CBSE Class 8 Mathematics Algebraic Expressions And Identities Worksheet Set 06

Chap. 9 ALGEBRAIC EXPRESSIONS AND IDENTITIES

1. Multiply the following.

Question 1(i). \(11x^2y\) and \(2x^2y^2\)
Answer: \(11x^2y \times 2x^2y^2\)
\(= (11 \times 2) \times (x^2 \cdot x^2) \times (y \cdot y^2)\)
\(= 22x^{2+2}y^{1+2}\)
\(= 22x^4y^3\)
In simple words: Multiply the numbers together and add the exponents of the same variables.

Exam Tip: Use the laws of exponents \(x^m \times x^n = x^{m+n}\) when multiplying terms with the same base.

Question 1(ii). \(3y^2\) and \(7y^5\)
Answer: \(3y^2 \times 7y^5\)
\(= (3 \times 7) \times (y^2 \cdot y^5)\)
\(= 21y^{2+5}\)
\(= 21y^7\)
In simple words: Multiply the coefficients 3 and 7, and add the powers of y.

Exam Tip: Ensure that only variables with the same base have their exponents added.

Question 1(iii). \(5x^3\) and \(4x^9\)
Answer: \(5x^3 \times 4x^9\)
\(= (5 \times 4) \times (x^3 \cdot x^9)\)
\(= 20x^{3+9}\)
\(= 20x^{12}\)
In simple words: Multiply 5 by 4 and add the exponents 3 and 9 of variable x.

Exam Tip: Double-check the addition of large exponents.

Question 1(iv). \(-9xy\) and \(4x^2z\)
Answer: \(-9xy \times 4x^2z\)
\(= (-9 \times 4) \times (x \cdot x^2) \times y \times z\)
\(= -36x^{1+2}yz\)
\(= -36x^3yz\)
In simple words: Multiply the coefficients to get -36, combine x terms to get \(x^3\), and write y and z as they are.

Question 2(i). \((-2xy^2)(5y)(-3z^2)\)
Answer: \(= (-2 \times 5 \times -3) \times x \times (y^2 \cdot y) \times z^2\)
\(= 30 \times x \times y^{2+1} \times z^2\)
\(= 30xy^3z^2\)
In simple words: Multiply all constants together, keeping track of negative signs, then combine like variables.

Exam Tip: The product of two negative numbers is positive: \((-2) \times (-3) = 6\).

Question 2(ii). \((ab)(bc)(ca)\)
Answer: \(= (a \cdot a) \times (b \cdot b) \times (c \cdot c)\)
\(= a^{1+1}b^{1+1}c^{1+1}\)
\(= a^2b^2c^2\)
In simple words: Group like variables together and add their exponents.

Exam Tip: Each variable appears twice, so each gets an exponent of 2.

Question 2(iii). \((6a^2b)(-2b^2c)(3ac^2)\)
Answer: \(= (6 \times -2 \times 3) \times (a^2 \cdot a) \times (b \cdot b^2) \times (c \cdot c^2)\)
\(= -36 \times a^{2+1} \times b^{1+2} \times c^{1+2}\)
\(= -36a^3b^3c^3\)
In simple words: Multiply numerical coefficients to get -36, and add exponents for a, b, and c.

Exam Tip: Ensure every variable has its exponent correctly updated from both factors.

Question 2(iv). \(\left(\frac{5}{9}ab\right)\left(\frac{9}{5}bc\right)\left(\frac{-7}{5}ca\right)\)
Answer: \(= \left(\frac{5}{9} \times \frac{9}{5} \times \frac{-7}{5}\right) \times (a \cdot a) \times (b \cdot b) \times (c \cdot c)\)
\(= \left(1 \times \frac{-7}{5}\right) \times a^2b^2c^2\)
\(= -\frac{7}{5}a^2b^2c^2\)
In simple words: Cancel out common factors in numerators and denominators of the fractions before multiplying.

Exam Tip: Cross-cancel numbers to simplify fraction multiplication quickly.

Question 3. Find the value of \((3p^2q) \times (8q^3)\), when \(p = 1\) and \(q = \left(-\frac{1}{4}\right)\).
Answer: First simplify the algebraic expression:
\((3p^2q) \times (8q^3) = (3 \times 8) \times p^2 \times (q \cdot q^3)\)
\(= 24p^2q^4\)
Now substitute \(p = 1\) and \(q = -\frac{1}{4}\):
\(= 24(1)^2 \left(-\frac{1}{4}\right)^4\)
\(= 24 \times 1 \times \frac{1}{256}\)
\(= \frac{24}{256} = \frac{3}{32}\)
In simple words: Simplify the expression first to make substitution easier, then plug in the given numbers.

Exam Tip: Always simplify the expression algebraically before substituting numerical values to avoid lengthy calculations.

Question 4. Find the value of \(\left(-8x^2y^3\right) \times \left(\frac{1}{4}xy^2\right)\), when \(x = -1\) and \(y = 2\).
Answer: Simplify the expression first:
\(\left(-8 \times \frac{1}{4}\right) \times (x^2 \cdot x) \times (y^3 \cdot y^2)\)
\(= -2x^3y^5\)
Now substitute \(x = -1\) and \(y = 2\):
\(= -2(-1)^3(2)^5\)
\(= -2(-1)(32)\)
\(= 2 \times 32 = 64\)
In simple words: Combine terms to get \(-2x^3y^5\), then substitute -1 for x and 2 for y.

Exam Tip: Be careful with negative signs when raising a negative number to an odd power: \((-1)^3 = -1\).

Question 5. Find the product of \((3a^2b^3), (-7a^2)\) and \((5a^2b^2)\), and then verify the result for \(a = 2\) and \(b = 3\).
Answer: Product:
\(= (3 \times -7 \times 5) \times (a^2 \cdot a^2 \cdot a^2) \times (b^3 \cdot b^2)\)
\(= -105a^{2+2+2}b^{3+2}\)
\(= -105a^6b^5\)
Verification:
LHS = Product of individual terms with values substituted:
\([3(2)^2(3)^3] \times [-7(2)^2] \times [5(2)^2(3)^2]\)
\(= [3 \times 4 \times 27] \times [-7 \times 4] \times [5 \times 4 \times 9]\)
\(= 324 \times (-28) \times 180 = -1632960\)
RHS = Value substituted in final product expression:
\(= -105(2)^6(3)^5\)
\(= -105 \times 64 \times 243 = -1632960\)
Since LHS = RHS, verified.
In simple words: Multiply the expressions to get the algebraic product, then check by putting the numbers into both the original product and the final answer.

Exam Tip: Verification requires showing that evaluating before and after multiplication yields the identical number.

Question 6. Find the product of \(\left(-\frac{3}{4}x^2y^2z\right)\) and \((-2z^2)\), then verify the result for \(x = 1, y = 2\) and \(z = 3\).
Answer: Product:
\(= \left(-\frac{3}{4} \times -2\right) \times x^2 \times y^2 \times (z \cdot z^2)\)
\(= \frac{3}{2}x^2y^2z^3\)
Verification:
LHS:
\(\left(-\frac{3}{4}(1)^2(2)^2(3)\right) \times (-2(3)^2)\)
\(= \left(-\frac{3}{4} \times 1 \times 4 \times 3\right) \times (-2 \times 9)\)
\(= (-9) \times (-18) = 162\)
RHS:
\(\frac{3}{2}(1)^2(2)^2(3)^3 = \frac{3}{2} \times 1 \times 4 \times 27 = 6 \times 27 = 162\)
Since LHS = RHS, verified.
In simple words: Multiply the terms to find the algebraic expression, then substitute the given values to confirm both sides match.

Exam Tip: Watch out for fraction multiplication signs and powers during substitution.

Question 7. Verify \(a^2b^2c^2 = (ab) \times (bc) \times (ca)\), for \(a = 3\) and \(b = 4\).
Answer: Note: The problem implies evaluating with \(a = 3, b = 4\). Let us assume \(c = 1\) or verify the identity algebraically.
Algebraic proof:
RHS = \((ab) \times (bc) \times (ca) = a \cdot a \cdot b \cdot b \cdot c \cdot c = a^2b^2c^2 =\) LHS.
Since LHS = RHS identically for all values of a, b, and c, it is verified.
In simple words: Re-arranging the terms on the right side shows it is identical to the left side regardless of values.

Exam Tip: Algebraic identities hold true for any real numbers.

8. Find the product and then verify the following for \(a = 2\) and \(b = -5\).

Question 8(i). \(a(a^2 - ab^2)\)
Answer: Product:
\(a(a^2 - ab^2) = a \cdot a^2 - a \cdot ab^2 = a^3 - a^2b^2\)
Verification:
LHS: \(2((2)^2 - (2)(-5)^2) = 2(4 - 2(25)) = 2(4 - 50) = 2(-46) = -92\)
RHS: \((2)^3 - (2)^2(-5)^2 = 8 - (4)(25) = 8 - 100 = -92\)
Since LHS = RHS, verified.
In simple words: Distribute a into both terms inside the parentheses, then check by substituting the values.

Exam Tip: Use distributive property \(a(b - c) = ab - ac\).

Question 8(ii). \(\frac{2}{7}a\left(ab - \frac{7}{6}ab^2\right)\)
Answer: Product:
\(= \left(\frac{2}{7}a \cdot ab\right) - \left(\frac{2}{7}a \cdot \frac{7}{6}ab^2\right)\)
\(= \frac{2}{7}a^2b - \frac{1}{3}a^2b^2\)
Verification for \(a = 2, b = -5\):
LHS:
\(\frac{2}{7}(2)\left((2)(-5) - \frac{7}{6}(2)(-5)^2\right)\)
\(= \frac{4}{7}\left(-10 - \frac{7}{6}(2)(25)\right)\)
\(= \frac{4}{7}\left(-10 - \frac{175}{3}\right) = \frac{4}{7}\left(\frac{-30 - 175}{3}\right) = \frac{4}{7}\left(\frac{-205}{3}\right) = \frac{-820}{21}\)
RHS:
\(\frac{2}{7}(2)^2(-5) - \frac{1}{3}(2)^2(-5)^2\)
\(= \frac{2}{7}(4)(-5) - \frac{1}{3}(4)(25)\)
\(= \frac{-40}{7} - \frac{100}{3} = \frac{-120 - 700}{21} = \frac{-820}{21}\)
Since LHS = RHS, verified.
In simple words: Multiply outside term with each term inside, simplify fractions, and substitute values.

Exam Tip: Take common denominators carefully when adding or subtracting fractions during verification.

9. Find the product of the following.

Question 9(i). \(2x(3x + y^2)\)
Answer: \(= (2x \cdot 3x) + (2x \cdot y^2)\)
\(= 6x^2 + 2xy^2\)
In simple words: Multiply \(2x\) by each term inside the bracket.

Exam Tip: Distribute monomial across binomial terms properly.

Question 9(ii). \((-3y)(x^2 + 3xy)\)
Answer: \(= (-3y \cdot x^2) + (-3y \cdot 3xy)\)
\(= -3x^2y - 9xy^2\)
In simple words: Distribute \(-3y\) to both terms inside.

Exam Tip: Pay attention to signs: negative times positive gives negative.

Question 9(iii). \(3a^2(4a - 5a^2)\)
Answer: \(= (3a^2 \cdot 4a) - (3a^2 \cdot 5a^2)\)
\(= 12a^3 - 15a^4\)
In simple words: Multiply coefficients and add exponents for variable a.

Exam Tip: Keep terms in standard descending order of powers if required.

Question 9(iv). \(-8a^2b(-3a^2 - 2b)\)
Answer: \(= (-8a^2b \cdot -3a^2) + (-8a^2b \cdot -2b)\)
\(= 24a^4b + 16a^2b^2\)
In simple words: Negative multiplied by negative makes positive.

Exam Tip: Double check sign changes when distributing a negative monomial.

Question 9(v). \(-\frac{5}{9}abc\left(\frac{18}{15}a^2bc - \frac{3}{10}abc^2\right)\)
Answer: \(= \left(-\frac{5}{9} \times \frac{18}{15}\right)a^{1+2}b^{1+1}c^{1} - \left(-\frac{5}{9} \times \frac{3}{10}\right)a^{1+1}b^{1+1}c^{1+2}\)
\(= \left(-\frac{2}{3}\right)a^3b^2c + \left(\frac{1}{6}\right)a^2b^2c^3\)
\(= -\frac{2}{3}a^3b^2c + \frac{1}{6}a^2b^2c^3\)
In simple words: Multiply coefficients fraction by fraction and apply laws of exponents to each variable.

Question 9(vi). \(7a(0.1a^2 - 0.5b)\)
Answer: \(= (7a \cdot 0.1a^2) - (7a \cdot 0.5b)\)
\(= 0.7a^3 - 3.5ab\)
In simple words: Multiply 7 by 0.1 and 0.5 respectively.

Exam Tip: Place decimal points correctly in the resulting products.

Question 10. Multiply \(\frac{5}{9}y^2z, \frac{7}{10}x^2\) and \((-3xz^2)\), and then verify the result for \(x = \frac{1}{2}, y = \frac{1}{3}\) and \(z = \frac{1}{4}\).
Answer: Product:
\(= \left(\frac{5}{9} \times \frac{7}{10} \times -3\right) \times x^2 \cdot x \times y^2 \times z \cdot z^2\)
\(= \left(-\frac{7}{6}\right)x^3y^2z^3\)
Verification:
LHS:
\(\left(\frac{5}{9}\left(\frac{1}{3}\right)^2\left(\frac{1}{4}\right)\right) \times \left(\frac{7}{10}\left(\frac{1}{2}\right)^2\right) \times \left(-3\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)^2\right)\)
\(= \left(\frac{5}{9} \times \frac{1}{9} \times \frac{1}{4}\right) \times \left(\frac{7}{10} \times \frac{1}{4}\right) \times \left(-3 \times \frac{1}{2} \times \frac{1}{16}\right)\)
\(= \frac{5}{324} \times \frac{7}{40} \times \left(-\frac{3}{32}\right) = -\frac{105}{414720} = -\frac{7}{27648}\)
RHS:
\(-\frac{7}{6}\left(\frac{1}{2}\right)^3\left(\frac{1}{3}\right)^2\left(\frac{1}{4}\right)^3\)
\(= -\frac{7}{6} \times \frac{1}{8} \times \frac{1}{9} \times \frac{1}{64} = -\frac{7}{27648}\)
Since LHS = RHS, verified.
In simple words: Multiply monomials together, then test with fractions.

Exam Tip: Careful arithmetic with fractions is essential for multi-variable verifications.

11. Find the following products and verify the results for \(x = -1\) and \(y = -2\).

Question 11(i). \((3x^2 + 2y^2)(x + y)\)
Answer: Product:
\(= 3x^2(x + y) + 2y^2(x + y)\)
\(= 3x^3 + 3x^2y + 2xy^2 + 2y^3\)
Verification for \(x = -1, y = -2\):
LHS: \((3(-1)^2 + 2(-2)^2)(-1 + (-2)) = (3(1) + 2(4))(-3) = (3 + 8)(-3) = 11(-3) = -33\)
RHS: \(3(-1)^3 + 3(-1)^2(-2) + 2(-1)(-2)^2 + 2(-2)^3\)
\(= 3(-1) + 3(1)(-2) + 2(-1)(4) + 2(-8)\)
\(= -3 - 6 - 8 - 16 = -33\)
Since LHS = RHS, verified.
In simple words: Multiply each term of the first binomial by each term of the second binomial, then check using values.

Exam Tip: FOIL method (First, Outer, Inner, Last) applies to multiplying two binomials.

Question 11(ii). \((x^2 - y^2)(x^2 + y^2)\)
Answer: Product:
\(= x^2(x^2 + y^2) - y^2(x^2 + y^2)\)
\(= x^4 + x^2y^2 - x^2y^2 - y^4\)
\(= x^4 - y^4\)
Verification for \(x = -1, y = -2\):
LHS: \(((-1)^2 - (-2)^2)((-1)^2 + (-2)^2) = (1 - 4)(1 + 4) = (-3)(5) = -15\)
RHS: \((-1)^4 - (-2)^4 = 1 - 16 = -15\)
Since LHS = RHS, verified.
In simple words: This is an application of \((a-b)(a+b) = a^2-b^2\).

Exam Tip: Recognize standard algebraic identities to shorten calculations.

Question 11(iii). \(\left(3x^2 + \frac{1}{3}y^2\right)(2y - 3x^2)\)
Answer: Product:
\(= 3x^2(2y - 3x^2) + \frac{1}{3}y^2(2y - 3x^2)\)
\(= 6x^2y - 9x^4 + \frac{2}{3}y^3 - x^2y^2\)
Verification for \(x = -1, y = -2\):
LHS: \(\left(3(-1)^2 + \frac{1}{3}(-2)^2\right)(2(-2) - 3(-1)^2)\)
\(= \left(3 + \frac{4}{3}\right)(-4 - 3) = \left(\frac{13}{3}\right)(-7) = -\frac{91}{3}\)
RHS: \(6(-1)^2(-2) - 9(-1)^4 + \frac{2}{3}(-2)^3 - (-1)^2(-2)^2\)
\(= 6(1)(-2) - 9(1) + \frac{2}{3}(-8) - (1)(4)\)
\(= -12 - 9 - \frac{16}{3} - 4 = -25 - \frac{16}{3} = \frac{-75 - 16}{3} = -\frac{91}{3}\)
Since LHS = RHS, verified.
In simple words: Expand carefully term by term and evaluate fractions.

Question 11(iv). \((x^4 - y^4)(x + y)\)
Answer: Product:
\(= x^4(x + y) - y^4(x + y)\)
\(= x^5 + x^4y - xy^4 - y^5\)
Verification for \(x = -1, y = -2\):
LHS: \(((-1)^4 - (-2)^4)(-1 + (-2)) = (1 - 16)(-3) = (-15)(-3) = 45\)
RHS: \((-1)^5 + (-1)^4(-2) - (-1)(-2)^4 - (-2)^5\)
\(= -1 + (1)(-2) - (-1)(16) - (-32)\)
\(= -1 - 2 + 16 + 32 = 45\)
Since LHS = RHS, verified.
In simple words: Multiply each term of the binomial into the polynomial.

Exam Tip: Pay special attention to signs when substituting negative values into high powers.

Question 11(v). \(\left(\frac{1}{3}x - y\right)\left(\frac{3}{5}x + y\right)\)
Answer: Product:
\(= \frac{1}{3}x\left(\frac{3}{5}x + y\right) - y\left(\frac{3}{5}x + y\right)\)
\(= \frac{1}{5}x^2 + \frac{1}{3}xy - \frac{3}{5}xy - y^2\)
\(= \frac{1}{5}x^2 + \left(\frac{1}{3} - \frac{3}{5}\right)xy - y^2\)
\(= \frac{1}{5}x^2 + \left(\frac{5 - 9}{15}\right)xy - y^2\)
\(= \frac{1}{5}x^2 - \frac{4}{15}xy - y^2\)
Verification for \(x = -1, y = -2\):
LHS: \(\left(\frac{1}{3}(-1) - (-2)\right)\left(\frac{3}{5}(-1) + (-2)\right)\)
\(= \left(-\frac{1}{3} + 2\right)\left(-\frac{3}{5} - 2\right) = \left(\frac{5}{3}\right)\left(-\frac{13}{5}\right) = -\frac{13}{3}\)
RHS: \(\frac{1}{5}(-1)^2 - \frac{4}{15}(-1)(-2) - (-2)^2\)
\(= \frac{1}{5}(1) - \frac{4}{15}(2) - 4 = \frac{1}{5} - \frac{8}{15} - 4 = \frac{3 - 8 - 60}{15} = \frac{-65}{15} = -\frac{13}{3}\)
Since LHS = RHS, verified.
In simple words: Combine like terms like \(xy\) after multiplying.

Exam Tip: Combine like middle terms by finding a common denominator.

Question 11(vi). \((0.7x - 0.6y)(2.3x - 2y)\)
Answer: Product:
\(= 0.7x(2.3x - 2y) - 0.6y(2.3x - 2y)\)
\(= 1.61x^2 - 1.4xy - 1.38xy + 1.2y^2\)
\(= 1.61x^2 - 2.78xy + 1.2y^2\)
In simple words: Multiply decimal coefficients carefully and add like terms \(-1.4xy\) and \(-1.38xy\).

Exam Tip: Double-check decimal multiplication placement.

12. Find the products of the following.

Question 12(i). \((2x - y)(5x^2 + 6x + 2)\)
Answer: \(= 2x(5x^2 + 6x + 2) - y(5x^2 + 6x + 2)\)
\(= 10x^3 + 12x^2 + 4x - 5x^2y - 6xy - 2y\)
In simple words: Distribute each term of the binomial to every term in the trinomial.

Exam Tip: A binomial times a trinomial yields 6 terms before combining any likes.

Question 12(ii). \((x^2 + y^2 + z^2)(xy + yz)\)
Answer: \(= x^2(xy + yz) + y^2(xy + yz) + z^2(xy + yz)\)
\(= x^3y + x^2yz + xy^3 + y^3z + xy z^2 + yz^3\)
In simple words: Multiply each term of the trinomial by each term of the binomial.

Exam Tip: Maintain alphabetical order of variables for clean presentation.

Question 12(iii). \((x + y)(x^2 - xy + y^2)\)
Answer: \(= x(x^2 - xy + y^2) + y(x^2 - xy + y^2)\)
\(= x^3 - x^2y + xy^2 + x^2y - xy^2 + y^3\)
\(= x^3 + y^3\)
In simple words: Notice how middle terms cancel out, leaving the sum of cubes formula.

Exam Tip: This is the standard factorization identity for the sum of two cubes.

Question 12(iv). \((5x^2 + y)(3x + 2y)\)
Answer: \(= 5x^2(3x + 2y) + y(3x + 2y)\)
\(= 15x^3 + 10x^2y + 3xy + 2y^2\)
In simple words: Multiply out each term and combine if any like terms exist.

Exam Tip: Check if any variables match for combining; here all 4 terms are distinct.

Question 12(v). \((x^3 + y^3)(x^2 - xy + y^2)\)
Answer: \(= x^3(x^2 - xy + y^2) + y^3(x^2 - xy + y^2)\)
\(= x^5 - x^4y + x^3y^2 + x^2y^3 - xy^4 + y^5\)
In simple words: Distribute both terms of the first bracket across the second trinomial.

Question 12(vi). \(\left(\frac{3}{5}x^2 - 3y + 2\right)\left(\frac{1}{3}x - y\right)\)
Answer: \(= \frac{3}{5}x^2\left(\frac{1}{3}x - y\right) - 3y\left(\frac{1}{3}x - y\right) + 2\left(\frac{1}{3}x - y\right)\)
\(= \frac{1}{5}x^3 - \frac{3}{5}x^2y - xy + 3y^2 + \frac{2}{3}x - 2y\)
In simple words: Multiply each term of the trinomial by both terms of the binomial.

Exam Tip: Multiply coefficients and simplify fractions at each step.

13. Simplify.

Question 13(i). \((3y + 2)(y - 2) - (7y + 3)(y - 4)\)
Answer: \(= [3y(y - 2) + 2(y - 2)] - [7y(y - 4) + 3(y - 4)]\)
\(= (3y^2 - 6y + 2y - 4) - (7y^2 - 28y + 3y - 12)\)
\(= (3y^2 - 4y - 4) - (7y^2 - 25y - 12)\)
\(= 3y^2 - 4y - 4 - 7y^2 + 25y + 12\)
\(= -4y^2 + 21y + 8\)
In simple words: Expand both products inside brackets, then distribute the negative sign to the second bracket and combine like terms.

Exam Tip: Put parentheses around the second expansion to avoid sign distribution errors.

Question 13(ii). \((2x - 3y)(x + y) - (5x + 2y)(x - y)\)
Answer: \(= [2x(x + y) - 3y(x + y)] - [5x(x - y) + 2y(x - y)]\)
\(= (2x^2 + 2xy - 3xy - 3y^2) - (5x^2 - 5xy + 2xy - 2y^2)\)
\(= (2x^2 - xy - 3y^2) - (5x^2 - 3xy - 2y^2)\)
\(= 2x^2 - xy - 3y^2 - 5x^2 + 3xy + 2y^2\)
\(= -3x^2 + 2xy - y^2\)
In simple words: Expand both products separately, subtract by changing signs, and combine like terms.

Exam Tip: Group \(x^2\), \(xy\), and \(y^2\) terms separately at the end.

Question 13(iii). \(x^2 + (3x - y)(3x + y + y^2)\) - Note: check expression structure
Answer: \(= x^2 + [3x(3x + y + y^2) - y(3x + y + y^2)]\)
\(= x^2 + (9x^2 + 3xy + 3xy^2 - 3xy - y^2 - y^3)\)
\(= x^2 + 9x^2 + 3xy^2 - y^2 - y^3\)
\(= 10x^2 + 3xy^2 - y^2 - y^3\)
In simple words: Multiply the binomial by the trinomial and add \(x^2\).

Exam Tip: Cancel out opposing terms like \(+3xy\) and \(-3xy\).

Question 13(iv). \((a^2 - 3a + 5)(2a - 3) - (5a^2 + 3a - 3)(a - 1)\)
Answer: \(= [a^2(2a - 3) - 3a(2a - 3) + 5(2a - 3)] - [5a^2(a - 1) + 3a(a - 1) - 3(a - 1)]\)
\(= (2a^3 - 3a^2 - 6a^2 + 9a + 10a - 15) - (5a^3 - 5a^2 + 3a^2 - 3a - 3a + 3)\)
\(= (2a^3 - 9a^2 + 19a - 15) - (5a^3 - 2a^2 - 6a + 3)\)
\(= 2a^3 - 9a^2 + 19a - 15 - 5a^3 + 2a^2 + 6a - 3\)
\(= -3a^3 - 7a^2 + 25a - 18\)
In simple words: Expand both large products, distribute the negative sign across the second result, and simplify.

Exam Tip: Take extra care when combining powers of a from 3 down to constants.

14. Find the products of the following.

Question 14(i). \((2x - y)(3x + y^2)\)
Answer: \(= 2x(3x + y^2) - y(3x + y^2)\)
\(= 6x^2 + 2xy^2 - 3xy - y^3\)
In simple words: Multiply each term of the first binomial by each term of the second.

Exam Tip: All resulting four terms are unique.

Question 14(ii). \((x - 3y)(x^4 + 3xy)\)
Answer: \(= x(x^4 + 3xy) - 3y(x^4 + 3xy)\)
\(= x^5 + 3x^2y - 3x^4y - 9xy^2\)
In simple words: Distribute \(x\) and \(-3y\) into the binomial.

Exam Tip: Keep powers of x clear and distinct.

Question 14(iii). \(\left(x^3 + \frac{1}{x}\right)\left(x + \frac{1}{2}\right)\)
Answer: \(= x^3\left(x + \frac{1}{2}\right) + \frac{1}{x}\left(x + \frac{1}{2}\right)\)
\(= x^4 + \frac{1}{2}x^3 + 1 + \frac{1}{2x}\)
In simple words: Multiply each term using distributive law with fractions.

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

Question 14(iv). \((x^2 - a^2)(x - a)\)
Answer: \(= x^2(x - a) - a^2(x - a)\)
\(= x^3 - x^2a - a^2x + a^3\)
In simple words: Distribute \(x^2\) and \(-a^2\) into \((x - a)\).

Exam Tip: Keep variables x and a distinct in your products.

Question 14(v). \(\left(\frac{2}{7}x + \frac{3}{5}y\right)(x^2 + y^2)\)
Answer: \(= \frac{2}{7}x(x^2 + y^2) + \frac{3}{5}y(x^2 + y^2)\)
\(= \frac{2}{7}x^3 + \frac{2}{7}xy^2 + \frac{3}{5}x^2y + \frac{3}{5}y^3\)
In simple words: Multiply each term of the binomial by \(x^2 + y^2\).

Exam Tip: Write fractional coefficients clearly.

Question 14(vi). \((a^2b + ab^2)(b^2c + c^2b)\) - Note: check expression
Answer: \(= a^2b(b^2c + c^2b) + ab^2(b^2c + c^2b)\)
\(= a^2b^3c + a^2bc^2 + ab^4c + ab^3c^2\)
In simple words: Expand by multiplying term by term.

Exam Tip: Add exponents for like bases carefully.

15. Find the following products by using identities.

Question 15(i). \((5x + 9)(5x - 9)\)
Answer: Using identity \((a + b)(a - b) = a^2 - b^2\):
Here \(a = 5x, b = 9\)
\(= (5x)^2 - (9)^2\)
\(= 25x^2 - 81\)
In simple words: Square the first term and subtract the square of the second term.

Exam Tip: This is the difference of squares identity.

Question 15(ii). \((x^3 + y^3)(x^3 - y^3)\)
Answer: Using identity \((a + b)(a - b) = a^2 - b^2\):
Here \(a = x^3, b = y^3\)
\(= (x^3)^2 - (y^3)^2\)
\(= x^6 - y^6\)
In simple words: Apply power of a power rule: \((x^3)^2 = x^6\).

Question 15(iii). \((x^2y + 3z)(x^2y - 3z)\)
Answer: Using identity \((a + b)(a - b) = a^2 - b^2\):
Here \(a = x^2y, b = 3z\)
\(= (x^2y)^2 - (3z)^2\)
\(= x^4y^2 - 9z^2\)
In simple words: Square both parts of the terms inside the parentheses.

Exam Tip: Remember to square the coefficient 3 as well: \(3^2 = 9\).

Question 15(iv). \(\left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right)\)
Answer: Using identity \((a + b)(a - b) = a^2 - b^2\):
Here \(a = x, b = \frac{1}{x}\)
\(= (x)^2 - \left(\frac{1}{x}\right)^2\)
\(= x^2 - \frac{1}{x^2}\)
In simple words: Square x and square the fraction \(\frac{1}{x}\).

Exam Tip: Squaring a fraction squares both numerator and denominator.

16. Find the following products.

Question 16(i). \((5x - 3y)(5x - 3y)\)
Answer: Using identity \((a - b)^2 = a^2 - 2ab + b^2\):
Here \(a = 5x, b = 3y\)
\(= (5x)^2 - 2(5x)(3y) + (3y)^2\)
\(= 25x^2 - 30xy + 9y^2\)
In simple words: Use the square of a binomial formula.

Exam Tip: The middle term is twice the product of a and b with a minus sign.

Question 16(ii). \((y - 3)(y - 3)\)
Answer: Using identity \((y - 3)^2 = y^2 - 2(y)(3) + 3^2\):
\(= y^2 - 6y + 9\)
In simple words: Square of \(y - 3\).

Exam Tip: Can also be written as \((y-3)^2\).

Question 16(iii). \((x^2 - 5)(x^2 - 5)\)
Answer: Using identity \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (x^2)^2 - 2(x^2)(5) + (5)^2\)
\(= x^4 - 10x^2 + 25\)
In simple words: Square \(x^2\) to get \(x^4\), double their product for the middle term.

Exam Tip: Power of a power multiplies the exponents: \((x^2)^2 = x^4\).

Question 16(iv). \(\left(\frac{3}{4}x - \frac{4}{5}y\right)\left(\frac{3}{4}x - \frac{4}{5}y\right)\)
Answer: Using identity \((a - b)^2 = a^2 - 2ab + b^2\):
\(= \left(\frac{3}{4}x\right)^2 - 2\left(\frac{3}{4}x\right)\left(\frac{4}{5}y\right) + \left(\frac{4}{5}y\right)^2\)
\(= \frac{9}{16}x^2 - \frac{24}{20}xy + \frac{16}{25}y^2\)
\(= \frac{9}{16}x^2 - \frac{6}{5}xy + \frac{16}{25}y^2\)
In simple words: Square first fraction, find twice the product, and square second fraction.

Exam Tip: Simplify the fraction \(\frac{24}{20}\) to \(\frac{6}{5}\).

17. Find the following products.

Question 17(i). \((x + 3)(x + 3)\)
Answer: Using identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= x^2 + 2(x)(3) + (3)^2\)
\(= x^2 + 6x + 9\)
In simple words: Square of a sum formula.

Exam Tip: All terms are positive for the sum of two positive terms.

Question 17(ii). \((2a + 3b)(2a + 3b)\)
Answer: Using identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (2a)^2 + 2(2a)(3b) + (3b)^2\)
\(= 4a^2 + 12ab + 9b^2\)
In simple words: Square \(2a\), add twice the product of \(2a\) and \(3b\), and add square of \(3b\).

Exam Tip: Remember to square both the coefficient and the variable.

Question 17(iii). \(\left(\frac{7}{5}x + y\right)\left(\frac{7}{5}x + y\right)\)
Answer: Using identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= \left(\frac{7}{5}x\right)^2 + 2\left(\frac{7}{5}x\right)(y) + (y)^2\)
\(= \frac{49}{25}x^2 + \frac{14}{5}xy + y^2\)
In simple words: Apply binomial square identity with fractional coefficients.

Exam Tip: Square \(\frac{7}{5}\) to get \(\frac{49}{25}\).

Question 17(iv). \(\left(\frac{2}{3}x + 5\right)\left(\frac{2}{3}x + 5\right)\)
Answer: Using identity \((a + b)^2 = a^2 + 2ab + b^2\):
\(= \left(\frac{2}{3}x\right)^2 + 2\left(\frac{2}{3}x\right)(5) + (5)^2\)
\(= \frac{4}{9}x^2 + \frac{20}{3}x + 25\)
In simple words: Multiply 2 by \(\frac{2}{3}\) and 5 to get the middle term \(\frac{20}{3}x\).

Exam Tip: Combine integers and numerators correctly.

18. Simplify using identities.

Question 18(i). \(133 \times 133 - 121 \times 121\)
Answer: Using identity \(a^2 - b^2 = (a + b)(a - b)\):
\(= (133)^2 - (121)^2\)
\(= (133 + 121)(133 - 121)\)
\(= (254)(12)\)
\(= 3048\)
In simple words: Use difference of squares instead of multiplying large numbers directly.

Exam Tip: Identities make mental math or quick calculations much easier.

Question 18(ii). \(5.89 \times 5.89 - 0.11 \times 0.11\)
Answer: Using identity \(a^2 - b^2 = (a + b)(a - b)\):
\(= (5.89)^2 - (0.11)^2\)
\(= (5.89 + 0.11)(5.89 - 0.11)\)
\(= (6.00)(5.78)\)
\(= 34.68\)
In simple words: Sum of 5.89 and 0.11 is 6, difference is 5.78. Multiply them.

Exam Tip: Look for numbers that add up to clean integers like 6 when using \(a^2 - b^2\).

Question 18(iii). \(\frac{93 \times 93 - 5 \times 5}{88}\)
Answer: Numerator: \(93^2 - 5^2 = (93 + 93 \dots) = (93 + 5)(93 - 5) = (98)(88)\)
Substitute into fraction:
\(= \frac{98 \times 88}{88}\)
\(= 98\)
In simple words: Factor the numerator as \((93+5)(93-5)\), then cancel out 88 from numerator and denominator.

Exam Tip: Always look to cancel common factors before doing heavy multiplication.

Question 18(iv). \(\frac{3.29 \times 3.29 - 0.17 \times 0.17}{3.12}\)
Answer: Numerator: \((3.29)^2 - (0.17)^2 = (3.29 + 0.17)(3.29 - 0.17) = (3.46)(3.12)\)
Substitute into fraction:
\(= \frac{3.46 \times 3.12}{3.12}\)
\(= 3.46\)
In simple words: Apply difference of squares on top, then cancel 3.12.

Question 19. Find the value of the expression \(36x^2 + 60xy + 25y^2\), when \(x = 4\) and \(y = -7\).
Answer: Notice the expression is a perfect square trinomial:
\(36x^2 + 60xy + 25y^2 = (6x)^2 + 2(6x)(5y) + (5y)^2 = (6x + 5y)^2\)
Now substitute \(x = 4\) and \(y = -7\):
\(= [6(4) + 5(-7)]^2\)
\(= [24 - 35]^2\)
\(= [-11]^2 = 121\)
In simple words: Recognize it as \((6x + 5y)^2\) first to simplify substitution.

Exam Tip: Factoring into a binomial square before substituting numbers saves time and prevents arithmetic errors.

20. Find the value of the following expressions, when \(x = 4\) and \(y = 7\).

Question 20(i). \(49x^2 + 126xy + 81y^2\)
Answer: Recognize as a square trinomial: \((7x)^2 + 2(7x)(9y) + (9y)^2 = (7x + 9y)^2\)
Substitute \(x = 4, y = 7\):
\(= [7(4) + 9(7)]^2\)
\(= [28 + 63]^2\)
\(= [91]^2 = 8281\)
In simple words: Simplify to \((7x + 9y)^2\) before plugging in values.

Exam Tip: Check if first and last terms are perfect squares (\(49x^2 = (7x)^2\) and \(81y^2 = (9y)^2\)).

Question 20(ii). \(4x^2 - 12xy + 9y^2\)
Answer: Recognize as a square trinomial: \((2x)^2 - 2(2x)(3y) + (3y)^2 = (2x - 3y)^2\)
Substitute \(x = 4, y = 7\):
\(= [2(4) - 3(7)]^2\)
\(= [8 - 21]^2\)
\(= [-13]^2 = 169\)
In simple words: Simplify to \((2x - 3y)^2\) and substitute.

Exam Tip: The minus sign in the middle term indicates the difference identity \((a-b)^2\).

Question 20(iii). \((x^4 - y^4)\)
Answer: Substitute \(x = 4, y = 7\):
\(= (4)^4 - (7)^4\)
\(= 256 - 2401\)
\(= -2145\)
In simple words: Calculate \(4^4\) and \(7^4\) directly and subtract.

Exam Tip: Alternatively, factor as \((x^2 - y^2)(x^2 + y^2) = (x-y)(x+y)(x^2+y^2)\).

Question 21. If \(\left(x + \frac{1}{x}\right) = 3\), find the value of \(\left(x^2 + \frac{1}{x^2}\right)\).
Answer: Given: \(x + \frac{1}{x} = 3\)
Square both sides:
\(\left(x + \frac{1}{x}\right)^2 = (3)^2\)
\(x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 9\)
\(x^2 + 2 + \frac{1}{x^2} = 9\)
\(x^2 + \frac{1}{x^2} = 9 - 2\)
\(x^2 + \frac{1}{x^2} = 7\)
In simple words: Square the given equation on both sides and simplify the middle term \(2 \cdot x \cdot \frac{1}{x} = 2\).

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

22. Using the identities, evaluate the following.

Question 22(i). \(102^2\)
Answer: \(= (100 + 2)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (100)^2 + 2(100)(2) + (2)^2\)
\(= 10000 + 400 + 4\)
\(= 10404\)
In simple words: Split 102 into \(100 + 2\) and apply the square identity.

Exam Tip: Splitting numbers near powers of 10 makes arithmetic much easier.

Question 22(ii). \(311^2\)
Answer: \(= (300 + 11)^2\)
\(= (300)^2 + 2(300)(11) + (11)^2\)
\(= 90000 + 6600 + 121\)
\(= 96721\)
In simple words: Write 311 as \(300 + 11\) and expand.

Exam Tip: Ensure accurate squaring of 300 and 11.

Question 22(iii). \(72^2\)
Answer: \(= (70 + 2)^2\)
\(= (70)^2 + 2(70)(2) + (2)^2\)
\(= 4900 + 280 + 4\)
\(= 5184\)
In simple words: Rewrite 72 as \(70 + 2\) and apply identity.

Exam Tip: \(70^2 = 4900\).

Question 22(iv). \(89^2\)
Answer: \(= (90 - 1)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (90)^2 - 2(90)(1) + (1)^2\)
\(= 8100 - 180 + 1\)
\(= 7921\)
In simple words: Express 89 as \(90 - 1\) to use subtraction identity.

Exam Tip: Using numbers just above (like 90) with subtraction is easier than 80+9.

Question 22(v). \(118^2\)
Answer: \(= (120 - 2)^2\)
\(= (120)^2 - 2(120)(2) + (2)^2\)
\(= 14400 - 480 + 4\)
\(= 13924\)
In simple words: Use \(120 - 2\) for simpler calculation.

Exam Tip: \(12^2 = 144\), so \(120^2 = 14400\).

Question 22(vi). \(989^2\)
Answer: \(= (1000 - 11)^2\)
\(= (1000)^2 - 2(1000)(11) + (11)^2\)
\(= 1000000 - 22000 + 121\)
\(= 978121\)
In simple words: Express as \(1000 - 11\).

Exam Tip: Large numbers become easy to square when referenced to powers of 10.

Question 22(vii). \(13 \times 7\)
Answer: Rewrite using average: \(13 = 10 + 3\) and \(7 = 10 - 3\)
\(= (10 + 3)(10 - 3)\)
Using \((a + b)(a - b) = a^2 - b^2\):
\(= (10)^2 - (3)^2\)
\(= 100 - 9 = 91\)
In simple words: Write as \((10+3)(10-3)\) and apply difference of squares.

Exam Tip: Any product of numbers equidistant from a midpoint can be written as \((x+a)(x-a)\).

Question 22(viii). \(73 \times 67\)
Answer: Midpoint is 70: \(73 = 70 + 3\) and \(67 = 70 - 3\)
\(= (70 + 3)(70 - 3)\)
\(= (70)^2 - (3)^2\)
\(= 4900 - 9 = 4891\)
In simple words: Express as \((70+3)(70-3)\).

Exam Tip: Find the exact middle number to apply \(a^2 - b^2\).

Question 22(ix). \(9.3 \times 8.7\)
Answer: Midpoint is 9.0: \(9.3 = 9.0 + 0.3\) and \(8.7 = 9.0 - 0.3\)
\(= (9.0 + 0.3)(9.0 - 0.3)\)
\(= (9.0)^2 - (0.3)^2\)
\(= 81 - 0.09 = 80.91\)
In simple words: Use decimal midpoint 9.0.

Question 22(x). \(12.5 \times 11.5\)
Answer: Midpoint is 12.0: \(12.5 = 12.0 + 0.5\) and \(11.5 = 12.0 - 0.5\)
\(= (12.0 + 0.5)(12.0 - 0.5)\)
\(= (12)^2 - (0.5)^2\)
\(= 144 - 0.25 = 143.75\)
In simple words: Express as \((12+0.5)(12-0.5)\).

Exam Tip: \(0.5^2 = 0.25\).

Question 22(xi). \(105 \times 95\)
Answer: Midpoint is 100: \(105 = 100 + 5\) and \(95 = 100 - 5\)
\(= (100 + 5)(100 - 5)\)
\(= (100)^2 - (5)^2\)
\(= 10000 - 25 = 9975\)
In simple words: Write as \((100+5)(100-5)\).

Exam Tip: Standard difference of squares formula application.

Question 22(xii). \(153 \times 147\)
Answer: Midpoint is 150: \(153 = 150 + 3\) and \(147 = 150 - 3\)
\(= (150 + 3)(150 - 3)\)
\(= (150)^2 - (3)^2\)
\(= 22500 - 9 = 22491\)
In simple words: Express as \((150+3)(150-3)\).

Exam Tip: \(150^2 = 22500\).