RS Aggarwal Class 7 Mathematics Solutions Chapter 6 Algebraic Expressions

Algebraic Expressions

Exercise 6A

Question 1. Add the following expressions by collecting like terms:
(i) \( 5x + 7x + (-6x) \)
(ii) \( \frac{3}{5}x + \frac{2}{3}x + \left(\frac{-4}{5}x\right) \)
(iii) \( 5a^2b + (-8a^2b) + 7a^2b \)
(iv) \( \frac{3}{4}x^2 + 5x^2 + (-3x^2) + \left(\frac{-1}{4}x^2\right) \)
(v) \( x - 3y + 4z + y - 2x - 8z + 5x - 2y - 3z \)
(vi) \( 2x^2 - 3y^2 + 5x^2 + 6y^2 + (-3x^2 - 4y^2) \)
(vii) \( 5x - 2x^2 - 8 + 8x^2 - 7x - 9 + 3 + 7x^2 - 2x \)
(viii) \( \frac{2}{3}a - \frac{4}{5}b + \frac{3}{5}c + \left(\frac{-3}{4}a - \frac{5}{2}b + \frac{2}{3}c\right) + \frac{5}{2}a + \frac{7}{4}b - \frac{5}{6}c \)
(ix) \( \frac{8}{5}x + \frac{11}{7}y + \frac{9}{4}xy + \left(\frac{-3}{2}x - \frac{5}{3}y - \frac{9}{5}xy\right) \)
(x) \( \frac{3}{2}x^3 - \frac{1}{4}x^2 + \frac{5}{3} + \left(\frac{-5}{4}x^3 + \frac{3}{5}x^2 - x + \frac{1}{5}\right) + \left(-x^2 + \frac{3}{8}x - \frac{8}{15}\right) \)
Answer:
(i) \( 5x + 7x - 6x = 6x \)
(ii) \( \frac{9x + 10x - 12x}{15} = \frac{7x}{15} \)
(iii) \( 5a^2b - 8a^2b + 7a^2b = 4a^2b \)
(iv) \( \left(\frac{3}{4} - \frac{1}{4}\right)x^2 + 5x^2 - 3x^2 = \frac{1}{2}x^2 + 2x^2 = \frac{5}{2}x^2 \)
(v) Gathering the x, y and z terms separately: \( x - 2x + 5x - 3y + y - 2y + 4z - 8z - 3z = 4x - 4y - 7z \)
(vi) \( 2x^2 + 5x^2 - 3x^2 - 3y^2 + 6y^2 - 4y^2 = 4x^2 - y^2 \)
(vii) \( -2x^2 + 8x^2 + 7x^2 + 5x - 7x - 2x - 8 - 9 + 3 = 13x^2 - 4x - 14 \)
(viii) Gathering the a, b and c terms and using the LCM of the denominators: \( \frac{(8 - 9 + 30)a}{12} + \frac{(-16 - 50 + 35)b}{20} + \frac{(18 + 20 - 25)c}{30} = \frac{29}{12}a - \frac{31}{20}b + \frac{13}{30}c \)
(ix) \( \frac{8}{5}x - \frac{3}{2}x + \frac{11}{7}y - \frac{5}{3}y + \frac{9}{4}xy - \frac{9}{5}xy = \frac{1}{10}x - \frac{2}{21}y + \frac{9}{20}xy \)
(x) \( \frac{3}{2}x^3 - \frac{5}{4}x^3 - \frac{1}{4}x^2 + \frac{3}{5}x^2 - x^2 - x + \frac{3}{8}x + \frac{5}{3} + \frac{1}{5} - \frac{8}{15} = \frac{1}{4}x^3 - \frac{13}{20}x^2 - \frac{5}{8}x + \frac{4}{3} \)

Exam Tip: Group terms with the same variable and power together first, then add or subtract their coefficients - never combine unlike terms such as x and x².

Question 2. Subtract:
(i) \( -8xy \) from \( 7xy \)
(ii) \( x^2 \) from \( -3x^2 \)
(iii) \( (x - y) \) from \( (4y - 5x) \)
(iv) \( (a^2 + b^2 - 2ab) \) from \( (a^2 + b^2 + 2ab) \)
(v) \( (x^2 - y^2) \) from \( (2x^2 - 3y^2 + 6xy) \)
(vi) \( (x - y + 3z) \) from \( (2z - x - 3y) \)
Answer:
(i) \( 7xy - (-8xy) = 7xy + 8xy = 15xy \)
(ii) \( -3x^2 - x^2 = -4x^2 \)
(iii) \( (4y - 5x) - (x - y) = 4y - 5x - x + y = 5y - 6x \)
(iv) \( (a^2 + b^2 + 2ab) - (a^2 + b^2 - 2ab) = a^2 - a^2 + b^2 - b^2 + 2ab + 2ab = 4ab \)
(v) \( (2x^2 - 3y^2 + 6xy) - (x^2 - y^2) = x^2 - 2y^2 + 6xy \)
(vi) \( (2z - x - 3y) - (x - y + 3z) = -z - 2x - 2y \)

Exam Tip: When subtracting an expression in brackets, flip the sign of every term inside those brackets before combining like terms.

Question 4. Add \( 8m - 7n + 6p^2 \) and \( -3m - 4n - p^2 \). Also add \( 2m + 4n - 3p^2 \) and \( -m - n - p^2 \). Then subtract the second sum from the first sum.
Answer:
\( (8m - 7n + 6p^2) + (-3m - 4n - p^2) = 8m - 3m - 7n - 4n + 6p^2 - p^2 = 5m - 11n + 5p^2 \)
\( (2m + 4n - 3p^2) + (-m - n - p^2) = 2m - m + 4n - n - 3p^2 - p^2 = m + 3n - 4p^2 \)
So, \( (m + 3n - 4p^2) - (5m - 11n + 5p^2) = -4m + 14n - 9p^2 \)

Exam Tip: When a question involves two sums followed by a subtraction, work out each sum fully first, then treat the two results as a single subtraction problem.

Question 5. Add \( 8a - 6a^2 + 9 \) and \( -10a - 8 + 8a^2 \). Then subtract the sum obtained from \( -3 \).
Answer:
\( (8a - 6a^2 + 9) + (-10a - 8 + 8a^2) = 8a - 10a - 6a^2 + 8a^2 + 9 - 8 = -2a + 2a^2 + 1 \)
So, \( -3 - (-2a + 2a^2 + 1) = 2a - 2a^2 - 4 \)

Exam Tip: Treat a whole number like -3 as a one-term expression when subtracting from it - just flip the sign of every term in the bracket and combine.

Question 6. Simplify by collecting like terms:
(i) \( 5x + 7x - 9y - y \)
(ii) \( x^2 - \frac{3}{2}x^2 - x - \frac{1}{2}x + \frac{3}{2} \)
(iii) \( 7 + 7 - 2x - x - 5x + 5y + y - 3y \)
(iv) \( \frac{1}{3}y^2 + \frac{2}{3}y^2 - 2y^2 - \frac{4}{7}y - \frac{2}{7}y - \frac{1}{7}y + 5 - 2 + 3 \)
Answer:
(i) \( 5x + 7x - 9y - y = 12x - 10y \)
(ii) \( x^2 - \frac{3}{2}x^2 - x - \frac{1}{2}x + \frac{3}{2} = -\frac{1}{2}x^2 - \frac{3}{2}x + \frac{3}{2} \)
(iii) \( 7 + 7 - 2x - x - 5x + 5y + y - 3y = 14 - 8x + 3y \)
(iv) \( \left(\frac{1}{3} + \frac{2}{3} - 2\right)y^2 + \left(\frac{-4}{7} - \frac{2}{7} - \frac{1}{7}\right)y + (5 - 2 + 3) = -y^2 - y + 6 \)

Exam Tip: Sort the terms mentally into groups (x² terms, x terms, plain numbers, and so on) before adding the coefficients within each group.

Exercise 6B

Question 1. Multiply: \( 3a^2 \) and \( 8a^4 \)
Answer:
\( 3a^2 \times 8a^4 = (3 \times 8) \times (a^2 \times a^4) = 24 \times a^{2+4} = 24a^6 \)

Exam Tip: Multiply the coefficients directly, and add the exponents of the same base rather than multiplying them.

Question 2. Multiply: \( -6x^3 \) and \( 5x^2 \)
Answer:
\( -6x^3 \times 5x^2 = (-6 \times 5) \times (x^3 \times x^2) = -30 \times x^{3+2} = -30x^5 \)

Exam Tip: Keep track of the sign separately from the powers - a negative times a positive coefficient always gives a negative answer.

Question 3. Multiply: \( -4ab \) and \( -3a^2bc \)
Answer:
\( (-4ab) \times (-3a^2bc) = (-4 \times -3) \times (a \times a^2 \times b \times b \times c) = 12 \times a^3b^2c = 12a^3b^2c \)

Exam Tip: Two negative coefficients multiply to give a positive answer - always check the sign before finalising the product.

Question 4. Multiply: \( 2a^2b^3 \) and \( -3a^3b \)
Answer:
\( (2a^2b^3) \times (-3a^3b) = (2 \times -3) \times (a^2 \times a^3 \times b^3 \times b) = -6 \times a^{2+3}b^{3+1} = -6a^5b^4 \)

Exam Tip: Line up the like bases before adding their exponents - it helps avoid missing a factor when several letters are involved.

Question 5. Multiply: \( \frac{2}{3}x^2y \) and \( \frac{3}{5}xy^2 \)
Answer:
\( \frac{2}{3}x^2y \times \frac{3}{5}xy^2 = \left(\frac{2}{3} \times \frac{3}{5}\right) \times (x^2 \times x \times y \times y^2) = \frac{2}{5} \times x^{2+1}y^{1+2} = \frac{2}{5}x^3y^3 \)

Exam Tip: Multiply fractional coefficients the same way as ordinary fractions, then handle the letters separately by adding exponents.

Question 6. Multiply: \( \frac{-3}{4}ab^3 \) and \( \frac{-2}{3}a^2b^4 \)
Answer:
\( \left(\frac{-3}{4} \times \frac{-2}{3}\right) \times (a \times a^2 \times b^3 \times b^4) = \frac{1}{2} \times a^{1+2}b^{3+4} = \frac{1}{2}a^3b^7 \)

Exam Tip: Cancel common factors in fractional coefficients before multiplying, just as you would in ordinary fraction arithmetic.

Question 7. Multiply: \( \frac{-1}{27}a^2b^2c^2 \) and \( \frac{-9}{2}a^3b \)
Answer:
\( \left(\frac{-1}{27} \times \frac{-9}{2}\right) \times (a^2 \times a^3 \times b^2 \times b \times c^2) = \frac{1}{6} \times a^{2+3}b^{2+1}c^2 = \frac{1}{6}a^5b^3c^2 \)

Exam Tip: With three different letters, tackle one letter at a time and add its exponents across the two factors before moving to the next.

Question 8. Multiply: \( \frac{-13}{5}ab^2c \) and \( \frac{7}{3}a^2bc^2 \)
Answer:
\( \left(\frac{-13}{5} \times \frac{7}{3}\right) \times (a \times a^2 \times b^2 \times b \times c \times c^2) = \frac{-91}{15} \times a^{1+2}b^{2+1}c^{1+2} = \frac{-91}{15}a^3b^3c^3 \)

Exam Tip: Do not simplify the fraction until all the multiplying is finished - it is easier to spot common factors at the end.

Question 9. Multiply: \( \frac{-18}{5}x^2z \) and \( \frac{-25}{6}xyz^2 \)
Answer:
\( \left(\frac{-18}{5} \times \frac{-25}{6}\right) \times (x^2 \times x \times z \times z^2 \times y) = 15 \times x^{2+1}yz^{1+2} = 15x^3yz^3 \)

Exam Tip: Even when a letter such as y appears in only one factor, still carry it through unchanged into the final answer.

Question 10. Multiply: \( \frac{-3}{14}xy^4 \) and \( \frac{7}{6}x^3y \)
Answer:
\( \left(\frac{-3}{14} \times \frac{7}{6}\right) \times (x \times x^3 \times y^4 \times y) = \frac{-1}{4} \times x^{1+3}y^{4+1} = \frac{-1}{4}x^4y^5 \)

Exam Tip: Cancel the numerical fraction fully before writing the final answer, so it stays in its simplest form.

Question 11. Multiply: \( \frac{-7}{5}x^2y \), \( \frac{3}{2}xy^2 \) and \( \frac{-6}{5}x^3y^3 \)
Answer:
\( \left(\frac{-7}{5} \times \frac{3}{2} \times \frac{-6}{5}\right) \times (x^2 \times x \times x^3 \times y \times y^2 \times y^3) = \frac{63}{25} \times x^{2+1+3}y^{1+2+3} = \frac{63}{25}x^6y^6 \)

Exam Tip: With three monomials, multiply all three coefficients together first, then add up all three exponents for each letter.

Question 12. Multiply: \( 2a^2b \), \( -5ab^2c \) and \( -6bc^2 \)
Answer:
\( (2 \times -5 \times -6) \times (a^2 \times a \times b \times b^2 \times b \times c \times c^2) = 60 \times a^{2+1}b^{1+2+1}c^{1+2} = 60a^3b^4c^3 \)

Exam Tip: Track each letter across all three monomials separately - a letter missing from one factor simply contributes zero to that letter's exponent.

Question 13. Multiply: \( -4x^2y^2 \), \( -6xy \) and \( -3 \)
Answer:
\( (-4 \times -6 \times -3) \times (x^2 \times x \times y^2 \times y) = -72 \times x^{2+1}y^{2+1} = -72x^3y^3 \)

Exam Tip: A plain number like -3 in a product still counts as a factor for the coefficient, even though it has no letters of its own.

Question 14. Multiply: \( \frac{-3}{5}s^2tu \), \( \frac{15}{7}st^2 \) and \( \frac{7}{9}su^2 \)
Answer:
\( \left(\frac{-3}{5} \times \frac{15}{7} \times \frac{7}{9}\right) \times (s^2 \times s \times s \times t \times t^2 \times u \times u^2) = -1 \times s^{2+1+1}t^{1+2}u^{1+2} = -s^4t^3u^3 \)

Exam Tip: When several fractions multiply together, cancel across all of them at once rather than one pair at a time - it keeps the arithmetic simpler.

Question 15. Multiply: \( \frac{-2}{7}u^4v \), \( \frac{-14}{5}uv^3 \) and \( \frac{-3}{4}u^2v^3 \)
Answer:
\( \left(\frac{-2}{7} \times \frac{-14}{5} \times \frac{-3}{4}\right) \times (u^4 \times u \times u^2 \times v \times v^3 \times v^3) = \frac{-3}{5} \times u^{4+1+2}v^{1+3+3} = \frac{-3}{5}u^7v^7 \)

Exam Tip: Three negative coefficients multiplied together give a negative result - keep count of the negative signs as you go.

Question 16. Multiply: \( -3ab^2c \), \( -a^2b^2c^3 \) and \( -abc \)
Answer:
\( (-3 \times -1 \times -1) \times (a \times a^2 \times a \times b^2 \times b^2 \times b \times c \times c^3 \times c) = -3 \times a^{1+2+1}b^{2+2+1}c^{1+3+1} = -3a^4b^5c^5 \)

Exam Tip: When a coefficient is just -1 (written as a lone minus sign), still include it in the running product of the numbers.

Question 17. Multiply: \( \frac{4}{3}x^2yz \), \( \frac{1}{3}xy^2z \) and \( -6xyz^2 \)
Answer:
\( \left(\frac{4}{3} \times \frac{1}{3} \times -6\right) \times (x^2 \times x \times x \times y \times y^2 \times y \times z \times z \times z^2) = \frac{-8}{3} \times x^{2+1+1}y^{1+2+1}z^{1+1+2} = \frac{-8}{3}x^4y^4z^4 \)

Exam Tip: With three letters and three monomials, it helps to deal with one letter completely across all three factors before starting the next.

Question 18. Multiply \( \frac{-2}{3}a^2b \) and \( \frac{6}{5}a^3b^2 \), and verify the result for \( a = 2 \) and \( b = 3 \).
Answer:
\( \frac{-2}{3}a^2b \times \frac{6}{5}a^3b^2 = \left(\frac{-2}{3} \times \frac{6}{5}\right) \times (a^2 \times a^3 \times b \times b^2) = \frac{-4}{5} \times a^{2+3}b^{1+2} = \frac{-4}{5}a^5b^3 \)
When \( a = 2 \) and \( b = 3 \): L.H.S. \( = \frac{-2}{3}a^2b \times \frac{6}{5}a^3b^2 = -8 \times \frac{432}{5} = \frac{-3456}{5} \)
R.H.S. \( = \frac{-4}{5}a^5b^3 = \frac{-4}{5} \times 2^5 \times 3^3 = \frac{-3456}{5} \)
Since L.H.S. = R.H.S., the result is verified.

Exam Tip: To verify a product, work out the original expression and the simplified answer separately at the given values - if the method is correct they will always match.

Question 19. Multiply \( \frac{-8}{21}x^2y^3 \) and \( \frac{-7}{16}xy^2 \), and verify the result for \( x = 3 \) and \( y = 2 \).
Answer:
\( \frac{-8}{21}x^2y^3 \times \frac{-7}{16}xy^2 = \left(\frac{-8}{21} \times \frac{-7}{16}\right) \times (x^{2+1})(y^{3+2}) = \frac{1}{6}x^3y^5 \)
When \( x = 3 \) and \( y = 2 \): L.H.S. \( = \frac{-8}{21}x^2y^3 \times \frac{-7}{16}xy^2 = \frac{-192}{21} \times \frac{-21}{4} = 144 \)
R.H.S. \( = \frac{1}{6}x^3y^5 = \frac{1}{6} \times 3^3 \times 2^5 = 144 \)
Since L.H.S. = R.H.S., \( \frac{-8}{21}x^2y^3 \times \frac{-7}{16}xy^2 = \frac{1}{6}x^3y^5 \).

Exam Tip: Plugging in numbers is a reliable way to double-check algebra - if both sides give the same value, the simplification is correct.

Question 20. Multiply \( 2.3a^5b^2 \) and \( 1.2a^2b^2 \), and find its value when \( a = 1 \) and \( b = 0.5 \).
Answer:
\( (2.3 \times 1.2) \times (a^5 \times a^2 \times b^2 \times b^2) = 2.76 \times a^{5+2}b^{2+2} = 2.76a^7b^4 \)
When \( a = 1 \) and \( b = 0.5 \): \( 2.76a^7b^4 = 2.76 \times 1^7 \times 0.5^4 = 0.1725 \)

Exam Tip: Decimal coefficients multiply exactly like whole numbers - just keep the decimal point in the right place throughout.

Question 21. Multiply \( -8u^2v^6 \) and \( -20uv \), and find its value when \( u = 2.5 \) and \( v = 1 \).
Answer:
\( (-8 \times -20) \times (u^2 \times u \times v^6 \times v) = 160 \times u^{2+1}v^{6+1} = 160u^3v^7 \)
When \( u = 2.5 \) and \( v = 1 \): \( 160u^3v^7 = 160 \times 2.5^3 \times 1^7 = 2500 \)

Exam Tip: A power of 1, such as v raised to any power here, always equals 1 and can be dropped from the numeric calculation.

Question 22. Multiply \( \frac{2}{5}a^2b \), \( -15ab^2c \) and \( \frac{-1}{2}c^2 \), and verify the result for \( a = 1 \), \( b = 2 \) and \( c = 3 \).
Answer:
\( \left(\frac{2}{5} \times -15 \times \frac{-1}{2}\right) \times (a^2 \times a \times b \times b^2 \times c \times c^2) = 3 \times a^{2+1}b^{1+2}c^{1+2} = 3a^3b^3c^3 \)
When \( a = 1 \), \( b = 2 \), \( c = 3 \): \( \frac{2}{5}a^2b = \frac{4}{5} \), \( -15ab^2c = -180 \), \( \frac{-1}{2}c^2 = \frac{-9}{2} \)
L.H.S. \( = \frac{4}{5} \times -180 \times \frac{-9}{2} = 648 \). R.H.S. \( = 3a^3b^3c^3 = 3 \times 1 \times 8 \times 27 = 648 \)
Since L.H.S. = R.H.S., the result is verified.

Exam Tip: With three monomials, substitute the given values into each one separately before multiplying, to keep the arithmetic manageable.

Question 23. Multiply \( \frac{1}{4}abc \), \( -6b^2c \) and \( \frac{-1}{3}c^3 \), and verify the result for \( a = 1 \), \( b = 2 \) and \( c = 3 \).
Answer:
\( \left(\frac{1}{4} \times -6 \times \frac{-1}{3}\right) \times (a \times b \times b^2 \times c \times c \times c^3) = \frac{1}{2} \times ab^{1+2}c^{1+1+3} = \frac{1}{2}ab^3c^5 \)
When \( a = 1 \), \( b = 2 \), \( c = 3 \): \( \frac{1}{4}abc = \frac{3}{2} \), \( -6b^2c = -72 \), \( \frac{-1}{3}c^3 = -9 \)
L.H.S. \( = \frac{3}{2} \times -72 \times -9 = 972 \). R.H.S. \( = \frac{1}{2}ab^3c^5 = \frac{1}{2} \times 1 \times 8 \times 243 = 972 \)
Since L.H.S. = R.H.S., the result is verified.

Exam Tip: Once each monomial's numeric value is found, multiply those three plain numbers together to get the left-hand side quickly.

Question 24. Multiply \( \frac{4}{9}abc^3 \), \( \frac{-27}{5}b^2c \) and \( -8a^3b^3 \), and verify the result for \( a = 1 \), \( b = 2 \) and \( c = 3 \).
Answer:
\( \left(\frac{4}{9} \times \frac{-27}{5} \times -8\right) \times (a \times a^3 \times b \times b^2 \times b^3 \times c^3 \times c) = \frac{96}{5} \times a^{1+3}b^{1+2+3}c^{3+1} = \frac{96}{5}a^4b^6c^4 \)
When \( a = 1 \), \( b = 2 \), \( c = 3 \): L.H.S. \( = \left(\frac{4}{9} \times \frac{-27}{5} \times -8\right) \times (1 \times 1^3 \times 2 \times 2^2 \times 2^3 \times 3^3 \times 3) = \frac{497664}{5} \)
R.H.S. \( = \frac{96}{5}a^4b^6c^4 = \frac{96}{5}(1^4 \times 2^6 \times 3^4) = \frac{497664}{5} \)
Since L.H.S. = R.H.S., the result is verified.

Exam Tip: For a lengthy verification, compute the coefficient product and the letter-substitution product as two separate mini-calculations before combining them.

Question 25. Multiply \( \frac{-4}{7}a^2bc \), \( \frac{-2}{3}ab^2c \) and \( \frac{-7}{6}c \), and verify the result for \( a = 1 \), \( b = 2 \) and \( c = 3 \).
Answer:
\( \left(\frac{-4}{7} \times \frac{-2}{3} \times \frac{-7}{6}\right) \times (a^2 \times a \times b \times b^2 \times c \times c \times c) = \frac{-4}{9} \times a^{2+1}b^{1+2}c^3 = \frac{-4}{9}a^3b^3c^3 \)
When \( a = 1 \), \( b = 2 \), \( c = 3 \): L.H.S. \( = \left(\frac{-4}{7} \times \frac{-2}{3} \times \frac{-7}{6}\right) \times (1^2 \times 1 \times 2 \times 2^2 \times 3 \times 3 \times 3) = -96 \)
R.H.S. \( = \frac{-4}{9}a^3b^3c^3 = \frac{-4}{9} \times 1^3 \times 2^3 \times 3^3 = -96 \)
Since L.H.S. = R.H.S., the result is verified.

Exam Tip: Three negative fractional coefficients multiply to give a negative overall coefficient - double check the sign before substituting numbers.

Exercise 6C

Question 1. Multiply: \( 4a(3a + 7b) \)
Answer:
\( 4a \times 3a + 4a \times 7b = 12a^2 + 28ab \)

Exam Tip: Multiply the term outside the bracket by every term inside it separately, then add the results together - this is the distributive law.

Question 2. Multiply: \( 5a(6a - 3b) \)
Answer:
\( 5a \times 6a - 5a \times 3b = 30a^2 - 15ab \)

Exam Tip: Keep track of the minus sign inside the bracket - it carries through to the corresponding term in the answer.

Question 3. Multiply: \( 8a^2(2a + 5b) \)
Answer:
\( 8a^2 \times 2a + 8a^2 \times 5b = 16a^3 + 40a^2b \)

Exam Tip: When the outside term already has a power, add its exponent to the exponent of the matching letter inside the bracket.

Question 4. Multiply: \( 9x^2(5x + 7) \)
Answer:
\( 9x^2 \times 5x + 9x^2 \times 7 = 45x^3 + 63x^2 \)

Exam Tip: A plain number term inside the bracket, like the 7 here, only affects the coefficient - the power of x stays the same as outside.

Question 5. Multiply: \( ab(a^2 - b^2) \)
Answer:
\( ab \times a^2 - ab \times b^2 = a^3b - ab^3 \)

Exam Tip: Distribute carefully when two different letters are involved - each term inside picks up both letters from outside.

Question 6. Multiply: \( 2x^2(3x - 4x^2) \)
Answer:
\( 2x^2 \times 3x - 2x^2 \times 4x^2 = 6x^3 - 8x^4 \)

Exam Tip: Watch the exponents carefully when both terms share the same base - adding them correctly avoids a common slip-up here.

Question 7. Multiply: \( \frac{3}{5}m^2n(m + 5n) \)
Answer:
\( \frac{3}{5}m^2n \times m + \frac{3}{5}m^2n \times 5n = \frac{3}{5}m^3n + 3m^2n^2 \)

Exam Tip: Multiply the fraction outside by the whole number inside as usual - cancelling early keeps the numbers small.

Question 8. Multiply: \( -17x^2(3x - 4) \)
Answer:
\( -17x^2 \times 3x - (-17x^2 \times 4) = -51x^3 + 68x^2 \)

Exam Tip: A negative times a negative gives a positive - this often trips students up in the second term of a distribution.

Question 9. Multiply: \( \frac{7}{2}x^2\left(\frac{4}{7}x + 2\right) \)
Answer:
\( \frac{7}{2}x^2 \times \frac{4}{7}x + \frac{7}{2}x^2 \times 2 = 2x^3 + 7x^2 \)

Exam Tip: Cancel matching numbers between the fractions before multiplying - here the 7s cancel neatly in the first term.

Question 10. Multiply: \( -4x^2y(3x^2 - 5y) \)
Answer:
\( -4x^2y \times 3x^2 - (-4x^2y \times 5y) = -12x^4y + 20x^2y^2 \)

Exam Tip: Add exponents of x carefully in the first term since both the outside term and the bracket term contain x squared.

Question 11. Multiply: \( \frac{-4}{27}xyz\left(\frac{9}{2}x^2yz - \frac{3}{4}xyz^2\right) \)
Answer:
\( \frac{-4}{27}xyz \times \frac{9}{2}x^2yz - \frac{-4}{27}xyz \times \frac{3}{4}xyz^2 = \frac{-2}{3}x^3y^2z^2 + \frac{1}{9}x^2y^2z^3 \)

Exam Tip: With three letters in every term, work through the exponents one letter at a time to avoid losing track partway.

Question 12. Multiply: \( 9t^2(t + 7t^3) \)
Answer:
\( 9t^2 \times t + 9t^2 \times 7t^3 = 9t^3 + 63t^5 \)

Exam Tip: Add the exponent of t from outside the bracket to the exponent of t in each term inside.

Question 13. Multiply: \( 10a^2(0.1a - 0.5b) \)
Answer:
\( 10a^2 \times 0.1a - 10a^2 \times 0.5b = a^3 - 5a^2b \)

Exam Tip: A decimal coefficient multiplying a whole number can often simplify to a clean whole number - check for this shortcut.

Question 14. Multiply: \( 1.5a(10a^2b - 100ab^2) \)
Answer:
\( 1.5a \times 10a^2b - 1.5a \times 100ab^2 = 15a^3b - 150a^2b^2 \)

Exam Tip: Multiply the decimal and whole-number coefficients together first, then handle the letters and their powers separately.

Question 15. Multiply: \( \frac{2}{3}abc(a^2 + b^2 - 3c^2) \)
Answer:
\( \frac{2}{3}abc \times a^2 + \frac{2}{3}abc \times b^2 - \frac{2}{3}abc \times 3c^2 = \frac{2}{3}a^3bc + \frac{2}{3}ab^3c - 2abc^3 \)

Exam Tip: Distribute across all three terms inside the bracket, including the negative one, before simplifying each product.

Question 16. Expand \( 24x^2(1 - 2x) \) and verify the result for \( x = 2 \).
Answer:
\( 24x^2 \times 1 - 24x^2 \times 2x = 24x^2 - 48x^3 \)
When \( x = 2 \): L.H.S. \( = 24x^2(1-2x) = 24 \times 4 \times (1-4) = 96 \times -3 = -288 \)
R.H.S. \( = 24x^2 - 48x^3 = 96 - 384 = -288 \)
Since L.H.S. = R.H.S., \( 24x^2(1-2x) = 24x^2 - 48x^3 \).

Exam Tip: To verify an expansion, substitute the given value into both the original bracketed form and the expanded form separately.

Question 17. Expand \( ab(a^2 + b^2) \) and verify the result for \( a = 2 \) and \( b = \frac{1}{2} \).
Answer:
\( ab \times a^2 + ab \times b^2 = a^3b + ab^3 \)
When \( a = 2 \), \( b = \frac{1}{2} \): L.H.S. \( = ab(a^2+b^2) = 2 \times \frac{1}{2} \left(2^2 + \frac{1}{2^2}\right) = 1 \times \left(4 + \frac{1}{4}\right) = \frac{17}{4} \)
R.H.S. \( = a^3b + ab^3 = 2^3 \times \frac{1}{2} + 2 \times \left(\frac{1}{2}\right)^3 = 4 + \frac{1}{4} = \frac{17}{4} \)
Since L.H.S. = R.H.S., the result is verified.

Exam Tip: When a value is a fraction like 1/2, keep it as a fraction throughout the substitution rather than converting to a decimal, to avoid rounding errors.

Question 18. Expand \( s(s^2 - st) \) and verify the result for \( s = 2 \) and \( t = 3 \).
Answer:
\( s \times s^2 - s \times st = s^3 - s^2t \)
When \( s = 2 \), \( t = 3 \): L.H.S. \( = s(s^2-st) = 2(2^2 - 2 \times 3) = 2(4-6) = -4 \)
R.H.S. \( = s^3 - s^2t = 2^3 - 2^2 \times 3 = 8 - 12 = -4 \)
Since L.H.S. = R.H.S., \( s(s^2-st) = s^3-s^2t \).

Exam Tip: Work out the value inside the bracket first when checking the original expression, then multiply by the outside term.

Question 19. Expand \( -3y(xy + y^2) \) and verify the result for \( x = 4 \) and \( y = 5 \).
Answer:
\( -3y \times xy - 3y \times y^2 = -3xy^2 - 3y^3 \)
When \( x = 4 \), \( y = 5 \): L.H.S. \( = -3y(xy+y^2) = -3 \times 5 \times (4 \times 5 + 5^2) = -15 \times (20+25) = -15 \times 45 = -675 \)
R.H.S. \( = -3xy^2 - 3y^3 = -3 \times 4 \times 25 - 3 \times 125 = -300 - 375 = -675 \)
Since L.H.S. = R.H.S., the result is verified.

Exam Tip: With a negative multiplier outside the bracket, every term produced inside also comes out negative.

Question 20. Simplify: \( a(b-c) + b(c-a) + c(a-b) \)
Answer:
\( ab - ac + bc - ba + ca - cb = ab - ab - ac + ac + bc - bc = 0 \)

Exam Tip: This is a classic identity that always simplifies to zero, no matter what values a, b and c take - a good pattern to recognise.

Question 21. Simplify: \( a(b-c) - b(c-a) - c(a-b) \)
Answer:
\( ab - ac - bc + ab - ca + cb = 2ab - 2ac = 2a(b-c) \)

Exam Tip: After expanding fully, look for a common factor across the surviving terms so the answer can be written compactly.

Question 22. Simplify: \( 3x^2 + 2(x+2) - 3x(2x+1) \)
Answer:
\( 3x^2 + 2x + 4 - 6x^2 - 3x = -3x^2 - x + 4 \)

Exam Tip: Expand every bracket fully before collecting like terms - trying to combine terms while brackets are still open leads to mistakes.

Question 23. Simplify: \( x(x+4) + 3x(2x^2-1) + 4x^2 + 4 \)
Answer:
\( x^2 + 4x + 6x^3 - 3x + 4x^2 + 4 = 6x^3 + 5x^2 + x + 4 \)

Exam Tip: Once every bracket is expanded, arrange the like terms from the highest power down to the constant for a tidy final answer.

Question 24. Simplify: \( 2x^2 + 3x(1-2x^3) + x(x+1) \)
Answer:
\( 2x^2 + 3x - 6x^4 + x^2 + x = -6x^4 + 3x^2 + 4x \)

Exam Tip: Watch for the highest power term, here x⁴ - it is easy to overlook when it appears only once, buried inside a bracket.

Question 25. Simplify: \( a^2b(a-b^2) + ab^2(4ab-2a^2) - a^3b(1-2b) \)
Answer:
\( a^3b - a^2b^3 + 4a^2b^3 - 2a^3b^2 - a^3b + 2a^3b^2 = 3a^2b^3 \)

Exam Tip: In a longer simplification, many terms cancel out completely - it is worth double-checking that the surviving term is correct on its own.

Question 26. Simplify: \( 4st(s-t) - 6s^2(t-t^2) - 3t^2(2s^2-s) + 2st(s-t) \)
Answer:
\( 4s^2t - 4st^2 - 6s^2t + 6s^2t^2 - 6s^2t^2 + 3st^2 + 2s^2t - 2st^2 = -3st^2 \)

Exam Tip: With four brackets to expand, work through them one at a time and keep a running tally of each type of term to avoid errors.