^2 +36xy +81y^2).\)
  8. \(16a^3 - 128b^3 = 16(a^3 -8b^3) = 16(a-2b)(a^2 +2ab +4b^2).\)

Simplifications (examples from book):

  1. \((x+y)^3 - (x-y)^3 = 6x^2y + 2y^3.\)
  2. \((3a+5b)^3 - (3a-5b)^3 = 270a^2b + 250b^3.\)
  3. \((a+b)^3 - a^3 - b^3 = 3ab(a+b).\)
  4. \(p^3 - (p+1)^3 = -3p^2 -3p -1.\)
  5. \((3xy-2ab)^3 - (3xy+2ab)^3 = -108x^2y^2 ab - 16 a^3 b^3.\)

Practice Set 6.4 — Rational algebraic expressions (best-effort)

The textbook Practice Set 6.4 contains several rational expression simplifications (fractions of algebraic expressions). A few items in your pasted text had OCR errors (digits and fractions stuck). I solved the clearly readable examples and illustrate method for simplifying rational algebraic expressions:

Method summary: Factor numerators and denominators, cancel common factors (watch domain: denominators ≠ 0). Example steps below.
  1. Example (from the book text earlier):
    Given (messy in paste), solved illustration: \( \dfrac{a^2}{a^3}\cdot\dfrac{a^5}{a^6} \) etc. (If you send the exact typed expression I will show step-by-step.)
  2. Example 2 (clearer): Simplify \(\dfrac{x^2 - 9}{x-3}\).
    Factor numerator: \((x-3)(x+3)\). Cancel \((x-3)\) to get \(x+3\) (for \(x\ne3\)).
  3. Example 3: \(\dfrac{a^3 - b^3}{a-b} = a^2 + ab + b^2\) (because \(a^3-b^3=(a-b)(a^2+ab+b^2)\)).
  4. Example 4: \(\dfrac{(x^2-4)}{(x-2)} = x+2\) (since \(x^2-4=(x-2)(x+2)\), for \(x\ne2\)).
The scanned text for Practice Set 6.4 in your message included several multi-line algebraic fraction problems which were partially garbled (line-breaks / superscripts). I solved the clear examples and gave general method + common textbook identities. If you paste the exact lines (or upload a photo/scan of the page), I will produce step-by-step solutions for every single item (6.4) and replace the ambiguous placeholders with precise calculations in the same style.
Final checks & tips for students:
  • Always try to factor out the greatest common factor (GCF) first.
  • Recognise patterns: \(a^2-b^2,\; a^3\pm b^3,\; (ax+b)(cx+d)\) possibilities.
  • For quadratics \(ax^2+bx+c\) either split the middle term or use AC method (find two numbers with product \(ac\) and sum \(b\)).
  • After factorising, expand to verify — this prevents sign mistakes.
  • When simplifying rational algebraic expressions, always state excluded values (denominator ≠ 0).