CBSE Class 8 Mathematics Factorisation Assignment Set 09

Question. The factors of x²/4 - y²/9 are ____.
(A) (x/4 + y/9) (x/4 - y/9)
(B) (x/2 + y/9) (x/2 - y/9)
(C) (x/2 + y/3) (x/2 - y/3)
(D) Both (A) and (B)
Answer : C

Question. The factors of 15x² – 26x + 8 are ____.
(A) (3x – 4) (5x + 2)
(B) (3x – 4) (5x – 2)
(C) (3x + 4) (5x – 2)
(D) (3x + 4) (5x + 2)
Answer : B

Question. One of the factors of (p + q)² – (a – b)² + p + q – a + b is
(A) (p + q + a + b)
(B) (p + q – a + b)
(C) (p – q + a – b)
(D) (p – q + a + b)
Answer : B

Question. If (x² + 3x + 5) (x² – 3x + 5) = m² – n², then m =____.
(A) x² – 3x
(B) 3x
(C) x² + 5
(D) Both (A) and (B)
Answer : C

Question. Which of the following is the factor of 12(a² + 7a)² – 8(a² + 7a)(²a – 1) – 15 (2a – 1)2 ?
(i) (2a² + 8a + 3) (ii) (6a² + 52a – 5)
(iii) (3a + 5)
(A) Only (i)
(B) Both (i) and (ii)
(C) Only (ii)
(D) All (i), (ii) and (iii)
Answer : B

Question. The factors of x² – 16 are ____.
(A) (x² + 2)(x² – 2)
(B) (x + 4)(x – 4)
(C) (x + 2)(x – 2)
(D) Does not exist
Answer : B

Question. The factors of √3x² + 11x + 6√3 are ____.
(A) (x − 3√3) (√3x + 2)
(B) (x − 3√3) (√3x − 2)
(C) (x + 3√3) (√3x − 2)
(D) (x + 3√3) (√3x + 2)
Answer : D

Question. Factorisation of xy – pq + qy – px is ____.
(A) (y – p) (x + q)
(B) (y – p) (x – q)
(C) (y + p) (x + q)
(D) (y + p) (x – q)
Answer : A

Question. The factors of x⁴ + y⁴ + x²y² are ____.
(A) (x² + y²)(x² + y² – xy)
(B) (x² + y²)(x² – y²)
(C) (x² + y² + xy) (x² + y² – xy)
(D) Factorisation is not possible
Answer : C

Question. For x² + 2x + 5 to be a factor of x⁴ + px² + q, the values of p and q must be ____.
(A) – 2, 5
(B) 5, 25
(C) 10, 20
(D) 6, 25
Answer : D

Question. One of the factors of 4(x + y)(3a – b) + 6 (x + y)(2b – 3a) is
(A) (2b – 3a)
(B) (3a – b)
(C) (4a – 3b)
(D) (– 3a + 4b)
Answer : D

Question. Divide (32x⁴y³ – 16x³y⁴) by (–8x²y)
(A) 4x³y² + 2xy³
(B) 4x³y – 2xy³
(C) – 4x²y² + 2xy³
(D) –4xy² + 2xy³
Answer : C

Question. Factorise: (2x + 3y)² – 5(2x + 3y) – 14.
(A) 4(2x + 3y)(x + y – 2)
(B) 4(2x + 3y)(x + y + 2)
(C) (2x – 3y + 7)(2x – 3y + 2)
(D) (2x + 3y – 7)(2x + 3y + 2)
Answer : D

Question. Simplify : −14x¹²y + 8x⁵z / 2x²
(A) x³(– 7x7y + 4z)
(B) x²(7x7y – 4z)
(C) x²(–7x6y + 2z)
(D) x³(7x7y + 4z)
Answer : A

Question. Do as directed.
(i) Factorise: x² + 1/x² − 3
(ii) Find the greatest common factors of 14x²y³, 21x³y² and 35x⁴y⁵z.
(iii) Divide z(5z² – 80) by 5z(z + 4).

Answer : C

Question. Fill in the blanks.

Answer : D

Question. Which of the following statements is CORRECT?
(A) The factors of an expression are always either algebraic variable or algebraic expression.
(B) An irreducible factor is a factor that cannot be expressed further as a product of factors.
(C) Every binomial expression can be factorised into two monomial expression.
(D) The process of writing a given expression as the product of two or more factors is called multiplication of factors.
Answer : B

Question. Match the expression given in Column-I to one of their factors given in Column-II.
         Column-I               Column-II
P. 9x² + 24x + 16           (i) (2x – 4)
Q. 25x² + 30x + 9          (ii) (4x + 1)
R. 40x² + 14x + 1          (iii) (5x + 3)
S. 4x² – 16x + 16          (iv) (3x + 4)
(A) P → (iv); Q → (iii); R → (ii); S → (i)
(B) P → (iii); Q → (i); R → (iv); S → (ii)
(C) P → (ii); Q → (i); R → (iv); S → (iii)
(D) P → (iv); Q → (iii); R → (i); S → (ii)
Answer : A

Question. Factors of x⁴ – (x – z)⁴ are ____.
(A) (2x + z)(2x³ + z³ – 2x²)
(B) z(x + 2z)(x² + z² – x²)
(C) z(2x – z)(2x² – 2xz + z²)
(D) z(x – 2z)(2z² – 2xz + x²)
Answer : C

Question. Factorising (x – y)² + 4xy – z², we get
(A) (x + y + z)(x + y – z)
(B) (x – y – z)(x + y – z)
(C) (x – y + z)(x + y – z)
(D) None of these
Answer : A

Question. Complete it : x² + 25x + 84 = (x + 21) × _______.
Answer : (x + 4)

Question. Find the product : (a – b – c) (a – b + c)
Answer : a² – 2ab + b² – c²

Question. Simplify : x² - 13x + 40 / (x - 8)
Answer : x – 5

Question. Find the value of 34 – 14 using the identity x⁴ – y⁴ = (x + y) (x – y) (x² + y²).
Answer : 80

Question. Express as product of two linear factors. x² – 17x + 16.
Answer : (x – 16) (x – 1)

Question. Find z if, (x + 4) (x – 5) – z = x(x – 1)
Answer : –20

Question. Factorize : x²(y + 5) –4(y + 5)
Answer : (y + 5) (x + 2) (x – 2)

Question. Divide 121x⁵y⁴z³ by 11x³y²z.
Answer : 11x²y²z²

Question. Find the common factor in 7x²y + 49xy² + 343x³y².
Answer : 7xy

Question. Factorize : x² – 3x + 2.
Answer : (x – 2) (x – 1)

Question. Find the value of  x³y xy³/xy as a product of two factors.
Answer : (x + y) (x – y).

Question. Complete it : (x + 3) (x – 5) = x² + ( )x + ( ).
Answer : –2, –15.

Question. Factorize : 16x² + 40xy + 25y²
Answer : (4x + 5y)²

Question. Simplify : (1.9)² - (1.1)² / 3
Answer : 0.8

Question. Find (x4 + y4) (x⁴ – y4) if x = 1, y = 1.
Answer : 0

Question. Find the value of 252 – 242.
Answer : 49

Question. Simplify : (0.87)² - (0.13)² / (0.87 - 0.13)
Answer : 1

Question. Find the quotient : z(4y² – 64) 4z(y – 4)
Answer : y + 4

Question. Find x and y that xy = 72, x + y = 17.
Answer : x = 9, y = 8.

Question. Simplify : 4m² - 169n² / 2m + 13n
Answer : 2m – 13n

Question. Find the value of : (3a + b)² – (3a – b)².
Answer : 12ab
 

Questions carrying 1 mark each :-

Q.1 Factorise x² + x.

Q.2 Is it correct : 3(x-4) = 3x -4.

Questions carrying 2 marks each :-

Q.3 Factorise 9a² - 16b²

Q.4 Factorise 24x⁴ + 18x³

Q.5 Factorise 2a - 32a⁵

Q.6 Evaluate(ax+ay+ax) a

Questions carrying 3 marks each :

Q.7 Factorise 1-a² - 2ab - b²

Q.8 Resolve into factors : 2x² - 3ax - 2a²

Q.9 Find the greatest common factor of - 4a³b², 12a³b³c, 16a⁵bc.

Q.10 Divide (4x²-9) by (2x-3).

Q.11 Factorise (x⁴) - (y-z)⁴

Multiple Choice Questions carrying 1mark each:-

Q.12 Division is opposite of

a) Subtraction

b) Addition

c) Multiplication

d) None of these

Multiple Choice Questions carrying 1mark each:-

Q.13 Division is opposite of
a) Subtraction
b) Addition
c) Multiplication
d) None of these

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

Exam Tip: Always look for the greatest common divisor (GCD) of the numerical coefficients first.

Question 2. Factorise \(12a^2b + 15ab^2\).
Answer: \(12a^2b + 15ab^2 = 3ab(4a + 5b)\)
In simple words: Find the common numerical factor (3) and the lowest powers of common variables (\(a\) and \(b\)) to factor out \(3ab\).

Exam Tip: For variables with exponents, take the lowest power present in all terms as part of the common factor.

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: The common factor is \(2x^2\). Divide each term by \(2x^2\) to get the remaining expression inside the bracket.

Exam Tip: Check your factorization by expanding the brackets back out to ensure you get the original expression.

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

Exam Tip: Ensure all common variables present across all terms are completely factored out.

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 the terms in pairs, factor out \(x\) from the first pair and 8 from the second pair, then take out the common binomial factor \((x + y)\).

Exam Tip: Grouping by terms is useful when there is no single common factor for all terms.

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 factor by grouping.

Question 7. Factorise \(a^2 + bc + ab + ac\).
Answer: Rearranging terms: \(a^2 + ab + ac + bc = a(a + b) + c(a + b) = (a + c)(a + b)\)
In simple words: Rearrange the terms so that like terms can be grouped easily before factoring.

Exam Tip: Rearrangement is often necessary in grouping problems if the initial order does not show immediate common factors.

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: Write 36 as \(6^2\) and apply the difference of squares identity.

Exam Tip: Memorize standard algebraic identities such as difference of squares thoroughly.

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

Exam Tip: Watch out for negative signs when rearranging terms.

Question 10. Factorise \(5a(2x - 3y) + 2b(2x - 3y)\).
Answer: \(5a(2x - 3y) + 2b(2x - 3y) = (5a + 2b)(2x - 3y)\)
In simple words: Treat the binomial \((2x - 3y)\) as a single common factor and pull it out.

Exam Tip: Binomial expressions can act as single terms for factoring purposes.

Question 11. Factorise \(ax^2 + by^2 + bx^2 + ay^2\).
Answer: Rearranging 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 containing \(x^2\) together and terms containing \(y^2\) together.

Exam Tip: Grouping terms with identical variable parts simplifies the expression quickly.

Question 12. Factorise \(8(4x + 5y)^2 - 12(4x + 5y)\).
Answer: \(8(4x + 5y)^2 - 12(4x + 5y) = 4(4x + 5y)[2(4x + 5y) - 3] = 4(4x + 5y)(8x + 10y - 3)\)
In simple words: Factor out the common numerical factor 4 and the binomial factor \((4x + 5y)\).

Exam Tip: Always extract both the greatest common number and common binomial expression simultaneously.

Question 13. Factorise using identity \(x^2 + 10x + 25\).
Answer: Using identity \(a^2 + 2ab + b^2 = (a + b)^2\):
\(x^2 + 2(x)(5) + 5^2 = (x + 5)^2\)
In simple words: Recognize the expression as a perfect square trinomial.

Exam Tip: Check if the middle term equals \(2 \times \text{first term} \times \text{square root of last term}\).

Question 14. Factorise using identity \(4x^2 + 9y^2 + 12xy\).
Answer: Rearranging into standard form: \(4x^2 + 12xy + 9y^2 = (2x)^2 + 2(2x)(3y) + (3y)^2 = (2x + 3y)^2\)
In simple words: Rewrite as \((2x)^2 + 2(2x)(3y) + (3y)^2\) and apply the square of a sum identity.

Exam Tip: Reorder terms into descending powers before applying identities.

Question 15. Factorise \(ax^2y + bxy^2 + cxyz\).
Answer: \(ax^2y + bxy^2 + cxyz = xy(ax + by + cz)\)
In simple words: Identify \(xy\) as the common factor across all three terms and factor it out.

Question 16. Factorise \(15xy - 6x + 5y - 2\).
Answer: Grouping terms: \(3x(5y - 2) + 1(5y - 2) = (3x + 1)(5y - 2)\)
In simple words: Factor \(3x\) from the first two terms and 1 from the remaining terms.

Exam Tip: Ensure the resulting binomial expressions in both groups match exactly.

Question 17. Factorise \(6pq - 4q + 6 - 9p\).
Answer: Rearranging terms: \(6pq - 9p - 4q + 6 = 3p(2q - 3) - 2(2q - 3) = (3p - 2)(2q - 3)\)
In simple words: Rearrange and factor by grouping carefully handling negative signs.

Exam Tip: Factoring out a negative sign changes the signs of the terms inside the parentheses.

Question 18. Find the factors of \(x^2 - 7x + 12\).
Answer: Splitting the middle term: \(x^2 - 4x - 3x + 12 = x(x - 4) - 3(x - 4) = (x - 3)(x - 4)\)
In simple words: Find two numbers that multiply to 12 and add up to \(-7\), which are \(-4\) and \(-3\).

Exam Tip: Use the middle-term splitting method for quadratic trinomials of the form \(ax^2 + bx + c\).

Question 19. Find the factors of \(3x^2 + 9x + 6\).
Answer: Factoring out 3 first: \(3(x^2 + 3x + 2) = 3(x^2 + 2x + x + 2) = 3[x(x + 2) + 1(x + 2)] = 3(x + 1)(x + 2)\)
In simple words: First factor out the common coefficient 3, then factor the remaining quadratic expression.

Exam Tip: Always check for a common numerical factor before attempting to split the middle term.

Question 20. Factorise \((a + b)^2 - (a - b)^2\).
Answer: Using difference of squares \(x^2 - y^2 = (x - y)(x + y)\):
\([(a + b) - (a - b)][(a + b) + (a - b)] = [a + b - a + b][a + b + a - b] = (2b)(2a) = 4ab\)
In simple words: Apply the difference of squares identity directly using \((a+b)\) and \((a-b)\) as terms.

Exam Tip: Be careful with distributing negative signs when simplifying bracketed expressions.

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

Exam Tip: Grouping terms in pairs is the standard technique for four-term polynomials.

Question 22. Divide \(24x^2y^2z^2\) by \(6yz\).
Answer: \(\frac{24x^2y^2z^2}{6yz} = 4x^2y^{(2-1)}z^{(2-1)} = 4x^2yz\)
In simple words: Divide the coefficients and subtract the exponents of matching variables.

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

Question 23. Divide \((7a^2 + 14a)\) by \((a + 2)\).
Answer: \(\frac{7a^2 + 14a}{a + 2} = \frac{7a(a + 2)}{a + 2} = 7a\)
In simple words: Factor out \(7a\) from the numerator so that \((a + 2)\) cancels out with the denominator.

Exam Tip: Always factor the numerator completely before attempting cancellation.

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

Exam Tip: Factor algebraic expressions fully before reducing fractions.

Question 25. Divide \(x^2 + 7x + 10\) by \(x + 5\).
Answer: Factorising the numerator: \(x^2 + 7x + 10 = (x + 5)(x + 2)\)
\(\frac{(x + 5)(x + 2)}{x + 5} = x + 2\)
In simple words: Factor the quadratic numerator by splitting the middle term, then cancel the common factor \((x + 5)\).

Exam Tip: Splitting the middle term is an essential prerequisite for algebraic division.

Question 26. Divide \(12pq(9p^2 - 16q^2)\) by \(4pq(3p + 4q)\).
Answer: \(\frac{12pq(9p^2 - 16q^2)}{4pq(3p + 4q)} = \frac{12pq(3p - 4q)(3p + 4q)}{4pq(3p + 4q)} = 3(3p - 4q) = 9p - 12q\)
In simple words: Write \(9p^2 - 16q^2\) as \((3p)^2 - (4q)^2\) and apply the difference of squares identity.

Exam Tip: Cancel common monomial coefficients and identical binomial factors simultaneously.

Question 27. Divide \(39x^3(50x^2 - 98)\) by \(26x^2(5x + 7)\).
Answer: Factor numerator: \(39x^3 \cdot 2(25x^2 - 49) = 78x^3(5x - 7)(5x + 7)\)
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 2 from the bracket, factor using difference of squares, and divide coefficients and variables.

Question 28. Find the factors of \(25x^2 - 4y^2 + 28yz - 49z^2\).
Answer: Rearranging terms: \(25x^2 - (4y^2 - 28yz + 49z^2) = (5x)^2 - [(2y)^2 - 2(2y)(7z) + (7z)^2] = (5x)^2 - (2y - 7z)^2\)
Using difference of squares: \([5x - (2y - 7z)][5x + (2y - 7z)] = (5x - 2y + 7z)(5x + 2y - 7z)\)
In simple words: Group the last three terms into a perfect square trinomial and apply the difference of squares identity.

Exam Tip: Recognizing grouped perfect square trinomials is key for complex factorization 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: \(x^2\), \(xy\), and constant terms.

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

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: Subtract each term of the first expression from the second expression, remembering to distribute the negative sign.

Exam Tip: The phrase "Subtract A from B" means writing it as \(B - A\).

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 (FOIL method) and combine like terms.

Exam Tip: Combine middle terms \(-49xy\) and \(-12xy\) carefully.

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: Square the first term, subtract twice the product of both terms, and add the square of the second term.

Exam Tip: Ensure the middle term includes the negative sign prescribed by the identity.

Question 5. Using identity \((a - b)^2 = a^2 - 2ab + b^2\) find the value of \(98^2\).
Answer: \(98^2 = (100 - 2)^2 = 100^2 - 2(100)(2) + 2^2\)
\(= 10000 - 400 + 4 = 9604\)
In simple words: Express 98 as \((100 - 2)\) and expand using the square of a difference identity.

Exam Tip: Converting numbers to powers of 10 makes mental arithmetic much easier.

Question 6. Subtract \(3x(x - 4y + 5z)\) from \(4x(2x - 3y + 10z)\).
Answer: Expand first expression: \(4x(2x - 3y + 10z) = 8x^2 - 12xy + 40xz\)
Expand second expression: \(3x(x - 4y + 5z) = 3x^2 - 12xy + 15xz\)
Subtract: \((8x^2 - 12xy + 40xz) - (3x^2 - 12xy + 15xz) = 8x^2 - 12xy + 40xz - 3x^2 + 12xy - 15xz = 5x^2 + 25xz\)
In simple words: Multiply out both brackets completely and then subtract like terms.

Exam Tip: Distribute the monomial outside the bracket to every term inside before subtracting.

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

Exam Tip: Double-check signs and variable exponents during multiplication.

Question 8. Using suitable identity find \((6x^2 - \frac{5}{3})^2\).
Answer: Using \((a - b)^2 = a^2 - 2ab + b^2\):
\((6x^2)^2 - 2(6x^2)(\frac{5}{3}) + (\frac{5}{3})^2 = 36x^4 - 20x^2 + \frac{25}{9}\)
In simple words: Apply the square of a difference identity directly with fractional terms.

Exam Tip: Simplify fractions carefully when multiplying the middle term: \(2 \times 6 \times \frac{5}{3} = 20\).

Question 9. Simplify \((xy + yz)^2 - (xy - yz)^2\).
Answer: Using difference of squares \(A^2 - B^2 = (A - B)(A + B)\):
\([(xy + yz) - (xy - yz)][(xy + yz) + (xy - yz)] = [xy + yz - xy + yz][xy + yz + xy - yz]\)
\(= (2yz)(2xy) = 4xy^2z\)
In simple words: Use the difference of squares identity to factor the expression quickly.

Question 10. Using identity find the product of \((\frac{a}{2} + \frac{3b}{4})(\frac{a}{2} + \frac{3b}{4})\).
Answer: \((\frac{a}{2} + \frac{3b}{4})^2 = (\frac{a}{2})^2 + 2(\frac{a}{2})(\frac{3b}{4}) + (\frac{3b}{4})^2\)
\(= \frac{a^2}{4} + \frac{3ab}{4} + \frac{9b^2}{16}\)
In simple words: Since the expression is multiplied by itself, apply the square of a sum identity \((a + b)^2\).

Exam Tip: Square both the numerator and denominator of fractional terms.

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

Exam Tip: Remember that \(x \times \frac{1}{x} = 1\).

Question 12. Find the product of \((\frac{5}{3}x + \frac{3}{4}y)^2\) and \((\frac{5}{3}x + \frac{3}{4}y)^2\) and also evaluate it when \(x = +2\) and \(y = -1\).
Answer: Product: \((\frac{5}{3}x + \frac{3}{4}y)^2 \cdot (\frac{5}{3}x + \frac{3}{4}y)^2 = (\frac{5}{3}x + \frac{3}{4}y)^4\)
Substitute \(x = 2\) and \(y = -1\):
\(\frac{5}{3}(2) + \frac{3}{4}(-1) = \frac{10}{3} - \frac{3}{4} = \frac{40 - 9}{12} = \frac{31}{12}\)
Evaluating fourth power: \((\frac{31}{12})^4 = \frac{923521}{20736}\)
In simple words: Combine the identical terms using exponent rules, then substitute the given values of \(x\) and \(y\).

Exam Tip: Substitute values into the base expression before raising it to higher powers to keep calculations manageable.

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

Exam Tip: This identity is extremely useful for mental calculation of squares of numbers.

Question 14. If \(x + y = 12\) and \(xy = 32\). Find the value of \(x^2 + y^2\).
Answer: Using \((x + y)^2 = x^2 + 2xy + y^2\):
\(12^2 = x^2 + 2(32) + y^2\)
\(144 = x^2 + 64 + y^2\)
\(x^2 + y^2 = 144 - 64 = 80\)
In simple words: Square the equation \((x + y)\) and substitute the given value of \(xy\).

Exam Tip: Memorize the expansion of \((x+y)^2\) in terms of sum and product.

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: \(x^2y^2 + 2xy^2z + y^2z^2 - 2x^2y^2z\)
Substitute \(x = -1\), \(y = 1\), \(z = 2\):
\((-1)^2(1)^2 + 2(-1)(1)^2(2) + (1)^2(2)^2 - 2(-1)^2(1)^2(2)\)
\(= 1(1) + 2(-1)(1)(2) + 1(4) - 2(1)(1)(2)\)
\(= 1 - 4 + 4 - 4 = -3\)
In simple words: Expand the binomial square, combine terms if any, and substitute the given values.

Exam Tip: Be careful with negative signs when evaluating powers.

Question 16. Subtract \(4x^2 - 2y + 7z^3 - 3\) from \(3y + 7x^2 - 2z^3 + 4\).
Answer: \((3y + 7x^2 - 2z^3 + 4) - (4x^2 - 2y + 7z^3 - 3)\)
\(= 3y + 7x^2 - 2z^3 + 4 - 4x^2 + 2y - 7z^3 + 3\)
\(= (7x^2 - 4x^2) + (3y + 2y) + (-2z^3 - 7z^3) + (4 + 3)\)
\(= 3x^2 + 5y - 9z^3 + 7\)
In simple words: Distribute the negative sign across all terms in the first polynomial and group like terms.

Exam Tip: Align like terms with identical variable parts to prevent subtraction errors.

Question 17. Divide \(-72x^2yz\) by \(-12xyz\).
Answer: \(\frac{-72x^2yz}{-12xyz} = \frac{-72}{-12} \cdot \frac{x^2}{x} \cdot \frac{y}{y} \cdot \frac{z}{z} = 6x^{(2-1)}y^{(1-1)}z^{(1-1)} = 6x\)
In simple words: Divide the numerical coefficients and subtract exponents of like variables.

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 numerator separately by the monomial denominator.

Exam Tip: Split polynomial division into separate monomial division steps.

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. \(\frac{\frac{2}{3}a^2b^2c^2}{\frac{1}{2}abc} = \frac{2}{3} \times 2 \cdot a^{(2-1)}b^{(2-1)}c^{(2-1)} = \frac{4}{3}abc\)
2. \(\frac{\frac{4}{3}ab^2c^3}{\frac{1}{2}abc} = \frac{4}{3} \times 2 \cdot a^{(1-1)}b^{(2-1)}c^{(3-1)} = \frac{8}{3}bc^2\)
3. \(\frac{-\frac{1}{5}ab^3c^2}{\frac{1}{2}abc} = -\frac{1}{5} \times 2 \cdot a^{(1-1)}b^{(3-1)}c^{(2-1)} = -\frac{2}{5}bc\)
Combined Answer: \(\frac{4}{3}abc + \frac{8}{3}bc^2 - \frac{2}{5}bc^2\) (or grouped as \(\frac{4}{3}abc + (\frac{8}{3} - \frac{2}{5})bc^2 = \frac{4}{3}abc + \frac{34}{15}bc^2\))
In simple words: Divide each fraction coefficient by \(\frac{1}{2}\) (which means multiplying by 2) and subtract exponents.

Exam Tip: Dividing by a fraction is equivalent to multiplying by its reciprocal.

Question 20. Divide \(6 + x - 4x^2 + x^3\) by \(x - 3\) by long division method.
Answer: Arrange dividend in standard descending order: \(x^3 - 4x^2 + x + 6\)
Long division steps:
1. Divide leading term: \(\frac{x^3}{x} = x^2\). Multiply \(x^2(x - 3) = x^3 - 3x^2\). Subtract to get \(-x^2 + x\).
2. Divide leading term: \(\frac{-x^2}{x} = -x\). Multiply \(-x(x - 3) = -x^2 + 3x\). Subtract to get \(-2x + 6\).
3. Divide leading term: \(\frac{-2x}{x} = -2\). Multiply \(-2(x - 3) = -2x + 6\). Remainder is \(0\).
Quotient = \(x^2 - x - 2\), Remainder = \(0\)
In simple words: Use algebraic long division by matching leading terms at each step.

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

Question 21. Divide \(12x^3 - 8x^2 - 6x + 10\) by \(3x - 2\). Write the quotient and remainder.
Answer: Long division steps:
1. \(\frac{12x^3}{3x} = 4x^2\). Multiply: \(4x^2(3x - 2) = 12x^3 - 8x^2\). Subtract: remainder terms leave \(-6x + 10\).
2. \(\frac{-6x}{3x} = -2\). Multiply: \(-2(3x - 2) = -6x + 4\). Subtract: \((-6x + 10) - (-6x + 4) = 6\).
Quotient = \(4x^2 - 2\), Remainder = \(6\)
In simple words: Divide step-by-step using polynomial long division.

Exam Tip: Pay close attention when subtraction eliminates intermediate terms.

Question 22. Divide \(x^3 - 6x^2 + 11x - 6\) by \(x^2 - 4x + 3\).
Answer: Long division steps:
1. \(\frac{x^3}{x^2} = x\). Multiply: \(x(x^2 - 4x + 3) = x^3 - 4x^2 + 3x\).
2. Subtract: \((x^3 - 6x^2 + 11x - 6) - (x^3 - 4x^2 + 3x) = -2x^2 + 8x - 6\).
3. \(\frac{-2x^2}{x^2} = -2\). Multiply: \(-2(x^2 - 4x + 3) = -2x^2 + 8x - 6\).
4. Subtract: Remainder is \(0\).
Quotient = \(x - 2\), Remainder = \(0\)
In simple words: Divide the leading terms of the dividend and divisor successively until the degree of the remainder is less than the divisor.

Exam Tip: Verify your answer by multiplying the quotient by the divisor.

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

Exam Tip: The Factor Theorem states that \((x - a)\) is a factor if and only if \(P(a) = 0\).

Question 24. Divide the following \(63a^2b^4c\) by \(7a^2b^2c\).
Answer: \(\frac{63a^2b^4c}{7a^2b^2c} = \frac{63}{7} \cdot \frac{a^2}{a^2} \cdot \frac{b^4}{b^2} \cdot \frac{c}{c} = 9a^{(2-2)}b^{(4-2)}c^{(1-1)} = 9b^2\)
In simple words: Divide coefficients and subtract exponents of identical variables.

Exam Tip: Any variable raised to the power of 0 equals 1 (\(a^0 = 1\)).

Question 25. Divide the polynomial \(5x(x^2 - x + 1) - (9 + 4x^4)\) by \(4x - 1\). Write quotient and remainder.
Answer: Expand and rearrange dividend in standard form:
\(5x^3 - 5x^2 + 5x - 9 - 4x^4 = -4x^4 + 5x^3 - 5x^2 + 5x - 9\)
Performing long division by \(4x - 1\):
1. \(\frac{-4x^4}{4x} = -x^3\). Multiply: \(-x^3(4x - 1) = -4x^4 + x^3\). Subtract: \(4x^3 - 5x^2\).
2. \(\frac{4x^3}{4x} = x^2\). Multiply: \(x^2(4x - 1) = 4x^3 - x^2\). Subtract: \(-4x^2 + 5x\).
3. \(\frac{-4x^2}{4x} = -x\). Multiply: \(-x(4x - 1) = -4x^2 + x\). Subtract: \(4x - 9\).
4. \(\frac{4x}{4x} = 1\). Multiply: \(1(4x - 1) = 4x - 1\). Subtract: \(-9 - (-1) = -8\).
Quotient = \(-x^3 + x^2 - x + 1\), Remainder = \(-8\)
In simple words: Expand the polynomial, arrange it in descending order of powers, and apply long division.

Exam Tip: Be meticulous with signs when subtracting polynomials during long division.

Question 26. Divide the polynomial \(x^3 + 3x^2 - 5x + 4\) by \(x - 1\). Write quotient and remainder.
Answer: Long division steps:
1. \(\frac{x^3}{x} = x^2\). Multiply: \(x^2(x - 1) = x^3 - x^2\). Subtract: \(4x^2 - 5x\).
2. \(\frac{4x^2}{x} = 4x\). Multiply: \(4x(x - 1) = 4x^2 - 4x\). Subtract: \(-x + 4\).
3. \(\frac{-x}{x} = -1\). Multiply: \(-1(x - 1) = -x + 1\). Subtract: \(4 - 1 = 3\).
Quotient = \(x^2 + 4x - 1\), Remainder = \(3\)
In simple words: Divide step-by-step using long division.

Exam Tip: Check your division using the division algorithm: \(\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}\).

Question 27. Divide \(z(5z^2 - 80)\) by \(5z(z + 4)\).
Answer: \(\frac{z(5z^2 - 80)}{5z(z + 4)} = \frac{5z(z^2 - 16)}{5z(z + 4)} = \frac{5z(z - 4)(z + 4)}{5z(z + 4)} = z - 4\)
In simple words: Factor out 5 from the bracket, apply the difference of squares identity on \(z^2 - 16\), and cancel common factors.

Exam Tip: Always factor completely before cancelling terms in rational expressions.