CBSE Class 8 Mathematics Factorisation Worksheet Set 01

Question. Express y² – 9y + 20 as product of two expressions.
Answer : (y – 4) (y – 5)

Question. Simplify [102 – 18 × 10 + 81]
Answer : 1

Question. Find the remainder in the following : (x⁴ – a⁴) (x² + a²).
Answer : 0

Question. Find two numbers A and B so that A + B = 12 and AB = 27.
Answer : A = 9, B = 3.

Question. (x + 1)² – (x – 1)² = _________.
Answer : 4x

Question. Find Quotient if
Dividend = x², Divisor = x – 3, Remainder = 9.
Answer : x + 3

Question. Factorise x² + 12x + 35
Answer : (x + 5) (x + 7)

Question. Factorize : x² + 4x – 21.
Answer : (x + 7) (x – 3)

Question. Complete it (a³ + b³) (a³ – b³) = _________ .
Answer : a6 – b6

Question. Factorize : 4x² – 12xy + 9y².
Answer : (2x – 3y)²

Question. The sum and product of two numbers is same and both equal to 4. Find the numbers
Answer : 2, 2

Question. Factorize : 3x² – 10x + 8.
Answer : (x – 2) (3x – 4)

Question. Find the value of 81x² + 36xy + 4y² if x = 1/9, y = -1/2
Answer : 0

Question. Find x if x (y - z) = (4y² - 4x²) / (x + 4)
Answer : 4

Question. Find two numbers m and n so that
m + n = 19
mn = 70
Answer : m = 14, n = 5.

Question. Find y if, (2x + 7) (2x – 7) + y = 4x²
Answer : y = 49

Question. Find two numbers so that their product is 56 and difference is 10.
Answer : 14, 4

Question. x² – x – 12 = (x – 4) × __________.
Answer : x + 3

Question. Factorize 6x³ – 4x².
Answer : 2x²(3x – 2)

Question. Divide 100 abc (3a – 12) (6b – 36) by 9(a – 4) (b – 6).
Answer : 200abc

Question. Give the two factors of x² + 6x + 5.
Answer : (x + 1) (x + 5)

Question. Factorize : 6xy – 2y + 3 – 9x.
Answer : (3x – 1) (2y – 3)

Question 1. Factorise \(2x + 4\).
Answer: \(2x + 4 = 2(x + 2)\)
In simple words: Take the common factor 2 out from both terms.

Exam Tip: Always look for the highest common factor (HCF) among all terms first.

Question 2. Factorise \(12a^2b + 15ab^2\).
Answer: \(12a^2b + 15ab^2 = 3ab(4a + 5b)\)
In simple words: Find the common numerical and variable factors and factor them out.

Exam Tip: For variables with exponents, take the lowest power present in all terms.

Question 3. Factorise \(10x^2 - 14x^3 + 18x^4\).
Answer: \(10x^2 - 14x^3 + 18x^4 = 2x^2(5 - 7x + 9x^2)\)
In simple words: Factor out the common monomial \(2x^2\).

Exam Tip: Check each term inside the parenthesis to ensure no common factor remains.

Question 4. Factorise \(20x^2y + 30axy\).
Answer: \(20x^2y + 30axy = 10xy(2x + 3a)\)
In simple words: The HCF of 20 and 30 is 10, and the common variables are x and y.

Exam Tip: Verify by multiplying back: \(10xy \times 2x = 20x^2y\) and \(10xy \times 3a = 30axy\).

Question 5. Factorise \(x^2 + xy + 8x + 8y\).
Answer: \(x^2 + xy + 8x + 8y = x(x + y) + 8(x + y) = (x + 8)(x + y)\)
In simple words: Group terms in pairs and factor out the common binomial.

Exam Tip: Grouping by rearrangement helps when all terms do not share a single common factor.

Question 6. Factorise \(1 + a + ac + a^2c\).
Answer: \(1 + a + ac + a^2c = 1(1 + a) + ac(1 + a) = (1 + ac)(1 + a)\)
In simple words: Group the first two terms and the last two terms to find a common binomial factor.

Exam Tip: Ensure the binomials in both groups are identical before taking them common.

Question 7. Factorise \(a^2 + bc + ab + ac\).
Answer: Rearrange terms: \(a^2 + ab + ac + bc\)
\(= a(a + b) + c(a + b)\)
\(= (a + c)(a + b)\)
In simple words: Rearrange the terms so that grouping produces matching binomial factors.

Exam Tip: If grouping directly does not work, try rearranging the terms.

Question 8. Factorise \(x^2 - 36\) using identity.
Answer: Using identity \(a^2 - b^2 = (a + b)(a - b)\):
\(x^2 - 36 = x^2 - 6^2 = (x + 6)(x - 6)\)
In simple words: Express 36 as \(6^2\) and apply the difference of squares formula.

Exam Tip: Recognize standard squares like 4, 9, 16, 25, 36, 49, 64, 81, 100 instantly.

Question 9. Factorise \(xy - pq + qy - px\).
Answer: Rearrange terms: \(xy + qy - px - pq\)
\(= y(x + q) - p(x + q)\)
\(= (y - p)(x + q)\)
In simple words: Group terms containing common variables like y and p.

Exam Tip: Be careful with negative signs when factoring out a minus sign.

Question 10. Factorise \(5a(2x - 3y) + 2b(2x - 3y)\).
Answer: \(5a(2x - 3y) + 2b(2x - 3y) = (2x - 3y)(5a + 2b)\)
In simple words: Take the common binomial expression \((2x - 3y)\) outside.

Exam Tip: Treat the bracketed expression as a single variable during factorisation.

Question 11. Factorise \(ax^2 + by^2 + bx^2 + ay^2\).
Answer: Rearrange terms: \(ax^2 + bx^2 + ay^2 + by^2\)
\(= x^2(a + b) + y^2(a + b)\)
\(= (x^2 + y^2)(a + b)\)
In simple words: Group terms with common powers of x and y together.

Exam Tip: Grouping terms with similar variables simplifies the expression quickly.

Question 12. Factorise \(8(4x + 5y)^2 - 12(4x + 5y)\).
Answer: Find HCF of coefficients 8 and 12, which is 4, and common bracket \((4x + 5y)\):
\(8(4x + 5y)^2 - 12(4x + 5y) = 4(4x + 5y)[2(4x + 5y) - 3]\)
\(= 4(4x + 5y)(8x + 10y - 3)\)
In simple words: Factor out the common numbers and binomial term, then simplify inside the remaining bracket.

Exam Tip: Do not expand the square; taking the common bracket out is much faster and cleaner.

Question 13. Factorise using identity \(x^2 + 10x + 25\).
Answer: Using identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(x^2 + 10x + 25 = x^2 + 2(x)(5) + 5^2 = (x + 5)^2\)
In simple words: Match the expression with the square of a binomial formula.

Exam Tip: Check if the middle term equals twice the product of the square roots of the first and last terms.

Question 14. Factorise using identity \(4x^2 + 9y^2 + 12xy\).
Answer: Rewrite as: \((2x)^2 + (3y)^2 + 2(2x)(3y)\)
Using identity \((a + b)^2 = a^2 + b^2 + 2ab\):
\(= (2x + 3y)^2\)
In simple words: Identify the squared terms and verify the middle term matches \(2ab\).

Exam Tip: Order of terms does not matter as long as signs are positive.

Question 15. Factorise \(ax^2y + bxy^2 + cxyz\).
Answer: Identify the common variable in all terms, which is xy:
\(ax^2y + bxy^2 + cxyz = xy(ax + by + cz)\)
In simple words: Factor out the lowest power of common variables x and y.

Exam Tip: Check every term to ensure all common variables have been extracted.

Question 16. Factorise \(15xy - 6x + 5y - 2\).
Answer: Group terms: \((15xy - 6x) + (5y - 2)\)
\(= 3x(5y - 2) + 1(5y - 2)\)
\(= (3x + 1)(5y - 2)\)
In simple words: Group the first two terms and factor out \(3x\), leaving \((5y - 2)\).

Exam Tip: Always verify that both bracketed expressions resulting from grouping are identical.

Question 17. Factorise \(6pq - 4q + 6 - 9p\).
Answer: Rearrange terms: \(6pq - 9p - 4q + 6\)
\(= 3p(2q - 3) - 2(2q - 3)\)
\(= (3p - 2)(2q - 3)\)
In simple words: Rearrange terms to easily extract common factors from pairs.

Exam Tip: Factoring out \(-2\) changes the signs inside the bracket to match \((2q - 3)\).

Question 18. Find the factors of \(x^2 - 7x + 12\).
Answer: Split the middle term \(-7x\) into \(-3x - 4x\) (since \((-3) \times (-4) = 12\) and \(-3 + (-4) = -7\)):
\(x^2 - 7x + 12 = x^2 - 3x - 4x + 12\)
\(= x(x - 3) - 4(x - 3)\)
\(= (x - 4)(x - 3)\)
In simple words: Split the middle term into two numbers whose product is 12 and sum is -7.

Exam Tip: This is standard quadratic splitting; always check product and sum conditions.

Question 19. Find the factors of \(3x^2 + 9x + 6\).
Answer: Take out common factor 3:
\(3(x^2 + 3x + 2)\)
Split middle term \(3x\) into \(2x + x\):
\(= 3(x^2 + 2x + x + 2)\)
\(= 3[x(x + 2) + 1(x + 2)]\)
\(= 3(x + 1)(x + 2)\)
In simple words: First factor out the constant 3, then factor the remaining quadratic expression.

Exam Tip: Always extract the numerical HCF first to simplify calculations.

Question 20. Factorise \((a + b)^2 - (a - b)^2\).
Answer: Using identity \(X^2 - Y^2 = (X + Y)(X - Y)\) where \(X = (a + b)\) and \(Y = (a - b)\):
\(= [(a + b) + (a - b)][(a + b) - (a - b)]\)
\(= [a + b + a - b][a + b - a + b]\)
\(= [2a][2b] = 4ab\)
In simple words: Apply the difference of squares identity directly to the two bracketed terms.

Exam Tip: Be extremely careful with negative signs when subtracting the second binomial expression.

Question 21. Factorise the expression \(10ab + 4a + 5b + 2\).
Answer: Group terms: \((10ab + 4a) + (5b + 2)\)
\(= 2a(5b + 2) + 1(5b + 2)\)
\(= (2a + 1)(5b + 2)\)
In simple words: Group the first two terms and factor out \(2a\).

Exam Tip: Ensure the common binomial \((5b + 2)\) is correctly factored out.

Question 22. Divide \(24x^2y^2z^2\) by \(6yz\).
Answer: \(\frac{24x^2y^2z^2}{6yz} = \frac{24}{6} \times x^2 \times \frac{y^2}{y} \times \frac{z^2}{z}\)
\(= 4x^2yz\)
In simple words: Divide the coefficients and subtract the powers of matching variables in the denominator from the numerator.

Exam Tip: Use laws of exponents: \(\frac{y^2}{y^1} = y^{2-1} = y\).

Question 23. Divide \((7a^2 + 14a)\) by \((a + 2)\).
Answer: Factorise numerator: \(7a^2 + 14a = 7a(a + 2)\)
\(\frac{7a(a + 2)}{a + 2} = 7a\)
In simple words: Factor the numerator first, then cancel the common term \((a + 2)\).

Exam Tip: Never cancel terms across addition without factoring first.

Question 24. Divide \(x(5x^2 - 80)\) by \(5x(x + 4)\).
Answer: Factorise numerator: \(5x(x^2 - 16) = 5x(x - 4)(x + 4)\)
\(\frac{5x(x - 4)(x + 4)}{5x(x + 4)} = x - 4\)
In simple words: Factor out 5 from the bracket, use the difference of squares, and cancel common terms.

Exam Tip: \(x^2 - 16\) factors into \((x - 4)(x + 4)\).

Question 25. Divide \(x^2 + 7x + 10\) by \(x + 5\).
Answer: Factorise numerator by splitting middle term: \(x^2 + 5x + 2x + 10 = x(x + 5) + 2(x + 5) = (x + 2)(x + 5)\)
\(\frac{(x + 2)(x + 5)}{x + 5} = x + 2\)
In simple words: Factor the quadratic numerator into two brackets and cancel \((x + 5)\).

Exam Tip: Splitting middle term makes division by binomials straightforward.

Question 26. Divide \(12pq(9p^2 - 16q^2)\) by \(4pq(3p + 4q)\).
Answer: Factorise numerator: \(9p^2 - 16q^2 = (3p)^2 - (4q)^2 = (3p - 4q)(3p + 4q)\)
Numerator = \(12pq(3p - 4q)(3p + 4q)\)
Denominator = \(4pq(3p + 4q)\)
\(\frac{12pq(3p - 4q)(3p + 4q)}{4pq(3p + 4q)} = 3(3p - 4q) = 9p - 12q\)
In simple words: Factor the difference of squares in the numerator and cancel common factors with the denominator.

Exam Tip: Simplify numerical coefficients (\(12 \div 4 = 3\)) alongside variables.

Question 27. Divide \(39x^3(50x^2 - 98)\) by \(26x^2(5x + 7)\).
Answer: Factor out 2 from numerator: \(50x^2 - 98 = 2(25x^2 - 49) = 2(5x - 7)(5x + 7)\)
Numerator = \(39x^3 \times 2(5x - 7)(5x + 7) = 78x^3(5x - 7)(5x + 7)\)
Denominator = \(26x^2(5x + 7)\)
\(\frac{78x^3(5x - 7)(5x + 7)}{26x^2(5x + 7)} = 3x(5x - 7) = 15x^2 - 21x\)
In simple words: Take out common numerical factors, apply difference of squares, and cancel terms.

Exam Tip: Always look for common factors before applying algebraic identities.

Question 28. Find the factors of \(25x^2 - 4y^2 + 28yz - 49z^2\).
Answer: Group the last three terms: \(25x^2 - (4y^2 - 28yz + 49z^2)\)
Notice that \(4y^2 - 28yz + 49z^2 = (2y)^2 - 2(2y)(7z) + (7z)^2 = (2y - 7z)^2\)
Expression becomes: \(25x^2 - (2y - 7z)^2\)
Apply difference of squares \(A^2 - B^2 = (A - B)(A + B)\) where \(A = 5x\) and \(B = (2y - 7z)\):
\(= [5x - (2y - 7z)][5x + (2y - 7z)]\)
\(= (5x - 2y + 7z)(5x + 2y - 7z)\)
In simple words: Group three terms to form a complete square, then use the difference of squares identity.

Exam Tip: Recognizing grouped complete squares is key for advanced factorisation problems.

Most Important Questions

Question 1. Add \(4x^2 + 2xy - 4\) & \(7x^2 - 3xy + 4\).
Answer: \((4x^2 + 2xy - 4) + (7x^2 - 3xy + 4)\)
\(= (4 + 7)x^2 + (2 - 3)xy + (-4 + 4)\)
\(= 11x^2 - xy\)
In simple words: Combine the coefficients of like terms together.

Exam Tip: Group like terms by their variable parts carefully.

Question 2. Subtract \(7x - 3x^2\) from \(4x + 8x^2\).
Answer: \((4x + 8x^2) - (7x - 3x^2)\)
\(= 4x + 8x^2 - 7x + 3x^2\)
\(= (8 + 3)x^2 + (4 - 7)x\)
\(= 11x^2 - 3x\)
In simple words: Write the expression to be subtracted after a minus sign, distribute the minus sign, and combine like terms.

Exam Tip: "Subtract A from B" means write B - A.

Question 3. Find the product of \((7x - 4y)\) and \((3x - 7y)\).
Answer: \((7x - 4y)(3x - 7y) = 7x(3x - 7y) - 4y(3x - 7y)\)
\(= 21x^2 - 49xy - 12xy + 28y^2\)
\(= 21x^2 - 61xy + 28y^2\)
In simple words: Multiply each term of the first binomial by each term of the second binomial.

Exam Tip: Combine middle like terms \(-49xy - 12xy\) to get \(-61xy\).

Question 4. Using suitable identity find \((7x - 3y)^2\).
Answer: Using identity \((a - b)^2 = a^2 - 2ab + b^2\):
\((7x - 3y)^2 = (7x)^2 - 2(7x)(3y) + (3y)^2\)
\(= 49x^2 - 42xy + 9y^2\)
In simple words: Apply the square of binomial formula directly.

Exam Tip: Do not forget the middle term \(2ab\).

Question 5. Using identity \((a - b)^2 = a^2 - 2ab + b^2\) find the value of \(98^2\).
Answer: Rewrite 98 as \((100 - 2)\):
\(98^2 = (100 - 2)^2\)
\(= 100^2 - 2(100)(2) + 2^2\)
\(= 10000 - 400 + 4\)
\(= 9604\)
In simple words: Express numbers near powers of 10 as a difference or sum to compute squares easily.

Exam Tip: This method saves time compared to standard multiplication for numbers close to 100.

Question 6. Subtract \(3x(x - 4y + 5z)\) from \(4x(2x - 3y + 10z)\).
Answer: First expand both expressions:
\(3x(x - 4y + 5z) = 3x^2 - 12xy + 15xz\)
\(4x(2x - 3y + 10z) = 8x^2 - 12xy + 40xz\)
Now subtract:
\((8x^2 - 12xy + 40xz) - (3x^2 - 12xy + 15xz)\)
\(= 8x^2 - 12xy + 40xz - 3x^2 + 12xy - 15xz\)
\(= (8 - 3)x^2 + (-12 + 12)xy + (40 - 15)xz\)
\(= 5x^2 + 25xz\)
In simple words: Expand the products first, then subtract corresponding terms.

Exam Tip: Watch out for sign changes when distributing the negative sign during subtraction.

Question 7. Multiply \((a^2 + 2c^2)(3a - 3c)\).
Answer: \(= a^2(3a - 3c) + 2c^2(3a - 3c)\)
\(= 3a^3 - 3a^2c + 6ac^2 - 6c^3\)
In simple words: Distribute each term of the first bracket across the second bracket.

Exam Tip: Check if any resulting terms are like terms (here all terms have different variable parts).

Question 8. Using suitable identity find \((6x^2 - \frac{5}{3})^2\).
Answer: Using identity \((a - b)^2 = a^2 - 2ab + b^2\):
\((6x^2 - \frac{5}{3})^2 = (6x^2)^2 - 2(6x^2)(\frac{5}{3}) + (\frac{5}{3})^2\)
\(= 36x^4 - \frac{60}{3}x^2 + \frac{25}{9}\)
\(= 36x^4 - 20x^2 + \frac{25}{9}\)
In simple words: Substitute fractional terms into the binomial square formula.

Exam Tip: Simplify fractions carefully (\(6 \div 3 = 2\), so \(2 \times 2 \times 5 = 20\)).

Question 9. Simplify \((xy + yz)^2 - (xy - yz)^2\).
Answer: Using difference of squares \(A^2 - B^2 = (A - B)(A + B)\) where \(A = (xy + yz)\) and \(B = (xy - yz)\):
\(= [(xy + yz) - (xy - yz)][(xy + yz) + (xy - yz)]\)
\(= [xy + yz - xy + yz][xy + yz + xy - yz]\)
\(= [2yz][2xy]\)
\(= 4xy^2z\)
In simple words: Apply the difference of squares identity to avoid long expansions.

Exam Tip: Using the identity is much faster than expanding both squares individually.

Question 10. Using identity find the product of \((\frac{a}{2} + \frac{3b}{4})(\frac{a}{2} + \frac{3b}{4})\).
Answer: Since both terms are identical, it is \((\frac{a}{2} + \frac{3b}{4})^2\).
Using identity \((x + y)^2 = x^2 + 2xy + y^2\):
\(= (\frac{a}{2})^2 + 2(\frac{a}{2})(\frac{3b}{4}) + (\frac{3b}{4})^2\)
\(= \frac{a^2}{4} + \frac{3ab}{4} + \frac{9b^2}{16}\)
In simple words: Write as a square of binomial and apply the formula.

Exam Tip: Square both the numerator and denominator of fractions.

Question 11. If \(x + \frac{1}{x} = 6\), find \(x^2 + \frac{1}{x^2}\).
Answer: Square both sides of the given equation:
\((x + \frac{1}{x})^2 = 6^2\)
\(x^2 + 2(x)(\frac{1}{x}) + \frac{1}{x^2} = 36\)
\(x^2 + 2 + \frac{1}{x^2} = 36\)
\(x^2 + \frac{1}{x^2} = 36 - 2 = 34\)
In simple words: Square both sides of the given equation and simplify the middle term.

Exam Tip: This is a standard algebraic identity problem frequently asked in exams.

Question 12. Find the product of \((\frac{5}{3}x + \frac{3}{4}y)^2 - (\frac{5}{3}x + \frac{3}{4}y)^2\) and also evaluate it when \(x = +2\) and \(y = -1\).
Answer: Notice the expression is \(A^2 - A^2\), which simplifies directly to 0.
Therefore, the product/value is 0 for any values of x and y.
In simple words: Since the expression is subtracting an identical term from itself, the result is always 0.

Exam Tip: Always observe if terms cancel out completely before performing lengthy calculations.

Question 13. Using identity \(a^2 - b^2 = (a + b)(a - b)\) find \((1.02)^2 - (0.98)^2\).
Answer: \((1.02)^2 - (0.98)^2 = (1.02 + 0.98)(1.02 - 0.98)\)
\(= (2.00)(0.04)\)
\(= 0.08\)
In simple words: Apply the difference of squares identity with decimal numbers.

Exam Tip: Avoid calculating squares of decimals directly; use the identity for quick computation.

Question 14. If \(x + y = 12\) and \(xy = 32\). Find the value of \(x^2 + y^2\).
Answer: We know that \((x + y)^2 = x^2 + 2xy + y^2\)
Substitute the given values:\(\quad 12^2 = x^2 + 2(32) + y^2\)
\(144 = x^2 + 64 + y^2\)
\(x^2 + y^2 = 144 - 64 = 80\)
In simple words: Use the binomial square expansion formula and substitute the known values.

Exam Tip: Memorize basic expansion formulas like \((x + y)^2\) and \((x - y)^2\).

Question 15. Simplify \((xy + yz)^2 - 2x^2y^2z\) find the value when \(x = -1\), \(y = 1\) and \(z = 2\).
Answer: Expand \((xy + yz)^2 = x^2y^2 + 2xy^2z + y^2z^2\)
Expression becomes: \(x^2y^2 + 2xy^2z + y^2z^2 - 2x^2y^2z\) — wait, let's re-verify the terms. Substituting values directly:
\(((-1)(1) + (1)(2))^2 - 2(-1)^2(1)^2(2)\)
\(= (-1 + 2)^2 - 2(1)(1)(2)\)
\(= (1)^2 - 4 = 1 - 4 = -3\)
In simple words: Substitute the given numerical values of x, y, and z directly into the expression and evaluate.

Exam Tip: Be careful with negative signs when substituting values into exponents.

Question 16. Subtract \(4x^2 - 2y + 7z^3 - 3\) from \(3y + 7x^2 - 2z^3 + 4\).
Answer: \((3y + 7x^2 - 2z^3 + 4) - (4x^2 - 2y + 7z^3 - 3)\)
\(= 3y + 7x^2 - 2z^3 + 4 - 4x^2 + 2y - 7z^3 + 3\)
Group like terms:
\(= (7 - 4)x^2 + (3 + 2)y + (-2 - 7)z^3 + (4 + 3)\)
\(= 3x^2 + 5y - 9z^3 + 7\)
In simple words: Group like terms after distributing the negative sign.

Exam Tip: Check every term to ensure none are missed during subtraction.

Question 17. Divide \(-72x^2yz\) by \(-12xyz\).
Answer: \(\frac{-72x^2yz}{-12xyz} = \frac{-72}{-12} \times \frac{x^2}{x} \times \frac{y}{y} \times \frac{z}{z}\)
\(= 6x\)
In simple words: Divide the negative coefficients and cancel matching variables.

Exam Tip: Negative divided by negative gives a positive result.

Question 18. Divide \(9m^5 + 12m^4 - 6m^2\) by \(3m^2\).
Answer: \(\frac{9m^5 + 12m^4 - 6m^2}{3m^2} = \frac{9m^5}{3m^2} + \frac{12m^4}{3m^2} - \frac{6m^2}{3m^2}\)
\(= 3m^3 + 4m^2 - 2\)
In simple words: Divide each term in the polynomial by the single monomial denominator.

Exam Tip: Divide each term separately and subtract exponents.

Question 19. Divide \(\frac{2}{3}a^2b^2c^2 + \frac{4}{3}ab^2c^3 - \frac{1}{5}ab^3c^2\) by \(\frac{1}{2}abc\).
Answer: Divide each term by \(\frac{1}{2}abc\):
1) \(\frac{\frac{2}{3}a^2b^2c^2}{\frac{1}{2}abc} = \frac{2}{3} \times 2 \times abc = \frac{4}{3}abc\)
2) \(\frac{\frac{4}{3}ab^2c^3}{\frac{1}{2}abc} = \frac{4}{3} \times 2 \times bc^2 = \frac{8}{3}bc^2\)
3) \(\frac{-\frac{1}{5}ab^3c^2}{\frac{1}{2}abc} = -\frac{1}{5} \times 2 \times b^2c = -\frac{2}{5}b^2c\)
Result: \(\frac{4}{3}abc + \frac{8}{3}bc^2 - \frac{2}{5}b^2c\)
In simple words: Divide each term of the polynomial by the monomial divisor by multiplying by the reciprocal.

Exam Tip: Dividing by a fraction is the same as multiplying by its reciprocal.

Question 20. Divide \(6 + x - 4x^2 + x^3\) by \(x - 3\) by long division method.
Answer: Arrange dividend in descending powers of x: \(x^3 - 4x^2 + x + 6\).
Divide by \(x - 3\):
- Quotient = \(x^2 - x - 2\)
- Remainder = 0
In simple words: Perform polynomial long division by matching the leading term at each step.

Exam Tip: Always arrange polynomials in standard descending order of powers before long division.

Question 21. Divide \(12x^3 - 8x^2 - 6x + 10\) by \(3x - 2\). Write the quotient and the remainder.
Answer: Using long division of polynomials:
- Quotient = \(4x^2 - 2\)
- Remainder = 6
In simple words: Divide term by term using polynomial long division.

Exam Tip: Verify using division algorithm: Dividend = Divisor \(\times\) Quotient + Remainder.

Question 22. Divide \(x^3 - 6x^2 + 11x - 6\) by \(x^2 - 4x + 3\).
Answer: Using polynomial long division:
- Quotient = \(x - 2\)
- Remainder = 0
In simple words: Divide the cubic polynomial by the quadratic divisor using long division.

Exam Tip: Ensure correct subtraction signs at each step of the long division process.

Question 23. Find whether \(x + 1\) is factor of \(2x^2 + 5x + 4\) or not.
Answer: Using remainder theorem, put \(x = -1\):
\(P(-1) = 2(-1)^2 + 5(-1) + 4\)
\(= 2(1) - 5 + 4 = 2 - 5 + 4 = 1 \neq 0\)
Since remainder is not zero, \((x + 1)\) is NOT a factor.
In simple words: Substitute \(x = -1\) into the polynomial; if the result is not zero, it is not a factor.

Exam Tip: The remainder theorem is much faster than long division for checking factors.

Question 24. Divide the following \(63a^2b^4c\) by \(7a^2b^2c\).
Answer: \(\frac{63a^2b^4c}{7a^2b^2c} = \frac{63}{7} \times \frac{a^2}{a^2} \times \frac{b^4}{b^2} \times \frac{c}{c}\)
\(= 9b^2\)
In simple words: Divide coefficients and cancel out matching variables with equal powers.

Exam Tip: \(a^2 \div a^2 = 1\) and \(c \div c = 1\).

Question 25. Divide the polynomial \(5x(x^2 - x + 1) - (9 + 4x^4)\) by \(4x - 1\). Write quotient and remainder.
Answer: Expand dividend: \(5x^3 - 5x^2 + 5x - 9 - 4x^4\)
Arrange in standard form: \(-4x^4 + 5x^3 - 5x^2 + 5x - 9\)
Dividing by \(4x - 1\) using long division:
- Quotient = \(-x^3 + x^2 - x + 1\)
- Remainder = \(-8\)
In simple words: Expand and arrange the polynomial in standard form before performing long division.

Exam Tip: Pay close attention to negative signs and missing powers.

Question 26. Divide the polynomial \(x^3 + 3x^2 - 5x + 4\) by \(x - 1\). Write quotient and remainder.
Answer: Using polynomial long division by \((x - 1)\):
- Quotient = \(x^2 + 4x - 1\)
- Remainder = 3
In simple words: Divide the cubic polynomial by \(x - 1\) to find quotient and remainder.

Exam Tip: Check your answer by multiplying quotient by divisor and adding remainder.

Question 27. Divide \(z(5z^2 - 80)\) by \(5z(z + 4)\).
Answer: Factorise numerator: \(z \times 5(z^2 - 16) = 5z(z - 4)(z + 4)\)
Denominator = \(5z(z + 4)\)
\(\frac{5z(z - 4)(z + 4)}{5z(z + 4)} = z - 4\)
In simple words: Factor out constants and use the difference of squares, then cancel common terms.

Exam Tip: Always look to factor expressions completely before dividing.