CBSE Class 8 Mathematics Algebraic Expressions and Identities Assignment Set 01

Very Short Answer type Questions

Question. If a = 3, b = 4 and x = 5. Find the value of (x + a) (x + b),
Answer : 72

Question. Simplify 4x² - 9y² / 2x + 3y
Answer : 2x – 3y

Question. Find (3pq + 4q)² – (3pq – 4q)².
Answer : 48pq²

Question. Find the value of (2/3 m - 3/2n)² + 2mn
Answer : 4/9m² + 9/4n²

Question. What should be added to a² + b² to get (a – b)².
Answer : –2ab

Question. Find the product of (t + s²) (t² – s).
Answer : t³ + t²s² – st – s³

Question. Find the product of (4m + 3) (4m + 1)
Answer : 16m² + 16m + 3

Question. Simplify 9x² + 16y² + 24xy / 3x + 4y
Answer : 3x + 4y

Question. Find the product of (xyz – 3) (xyz – 2).
Answer : x²y²z² – 5xyz + 6

Question. Simplify : (2x + 5)² – (2x – 5)².
Answer : 40x

Question. Simplify : (m² + 9) (m² – 9) + 81.
Answer : m⁴

Question. Multiply : (2.5m – 0.5n) (2.5m + 0.5n).
Answer : 6.25m² – 0.25n²

Question. Multiply : (a + 1) (a + 2)
Answer : a² + 3a + 2

Question. Simplify 6.25m² - 2.25n² / 2.5m - 1.5n
Answer : 2.5m + 1.5n

Question. Find the value of (4.9)².
Answer : 24.01

Algebraic Expressions and Identities

Question 1. Using identity \((x - a)(x + a) = x^2 - a^2\) find \(6^2 - 5^2\).
Answer: Given expression: \(6^2 - 5^2\)
Comparing with \(x^2 - a^2 = (x - a)(x + a)\), where \(x = 6\) and \(a = 5\).
\(6^2 - 5^2 = (6 - 5)(6 + 5)\)
\(= (1)(11)\)
\(= 11\)
In simple words: Apply the difference of squares identity by subtracting and adding the two numbers, then multiply the results.

Exam Tip: Recognize that \(x^2 - a^2\) can be factored directly to save calculation time over squaring large numbers.

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

Exam Tip: Be careful with negative signs when distributing terms across subtraction.

Question 3. Using suitable identity find \((a + 3)(a + 2)\).
Answer: Using identity \((x + a)(x + b) = x^2 + (a + b)x + ab\)
Here, \(x = a\), \(a = 3\), and \(b = 2\).
\((a + 3)(a + 2) = a^2 + (3 + 2)a + (3)(2)\)
\(= a^2 + 5a + 6\)
In simple words: Use the algebraic identity for the product of two binomials with a common term.

Exam Tip: Memorize the standard product identity \((x + a)(x + b)\) to quickly expand binomial products.

Question 4. Using identity \((a + b)^2 = a^2 + 2ab + b^2\) find the value of \(103^2\).
Answer: Express \(103\) as \((100 + 3)\).
\(103^2 = (100 + 3)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\) with \(a = 100\) and \(b = 3\):
\(= (100)^2 + 2(100)(3) + (3)^2\)
\(= 10000 + 600 + 9\)
\(= 10609\)
In simple words: Split the number into a round base of ten and a small integer, then expand using the square of a binomial.

Exam Tip: Always choose the base as a multiple of 10 or 100 to make mental multiplication easy.

Question 5. Using identity \((a - b)^2 = a^2 - 2ab + b^2\) find the value of \(98^2\).
Answer: Express \(98\) as \((100 - 2)\).
\(98^2 = (100 - 2)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = 100\) and \(b = 2\):
\(= (100)^2 - 2(100)(2) + (2)^2\)
\(= 10000 - 400 + 4\)
\(= 9600 + 4\)
\(= 9604\)
In simple words: Write the number as a subtraction from a neat power of ten and apply the square formula.

Exam Tip: Double-check subtraction signs when expanding the middle term in the minus identity.

Question 6. Using identity find \((2x + 3)^2\).
Answer: Using identity \((a + b)^2 = a^2 + 2ab + b^2\) with \(a = 2x\) and \(b = 3\):
\((2x + 3)^2 = (2x)^2 + 2(2x)(3) + (3)^2\)
\(= 4x^2 + 12x + 9\)
In simple words: Square the first term, add twice the product of both terms, and square the last term.

Exam Tip: Remember to square both the coefficient and the variable when evaluating \((2x)^2\).

Question 7. Subtract \(7x - 3x^2\) from \(4x + 8x^2\).
Answer: Setup the subtraction expression: \((4x + 8x^2) - (7x - 3x^2)\)
Remove parentheses by distributing the negative sign:
\(= 4x + 8x^2 - 7x + 3x^2\)
Group like terms together:
\(= (8x^2 + 3x^2) + (4x - 7x)\)
\(= 11x^2 - 3x\)
In simple words: Place the second expression first, change all signs inside the subtracted expression, and combine like terms.

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

Question 8. Using suitable identity find \((7x - 3y)^2\).
Answer: Using identity \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = 7x\) and \(b = 3y\):
\((7x - 3y)^2 = (7x)^2 - 2(7x)(3y) + (3y)^2\)
\(= 49x^2 - 42xy + 9y^2\)
In simple words: Apply the binomial square subtraction identity carefully.

Exam Tip: Ensure the product term includes both variables \(x\) and \(y\).

Question 9. Add \(4x^2 + 2xy - 4\) and \(7x^2 - 3xy + 4\).
Answer: Write the addition expression: \((4x^2 + 2xy - 4) + (7x^2 - 3xy + 4)\)
Group like terms:
\(= (4x^2 + 7x^2) + (2xy - 3xy) + (-4 + 4)\)
\(= 11x^2 - xy + 0\)
\(= 11x^2 - xy\)
In simple words: Combine the coefficients of identical variable terms and constant terms.

Exam Tip: Constant terms that add up to zero disappear from the final expression.

Question 10. Find the product of \(4x\), \(7x^2\), \(-2x\).
Answer: Multiply the numerical coefficients and the variables separately:
\(= (4 \cdot 7 \cdot -2) \cdot (x \cdot x^2 \cdot x)\)
\(= -56 \cdot x^{1 + 2 + 1}\)
\(= -56x^4\)
In simple words: Multiply all numbers together and add the exponents of the variable \(x\).

Exam Tip: Use the law of exponents \(x^a \cdot x^b = x^{a+b}\) when multiplying identical bases.

Question 11. Find the product of \((x^2 - y^2)(2x + y)\).
Answer: Distribute each term of the first binomial across the second binomial:
\(= x^2(2x + y) - y^2(2x + y)\)
\(= (2x^3 + x^2y) - (2xy^2 + y^3)\)
\(= 2x^3 + x^2y - 2xy^2 - y^3\)
In simple words: Multiply every term in the first bracket by every term in the second bracket.

Exam Tip: Check that all four distinct product terms are fully simplified with no like terms remaining.

Question 12. Simplify: \((xy + yz)^2 - (xy - yz)^2\)
Answer: Method 1 (Using identity \(a^2 - b^2 = (a - b)(a + b)\)):
Let \(A = (xy + yz)\) and \(B = (xy - yz)\).
\(A^2 - B^2 = [ (xy + yz) - (xy - yz) ] [ (xy + yz) + (xy - yz) ]\)
\(= [ xy + yz - xy + yz ] [ xy + yz + xy - yz ]\)
\(= [ 2yz ] [ 2xy ]\)
\(= 4xy^2z\)
In simple words: Use the difference of two squares identity to factor and simplify the expression rapidly.

Exam Tip: Applying the difference of squares identity avoids the lengthy process of expanding both squares individually.

Question 13. Using identity find the product of \(\left(\frac{a}{2} + \frac{3b}{4}\right)\left(\frac{a}{2} + \frac{3b}{4}\right)\)
Answer: The expression can be written as \(\left(\frac{a}{2} + \frac{3b}{4}\right)^2\).
Using identity \((x + y)^2 = x^2 + 2xy + y^2\):
\(= \left(\frac{a}{2}\right)^2 + 2\left(\frac{a}{2}\right)\left(\frac{3b}{4}\right) + \left(\frac{3b}{4}\right)^2\)
\(= \frac{a^2}{4} + \frac{3ab}{4} + \frac{9b^2}{16}\)
In simple words: Square the binomial expression by applying the standard sum-of-squares formula with fractions.

Exam Tip: Remember to square both the numerator and the denominator of each fractional term.

Question 14. Multiply: \((a^2 + 2c^2)(3a - 3c)\)
Answer: Multiply each term of the first factor by each term of the second factor:
\(= a^2(3a - 3c) + 2c^2(3a - 3c)\)
\(= (3a^3 - 3a^2c) + (6ac^2 - 6c^3)\)
\(= 3a^3 - 3a^2c + 6ac^2 - 6c^3\)
In simple words: Distribute both terms across the second polynomial and collect the resulting terms.

Exam Tip: Ensure all variables maintain correct alphabetical ordering in each term.

Question 15. Simplify: \((x + y)(2x - 3y + z) - (2x - 3y)z\)
Answer: Expand the first product: \(x(2x - 3y + z) + y(2x - 3y + z)\)
\(= 2x^2 - 3xy + zx + 2xy - 3y^2 + yz\)
Expand the second part: \(-(2x - 3y)z = -2xz + 3yz\)
Combine all expanded parts:
\(= 2x^2 - 3xy + xz + 2xy - 3y^2 + yz - 2xz + 3yz\)
Combine like terms:
\(x^2\text{ term: } 2x^2\)
\(xy\text{ terms: } -3xy + 2xy = -xy\)
\(y^2\text{ term: } -3y^2\)
\(xz\text{ terms: } xz - 2xz = -xz\)
\(yz\text{ terms: } yz + 3yz = 4yz\)
Final expression: \(= 2x^2 - xy - 3y^2 - xz + 4yz\)
In simple words: Expand both parts completely and carefully combine all like terms.

Exam Tip: Group terms by variable combinations to avoid missing any terms during simplification.

Question 16. Subtract \(3x(x - 4y + 5z)\) from \(4x(2x - 3y + 10z)\).
Answer: First, expand both expressions:
First expression \(= 4x(2x - 3y + 10z) = 8x^2 - 12xy + 40xz\)
Second expression \(= 3x(x - 4y + 5z) = 3x^2 - 12xy + 15xz\)
Perform the subtraction: \((8x^2 - 12xy + 40xz) - (3x^2 - 12xy + 15xz)\)
\(= 8x^2 - 12xy + 40xz - 3x^2 + 12xy - 15xz\)
Combine like terms:
\(= (8x^2 - 3x^2) + (-12xy + 12xy) + (40xz - 15xz)\)
\(= 5x^2 + 0 + 25xz\)
\(= 5x^2 + 25xz\)
In simple words: Expand both products first, then subtract the second polynomial from the first by reversing signs.

Exam Tip: Notice that the \(-12xy\) and \(+12xy\) terms cancel each other out completely.

Question 17. Simplify: \((x^2 - y^2)^2\)
Answer: Using identity \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = x^2\) and \(b = y^2\):
\((x^2 - y^2)^2 = (x^2)^2 - 2(x^2)(y^2) + (y^2)^2\)
\(= x^4 - 2x^2y^2 + y^4\)
In simple words: Apply the square of a difference identity, remembering to multiply exponents when raising a power to a power.

Exam Tip: \((x^2)^2 = x^4\) by the power rule of exponents.

Question 18. Using suitable identity find the product of \(\left(\frac{2}{3}x - 5\right)\left(\frac{2}{3}x + 5\right)\).
Answer: Using identity \((a - b)(a + b) = a^2 - b^2\) with \(a = \frac{2}{3}x\) and \(b = 5\):
\(\left(\frac{2}{3}x - 5\right)\left(\frac{2}{3}x + 5\right) = \left(\frac{2}{3}x\right)^2 - (5)^2\)
\(= \frac{4}{9}x^2 - 25\)
In simple words: Use the difference of squares identity since the terms in both brackets are identical except for the middle sign.

Exam Tip: Square both the numerator and denominator of the fraction \(\frac{2}{3}\) to get \(\frac{4}{9}\).

Question 19. Simplify \(3a(4a - 5) + 3\) and find its value for \(a = 3\).
Answer: Expand the expression:
\(3a(4a - 5) + 3 = 12a^2 - 15a + 3\)
Substitute \(a = 3\) into the simplified expression:
\(= 12(3)^2 - 15(3) + 3\)
\(= 12(9) - 45 + 3\)
\(= 108 - 45 + 3\)
\(= 63 + 3\)
\(= 66\)
In simple words: Expand the bracket first, then substitute the given numerical value for the variable.

Exam Tip: Always simplify the algebraic expression before plugging in values to minimize arithmetic errors.

Question 20. Using suitable identity find \(\left(6x^2 - \frac{5}{3}\right)^2\).
Answer: Using identity \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = 6x^2\) and \(b = \frac{5}{3}\):
\(\left(6x^2 - \frac{5}{3}\right)^2 = (6x^2)^2 - 2(6x^2)\left(\frac{5}{3}\right) + \left(\frac{5}{3}\right)^2\)
\(= 36x^4 - 2(2x^2)(5) + \frac{25}{9}\)
\(= 36x^4 - 20x^2 + \frac{25}{9}\)
In simple words: Apply the binomial square subtraction formula, simplifying the middle term's fraction.

Exam Tip: Cancel common factors in the middle term before multiplying coefficients.

Question 21. Using identity \(a^2 - b^2 = (a + b)(a - b)\), find \((1.02)^2 - (0.98)^2\).
Answer: Let \(a = 1.02\) and \(b = 0.98\).
\((1.02)^2 - (0.98)^2 = (1.02 + 0.98)(1.02 - 0.98)\)
\(= (2.00)(0.04)\)
\(= 2 \cdot 0.04\)
\(= 0.08\)
In simple words: Factor the difference of squares into the sum and difference of the two decimals, then multiply.

Exam Tip: This identity turns a difficult decimal squaring problem into simple addition and subtraction.

Question 22. Using \((x + a)(x + b) = x^2 + (a + b)x + ab\) find \(105 \times 107\).
Answer: Express \(105 \times 107\) as \((100 + 5)(100 + 7)\).
Here, \(x = 100\), \(a = 5\), and \(b = 7\).
\((100 + 5)(100 + 7) = (100)^2 + (5 + 7)(100) + (5)(7)\)
\(= 10000 + (12)(100) + 35\)
\(= 10000 + 1200 + 35\)
\(= 11235\)
In simple words: Write both numbers as a base of one hundred plus their offsets, then apply the binomial product identity.

Exam Tip: Choosing 100 as the common term makes multiplication by powers of ten effortless.

Question 23. Using identity find the value of \((7.2)^2\).
Answer: Express \(7.2\) as \((7 + 0.2)\).
\((7.2)^2 = (7 + 0.2)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\) with \(a = 7\) and \(b = 0.2\):
\(= (7)^2 + 2(7)(0.2) + (0.2)^2\)
\(= 49 + 2.8 + 0.04\)
\(= 51.84\)
In simple words: Split the decimal into an integer and fractional part, then use the square of a sum identity.

Question 24. Using identity evaluate \(297 \times 303\).
Answer: Express numbers around a common base: \(297 = 300 - 3\) and \(303 = 300 + 3\).
\(297 \times 303 = (300 - 3)(300 + 3)\)
Using identity \((a - b)(a + b) = a^2 - b^2\) with \(a = 300\) and \(b = 3\):
\(= (300)^2 - (3)^2\)
\(= 90000 - 9\)
\(= 89991\)
In simple words: Write the expression as the product of a sum and difference from 300.

Exam Tip: Recognizing equidistant numbers allows direct application of the difference of squares identity.

Question 25. Using identity find the value of \((4.7)^2\).
Answer: Express \(4.7\) as \((5 - 0.3)\).
\((4.7)^2 = (5 - 0.3)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\) with \(a = 5\) and \(b = 0.3\):
\(= (5)^2 - 2(5)(0.3) + (0.3)^2\)
\(= 25 - 3 + 0.09\)
\(= 22 + 0.09\)
\(= 22.09\)
In simple words: Represent the decimal as a subtraction from a whole integer and apply the binomial square formula.

Exam Tip: Squaring \(0.3\) gives \(0.09\), not \(0.9\).

Question 26. Simplify \(\left(\frac{5}{3}x + \frac{3}{4}y\right)^2 - \left(\frac{5}{3}x - \frac{3}{4}y\right)^2\) and also evaluate it when \(x = 2\) and \(y = -1\).
Answer: Let \(A = \left(\frac{5}{3}x + \frac{3}{4}y\right)\) and \(B = \left(\frac{5}{3}x - \frac{3}{4}y\right)\).
Using \(A^2 - B^2 = (A - B)(A + B)\):
\(= \left[ \left(\frac{5}{3}x + \frac{3}{4}y\right) - \left(\frac{5}{3}x - \frac{3}{4}y\right) \right] \left[ \left(\frac{5}{3}x + \frac{3}{4}y\right) + \left(\frac{5}{3}x - \frac{3}{4}y\right) \right]\)
\(= \left[ \frac{5}{3}x + \frac{3}{4}y - \frac{5}{3}x + \frac{3}{4}y \right] \left[ \frac{5}{3}x + \frac{3}{4}y + \frac{5}{3}x - \frac{3}{4}y \right]\)
\(= \left[ \frac{6}{4}y \right] \left[ \frac{10}{3}x \right]\)
\(= \left( \frac{3}{2}y \right) \left( \frac{10}{3}x \right)\)
\(= \frac{3 \cdot 10}{2 \cdot 3}xy\)
\(= 5xy\)
Now, evaluate for \(x = 2\) and \(y = -1\):
\(= 5(2)(-1)\)
\(= -10\)
In simple words: Simplify the difference of two squared binomials using the algebraic difference of squares identity before substituting values.

Exam Tip: Factoring first reduces a massive expansion involving fractions into a simple single-term product.

Question 27. Simplify \((xy + yz)^2 - 2x^2y^2z\). Find the value when \(x = -1\), \(y = 1\) and \(z = 2\).
Answer: Expand \((xy + yz)^2\):
\((xy + yz)^2 = (xy)^2 + 2(xy)(yz) + (yz)^2\)
\(= x^2y^2 + 2xy^2z + y^2z^2\)
Substitute into the full expression:
\(= x^2y^2 + 2xy^2z + y^2z^2 - 2x^2y^2z\)
Notice there are no like terms to combine further, so the simplified expression is:
\(= x^2y^2 + 2xy^2z + y^2z^2 - 2x^2y^2z\)
Now substitute \(x = -1\), \(y = 1\), and \(z = 2\):
\(= (-1)^2(1)^2 + 2(-1)(1)^2(2) + (1)^2(2)^2 - 2(-1)^2(1)^2(2)\)
\(= (1)(1) + 2(-1)(1)(4) + (1)(4) - 2(1)(1)(2)\)
\(= 1 - 8 + 4 - 4\)
\(= -7 + 4 - 4\)
\(= -7\)
In simple words: Expand the binomial square, combine any like terms, and substitute the given values step by step.

Exam Tip: Pay close attention to negative signs when evaluating powers of negative numbers.

Question 28. Simplify: \((1.5x - 4y)(1.5x + 4y + 3) - 4.5x + 12y\)
Answer: Expand the first product: \(1.5x(1.5x + 4y + 3) - 4y(1.5x + 4y + 3)\)
\(= 2.25x^2 + 6xy + 4.5x - 6xy - 16y^2 - 12y\)
Bring down the remaining terms from the expression: \(- 4.5x + 12y\)
Combine all parts:
\(= 2.25x^2 + 6xy + 4.5x - 6xy - 16y^2 - 12y - 4.5x + 12y\)
Group and cancel like terms:
\(x^2\text{ term: } 2.25x^2\)
\(xy\text{ terms: } 6xy - 6xy = 0\)
\(x\text{ terms: } 4.5x - 4.5x = 0\)
\(y^2\text{ term: } -16y^2\)
\(y\text{ terms: } -12y + 12y = 0\)
Final simplified expression: \(= 2.25x^2 - 16y^2\)
In simple words: Multiply out all terms and watch how many terms cancel out completely to leave a clean binomial.

Exam Tip: Cancellation of multiple terms is a sign that the algebraic expansion is correct.

Most Important Questions

Question 1. What are algebraic expressions?
Answer: An algebraic expression is a combination of constants and variables connected by mathematical operations such as addition, subtraction, multiplication, and division.
In simple words: Mathematical phrases built using numbers, letters, and operation signs.

Exam Tip: Ensure you distinguish between algebraic expressions and algebraic equations (which contain an equals sign).

Question 2. Expressions consists of ___________ & ___________ .
Answer: Terms & Factors (or Constants & Variables)
In simple words: Expressions are made up of terms, which in turn consist of factors.

Exam Tip: Know the hierarchical breakdown of algebraic structures.

Question 3. T/F. The value of an expression changes with the value chosen for the variables it contains.
Answer: True
In simple words: Changing the number assigned to a variable changes the overall value of the expression.

Exam Tip: Variables represent values that can vary, hence the name.

Question 4. When numbers/literals are added or subtracted, they are called _________ .
Answer: Terms
In simple words: Parts of an expression separated by plus or minus signs are terms.

Exam Tip: Addition and subtraction separate terms, while multiplication and division form parts of a single term.

Question 5. When numbers/literals are multiplied, they are called ________.
Answer: Factors
In simple words: Individual multiplied components that form a term are called factors.

Exam Tip: Do not confuse terms (separated by \(+\) or \(-\)) with factors (multiplied together).

Question 6. The terms in the expression \(4ab + 5a(b + c)\) are:
Answer: First expand \(5a(b + c) = 5ab + 5ac\).
The terms in the expression are \(4ab\), \(5ab\), and \(5ac\) (or viewed in raw form as \(4ab\) and \(5a(b+c)\)).
In simple words: The distinct parts separated by plus or minus signs in the expression.

Exam Tip: Always expand bracketed expressions to correctly identify all individual terms.

Question 7. The factors in the term \(5a(b + c)\) are :
Answer: \(5\), \(a\), and \((b + c)\)
In simple words: The individual quantities multiplied together to form the term.

Exam Tip: Binomial expressions in parentheses count as a single factor unit when multiplied.

Question 8. A monomial is an expression in which ____________________ .
Answer: there is only one term.
In simple words: A single-term algebraic expression.

Exam Tip: Example: \(7xy^2\) is a monomial.

Question 9. A binomial is an expression in which ___________________ .
Answer: there are exactly two unlike terms.
In simple words: An expression made of two terms.

Exam Tip: Example: \(3x + 4y\) is a binomial.

Question 10. While multiplying two monomials, Coefficient of product = _________ X ___________.
Answer: Coefficient of first monomial \(\times\) Coefficient of second monomial
In simple words: Multiply the numerical numbers of both monomials to get the new coefficient.

Exam Tip: Coefficients multiply with coefficients, and variables multiply with variables.

Question 11. Identify the terms, their coefficients for the expression: \(0.75x + 0.44y + 1.56zx\)
Answer:
Term 1: \(0.75x\), Coefficient \(= 0.75\)
Term 2: \(0.44y\), Coefficient \(= 0.44\)
Term 3: \(1.56zx\), Coefficient \(= 1.56\)
In simple words: List each term and state the numerical number multiplying the variables.

Exam Tip: The coefficient includes its sign (positive or negative).

Question 12. Classify the following as binomials and trinomials: \(2a + 3b\), \(2x + 3y - 5\), \(a + 4\), \(12x + 13y + 17z\)
Answer:
Binomials (2 terms): \(2a + 3b\), \(a + 4\)
Trinomials (3 terms): \(2x + 3y - 5\), \(12x + 13y + 17z\)
In simple words: Count the number of terms to classify as binomial (two) or trinomial (three).

Exam Tip: Prefix "bi" means two, and "tri" means three.

Question 13. What are the polynomials? Give an example.
Answer: An expression containing one or more terms with non-negative integer exponents for the variables is called a polynomial.
Example: \(3x^3 - 5x^2 + 7x - 9\)
In simple words: Expressions with one or many terms where variables have whole number powers.

Exam Tip: Monomials, binomials, and trinomials are all special types of polynomials.

Question 14. What are like and unlike terms?
Answer: Like terms are terms that have the same variables with identical exponents. Unlike terms are terms that have different variables or different exponents on the same variables.
In simple words: Like terms match exact letter variables and powers; unlike terms do not.

Exam Tip: Only like terms can be added or subtracted together.

Question 15. Classify as like and unlike terms: \(2abc\) and \(bac\), \(x^2y^2z\) and \(y^2zx^2\), \(7x\) and \(3y\), \(xy + z\) and \(xyz\)
Answer:
1. \(2abc\) and \(bac\) - Like terms (order of multiplication does not change variables)
2. \(x^2y^2z\) and \(y^2zx^2\) - Like terms (same variable powers)
3. \(7x\) and \(3y\) - Unlike terms (different variables)\(x\) vs \(y\)
4. \(xy + z\) and \(xyz\) - Unlike terms (different structures and variable combinations)
In simple words: Check if variable letters and their exponents match regardless of order.

Exam Tip: Multiplication is commutative, so \(abc\) is identical to \(bac\).

Question 16. Add: \(2p^2q^2 - 3pq + 4\) & \(5 + 7pq - 3p^2q^2\)
Answer: Write the sum: \((2p^2q^2 - 3pq + 4) + (5 + 7pq - 3p^2q^2)\)
Group like terms:
\(= (2p^2q^2 - 3p^2q^2) + (-3pq + 7pq) + (4 + 5)\)
\(= -p^2q^2 + 4pq + 9\)
In simple words: Combine coefficients of matching variable terms and constants.

Exam Tip: Grouping terms with identical variable parts prevents arithmetic mistakes.

Question 17. Add: \(-8x^2 + 2y^2z - 11\) and \(-2x^2 + 9y^2z - 1\)
Answer: Write the sum: \((-8x^2 + 2y^2z - 11) + (-2x^2 + 9y^2z - 1)\)
Group like terms:
\(= (-8x^2 - 2x^2) + (2y^2z + 9y^2z) + (-11 - 1)\)
\(= -10x^2 + 11y^2z - 12\)
In simple words: Add coefficients of like terms together.

Exam Tip: Watch negative signs carefully when adding numbers with the same sign.

Question 18. Subtract \(4p^2q + 5pq + 5pq^2 - 8p + 7q - 10\) from \(18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q\).
Answer: Setup subtraction: \((18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q) - (4p^2q + 5pq + 5pq^2 - 8p + 7q - 10)\)
Distribute the negative sign to all terms in the second polynomial:
\(= 18 - 3p - 11q + 5pq - 2pq^2 + 5p^2q - 4p^2q - 5pq - 5pq^2 + 8p - 7q + 10\)
Group like terms:
Constants: \(18 + 10 = 28\)
\(p\text{ terms: } -3p + 8p = 5p\)
\(q\text{ terms: } -11q - 7q = -18q\)
\(pq\text{ terms: } 5pq - 5pq = 0\)
\(pq^2\text{ terms: } -2pq^2 - 5pq^2 = -7pq^2\)
\(p^2q\text{ terms: } 5p^2q - 4p^2q = p^2q\)
Final expression: \(= 28 + 5p - 18q - 7pq^2 + p^2q\)
In simple words: Reverse every sign in the polynomial being subtracted and combine like terms.

Exam Tip: Double-check that every term from the subtracted polynomial has its sign flipped.

Question 19. Find the product of : \(2z\), \(4y\), \(2y^2\) & \(6xyz\)
Answer: Multiply coefficients and variables:
Coefficient product \(= 2 \cdot 4 \cdot 2 \cdot 6 = 96\)
Variable product \(= z \cdot y \cdot y^2 \cdot xyz\)
\(= x^1 \cdot (y \cdot y^2 \cdot y) \cdot (z \cdot z)\)
\(= x \cdot y^4 \cdot z^2\)
Final product \(= 96xy^4z^2\)
In simple words: Multiply all numbers together and add exponents for each matching variable letter.

Exam Tip: List variables in alphabetical order in the final monomial product.

Question 20. Find : a) \(13mn \times 13np\) b) \(-4xy \times -7x^2y\)
Answer:
a) \(13mn \times 13np = (13 \cdot 13) \cdot (m) \cdot (n \cdot n) \cdot (p)\)
\(= 169mn^2p\)
b) \(-4xy \times -7x^2y = (-4 \cdot -7) \cdot (x \cdot x^2) \cdot (y \cdot y)\)
\(= 28x^3y^2\)
In simple words: Multiply coefficients and combine identical variables using exponent laws.

Exam Tip: Negative multiplied by negative gives a positive coefficient.

Question 21. State the distributive property.
Answer: The distributive property states that multiplication distributes over addition and subtraction: \(a(b + c) = ab + ac\) and \(a(b - c) = ab - ac\).
In simple words: Multiplying a number by a sum gives the same result as multiplying the number by each addend and then adding the products.

Exam Tip: This is a foundational property used heavily in polynomial expansion.

Question 22. T/F: \(n(4 + m) = 4n + nm\)
Answer: True
In simple words: Distributing \(n\) to both \(4\) and \(m\) gives \(4n + nm\).

Exam Tip: \(nm\) can also be written alphabetically as \(mn\).

Question 23. T/F : \(p(9 - p) = 9p - 2p\)
Answer: False
Correct expansion: \(p(9 - p) = 9p - p^2\).
In simple words: Multiplying \(p\) by \(p\) gives \(p^2\), not \(2p\).

Exam Tip: Multiplying identical variables adds exponents (\(p^1 \cdot p^1 = p^2\)), whereas adding them gives coefficients (\(p + p = 2p\)).

Question 24. Whenever we multiply a binomial by a binomial, we get _____terms in the product.
Answer: 4 terms (before combining like terms)
In simple words: Multiplying two terms by two terms initially produces four distinct product terms.

Exam Tip: These four terms may reduce to fewer if like terms combine.

Question 25. Whenever we multiply a binomial by a trinomial, we get ____terms in the product.
Answer: 6 terms (before combining like terms)
In simple words: Two terms multiplied by three terms initially yields six distinct products.

Exam Tip: Always check for like terms to simplify the expanded polynomial.

Question 26. Find : \(a^2(2ab - 5c)\)
Answer: Distribute \(a^2\) across the binomial:
\(= a^2(2ab) - a^2(5c)\)
\(= 2a^{2+1}b - 5a^2c\)
\(= 2a^3b - 5a^2c\)
In simple words: Multiply \(a^2\) into each term inside the bracket, adding exponents for matching variables.

Exam Tip: \(a^2 \cdot a = a^3\).

Question 27. Simplify \(x(x - 3) + 2\) and evaluate for \(x = 2\).
Answer: Expand the expression:
\(x(x - 3) + 2 = x^2 - 3x + 2\)
Substitute \(x = 2\):
\(= (2)^2 - 3(2) + 2\)
\(= 4 - 6 + 2\)
\(= -2 + 2\)
\(= 0\)
In simple words: Expand the bracket first, then plug in \(2\) for \(x\) and evaluate.

Exam Tip: Follow order of operations (exponents first, then multiplication, then addition/subtraction).

Question 28. Find: \(\left(-\frac{7}{3}pq\right) \times \left(\frac{6}{5}pqr\right)\)
Answer: Multiply coefficients and variables separately:
Coefficient product \(= \left(-\frac{7}{3}\right) \times \left(\frac{6}{5}\right) = \frac{-42}{15} = -\frac{14}{5}\)
Variable product \(= (pq) \times (pqr) = (p \cdot p) \cdot (q \cdot q) \cdot (r) = p^2q^2r\)
Final product \(= -\frac{14}{5}p^2q^2r\)
In simple words: Multiply fractions normally, simplify, and add exponents for matching variables.

Exam Tip: Cross-cancel numbers before multiplying fractions to keep calculations simple.

Question 29. Find : \((3x^6) \times (6x^{12}) \times (9x^{18})\)
Answer: Multiply coefficients and add exponents of \(x\):
Coefficient product \(= 3 \times 6 \times 9 = 18 \times 9 = 162\)
Exponent sum \(= 6 + 12 + 18 = 36\)
Final product \(= 162x^{36}\)
In simple words: Multiply all numbers together and add up all the exponents of \(x\).

Exam Tip: Use the exponent rule \(x^a \cdot x^b \cdot x^c = x^{a+b+c}\).

Question 30. Find using distributive property : \(125 \times 42\)
Answer: Rewrite \(42\) as \((40 + 2)\):
\(125 \times 42 = 125 \times (40 + 2)\)
Apply distributive property: \((125 \times 40) + (125 \times 2)\)
\(= 5000 + 250\)
\(= 5250\)
In simple words: Split one number into tens and units to multiply mentally using the distributive law.

Exam Tip: Splitting numbers into multiples of 10 makes mental arithmetic much faster.

Question 31. Find : \(ab(a^2 + bc + c^2)\)
Answer: Distribute \(ab\) to each term inside the bracket:
\(= ab(a^2) + ab(bc) + ab(c^2)\)
\(= a^{1+2}b + ab^2c + abc^2\)
\(= a^3b + ab^2c + abc^2\)
In simple words: Multiply \(ab\) across every term in the expression.

Exam Tip: Ensure variables are multiplied correctly by combining like bases.

Question 32. Find : \(5m(m^2 + m + 1)\) and evaluate for \(m = 1\).
Answer: Expand the expression:
\(5m(m^2 + m + 1) = 5m(m^2) + 5m(m) + 5m(1)\)
\(= 5m^3 + 5m^2 + 5m\)
Now evaluate for \(m = 1\):
\(= 5(1)^3 + 5(1)^2 + 5(1)\)
\(= 5(1) + 5(1) + 5(1)\)
\(= 5 + 5 + 5\)
\(= 15\)
In simple words: Distribute \(5m\) into the polynomial, then substitute \(1\) for \(m\).

Exam Tip: Powers of \(1\) always equal \(1\), making evaluation simple.

Question 33. Multiply : \((2.5a - 0.5b)\) and \((2.5a + 0.5b + c)\)
Answer: Distribute each term of the first binomial across the trinomial:
\(= 2.5a(2.5a + 0.5b + c) - 0.5b(2.5a + 0.5b + c)\)
\(= (6.25a^2 + 1.25ab + 2.5ac) - (1.25ab + 0.25b^2 + 0.5bc)\)
Remove parentheses and combine like terms:
\(= 6.25a^2 + 1.25ab + 2.5ac - 1.25ab - 0.25b^2 - 0.5bc\)
\(= 6.25a^2 + (1.25ab - 1.25ab) + 2.5ac - 0.25b^2 - 0.5bc\)
\(= 6.25a^2 + 2.5ac - 0.25b^2 - 0.5bc\)
In simple words: Multiply every term in the first bracket by every term in the second bracket and cancel opposing terms.

Exam Tip: The \(1.25ab\) and \(-1.25ab\) terms cancel out completely.

Question 34. Multiply : \((ab + 5)(a + c^2)(b + 6)\)
Answer: First multiply the binomials \((ab + 5)\) and \((b + 6)\) or rearrange factors. Let us multiply \((ab + 5)\) by \((b + 6)\):
\((ab + 5)(b + 6) = ab(b) + ab(6) + 5(b) + 5(6)\)
\(= ab^2 + 6ab + 5b + 30\)
Now multiply this result by \((a + c^2)\):
\(= (ab^2 + 6ab + 5b + 30)(a + c^2)\)
\(= a(ab^2 + 6ab + 5b + 30) + c^2(ab^2 + 6ab + 5b + 30)\)
\(= (a^2b^2 + 6a^2b + 5ab + 30a) + (abc^2 + 6abc^2 + 5bc^2 + 30c^2)\)
\(= a^2b^2 + 6a^2b + 5ab + 30a + abc^2 + 6abc^2 + 5bc^2 + 30c^2\)
In simple words: Multiply pairs of binomials sequentially using the distributive property.

Exam Tip: Break triple products into two-step multiplications to avoid errors.

Question 35. Simplify : \((m + n)(3m + n) + (m + 2n)(m - n)\)
Answer: Expand the first product \((m + n)(3m + n)\):
\(= m(3m + n) + n(3m + n)\)
\(= 3m^2 + mn + 3mn + n^2\)
\(= 3m^2 + 4mn + n^2\)
Expand the second product \((m + 2n)(m - n)\):
\(= m(m - n) + 2n(m - n)\)
\(= m^2 - mn + 2mn - 2n^2\)
\(= m^2 + mn - 2n^2\)
Add both expanded parts together:
\(= (3m^2 + 4mn + n^2) + (m^2 + mn - 2n^2)\)
Combine like terms:
\(m^2\text{ terms: } 3m^2 + m^2 = 4m^2\)
\(mn\text{ terms: } 4mn + mn = 5mn\)
\(n^2\text{ terms: } n^2 - 2n^2 = -n^2\)
Final expression: \(= 4m^2 + 5mn - n^2\)
In simple words: Expand each binomial product separately, then combine all resulting like terms.

Exam Tip: Keep brackets around each product until all expansions are finished.

Question 36. Simplify: \((3.5e - 4.5f)(1.5e + 4f + ef) - 4.5e + 10f\)
Answer: Expand the product \((3.5e - 4.5f)(1.5e + 4f + ef)\):
\(= 3.5e(1.5e + 4f + ef) - 4.5f(1.5e + 4f + ef)\)
\(= (5.25e^2 + 14ef + 3.5e^2f) - (6.75ef + 18f^2 - 4.5ef^2)\)
\(= 5.25e^2 + 14ef + 3.5e^2f - 6.75ef - 18f^2 - 4.5ef^2\)
Combine like terms inside the product: \(14ef - 6.75ef = 7.25ef\)
Expression becomes: \(5.25e^2 + 7.25ef + 3.5e^2f - 18f^2 - 4.5ef^2 - 4.5e + 10f\)
In simple words: Distribute all terms carefully and combine like variable combinations.

Exam Tip: Decimal multiplication requires careful placement of decimal points.

Question 37. Simplify : \((3.5e - 4.5f)(1.5e + 4f + ef) - 4.5e + 10f\)
Answer: (Identical to Question 36)
\(= 5.25e^2 + 7.25ef + 3.5e^2f - 18f^2 - 4.5ef^2 - 4.5e + 10f\)
In simple words: Identical expression evaluated following standard polynomial expansion steps.

Exam Tip: Verify each intermediate product to ensure no terms are missed.

Question 38. What is an identity.
Answer: An identity is an equality relation that is true for all possible values of the variables involved in it.
In simple words: An equation that holds true no matter what numbers you substitute for the variables.

Exam Tip: Distinguish between equations (true for specific values) and identities (true for all values).

Question 39. True/False. An equation, which is true for only certain values of the variable in it, is not an identity.
Answer: True
In simple words: If an equation works only for specific numbers, it is a conditional equation, not an identity.

Exam Tip: Identities work universally for any variable replacement.

Question 40. \((x + a)(x + b) = \text{________________}\)
Answer: \(x^2 + (a + b)x + ab\)
In simple words: The standard algebraic product identity for binomials with a common term.

Exam Tip: Memorize this identity for quick expansion of quadratic forms.

Question 41. \((a - b)^2 = \text{________________}\)
Answer: \(a^2 - 2ab + b^2\)
In simple words: The square of a binomial difference identity.

Exam Tip: Never forget the middle term \(-2ab\).

Question 42. \((x - a)(x + a) = \text{___________}\).
Answer: \(x^2 - a^2\)
In simple words: The difference of squares identity.

Exam Tip: The middle terms cancel out in this product.

Question 43. If \(x + \frac{1}{x} = 6\), find \(x^2 + \frac{1}{x^2}\).
Answer: Square both sides of the given equation \(\left(x + \frac{1}{x}\right) = 6\):
\(\left(x + \frac{1}{x}\right)^2 = 6^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\):
\(x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 36\)
\(x^2 + 2 + \frac{1}{x^2} = 36\)
\(x^2 + \frac{1}{x^2} = 36 - 2\)
\(x^2 + \frac{1}{x^2} = 34\)
In simple words: Square the given sum equation and simplify the middle reciprocal term to solve for the squares.

Exam Tip: The product \(x \cdot \frac{1}{x}\) always simplifies to \(1\).

Question 44. If \(x + y = 12\) and \(xy = 32\). Find the value of \(x^2 + y^2\).
Answer: Use the identity \((x + y)^2 = x^2 + 2xy + y^2\).
Substitute the known values \(x + y = 12\) and \(xy = 32\):
\((12)^2 = x^2 + 2(32) + y^2\)
\(144 = x^2 + 64 + y^2\)
\(x^2 + y^2 = 144 - 64\)
\(x^2 + y^2 = 80\)
In simple words: Square the sum equation, substitute the known product value, and isolate \(x^2 + y^2\).

Exam Tip: This is a standard algebraic substitution problem frequently tested in exams.

Question 45. Find using identities: \(106^2\)
Answer: Express \(106\) as \((100 + 6)\).
\(106^2 = (100 + 6)^2\)
Using \((a + b)^2 = a^2 + 2ab + b^2\):
\(= (100)^2 + 2(100)(6) + (6)^2\)
\(= 10000 + 1200 + 36\)
\(= 11236\)
In simple words: Split the number into \(100 + 6\) and apply the square formula.

Exam Tip: Using a base of 100 makes squaring and multiplication mental tasks.

Question 46. Find using identities: \((4.8)^2\)
Answer: Express \(4.8\) as \((5 - 0.2)\).
\((4.8)^2 = (5 - 0.2)^2\)
Using \((a - b)^2 = a^2 - 2ab + b^2\):
\(= (5)^2 - 2(5)(0.2) + (0.2)^2\)
\(= 25 - 2 + 0.04\)
\(= 23 + 0.04\)
\(= 23.04\)
In simple words: Write the decimal as \(5 - 0.2\) and use the subtraction identity.

Exam Tip: Squaring \(0.2\) gives \(0.04\).

Question 47. Find using identities: \((-p + q)(-p + q)\)
Answer: The expression is \((-p + q)^2\).
Using \((a + b)^2 = a^2 + 2ab + b^2\) with \(a = -p\) and \(b = q\):
\((-p + q)^2 = (-p)^2 + 2(-p)(q) + (q)^2\)
\(= p^2 - 2pq + q^2\)
In simple words: Apply the binomial square formula treating the first term as negative.

Exam Tip: \((-p)^2 = p^2\) because squaring a negative number yields a positive result.

Question 48. Find : \((2x + 5y)(2x + 3y)\)
Answer: Using identity \((x + a)(x + b) = x^2 + (a + b)x + ab\) with \(x = 2x\), \(a = 5y\), and \(b = 3y\):
\((2x + 5y)(2x + 3y) = (2x)^2 + (5y + 3y)(2x) + (5y)(3y)\)
\(= 4x^2 + (8y)(2x) + 15y^2\)
\(= 4x^2 + 16xy + 15y^2\)
In simple words: Apply the common first term binomial product identity.

Exam Tip: Remember to substitute \(2x\) everywhere \(x\) appears in the standard identity formula.

Question 49. Find : \((2x - y)(2x + y)(4x^2 + y^2)\)
Answer: First, multiply the first two factors using \((a - b)(a + b) = a^2 - b^2\):
\((2x - y)(2x + y) = (2x)^2 - (y)^2 = 4x^2 - y^2\)
Now multiply this result by the remaining factor \((4x^2 + y^2)\):
\(= (4x^2 - y^2)(4x^2 + y^2)\)
Apply the difference of squares identity again with \(a = 4x^2\) and \(b = y^2\):
\(= (4x^2)^2 - (y^2)^2\)
\(= 16x^4 - y^4\)
In simple words: Apply the difference of squares identity twice consecutively to simplify the expression.

Exam Tip: Recognizing consecutive difference of squares patterns saves considerable expansion time.

Question 50. Multiply : \([p^2 + (qr)^2][p^2 - (qr)^2]\) and evaluate for \(p = 1\), \(q = 2\), \(r = 3\).
Answer: Using identity \((a + b)(a - b) = a^2 - b^2\) with \(a = p^2\) and \(b = (qr)^2\):
\(= (p^2)^2 - ((qr)^2)^2\)
\(= p^4 - q^2r^2\)
Now, evaluate for \(p = 1\), \(q = 2\), \(r = 3\):
\(p^4 - (2)^2(3)^2 = (1)^4 - (4)(9)\)
\(= 1 - 36\)
\(= -35\)
In simple words: Simplify using the difference of squares identity first, then substitute the given numerical values.

Exam Tip: Simplify the algebraic expression completely before plugging in values.

Question 51. Derive the identity : \((x + a)(x + b) = x^2 + (a + b)x + ab\)
Answer: Start with the left-hand side product: \((x + a)(x + b)\)
Distribute \((x + a)\) across \((x + b)\):
\(= x(x + b) + a(x + b)\)
Expand each term:
\(= x \cdot x + x \cdot b + a \cdot x + a \cdot b\)
\(= x^2 + bx + ax + ab\)
Combine the middle like terms (\(bx + ax = (a + b)x\)):
\(= x^2 + (a + b)x + ab\)
This equals the right-hand side. Hence derived.
In simple words: Multiply out the binomials using distribution and combine the middle like terms.

Exam Tip: Showing every distributive step is essential for complete algebraic proofs.

Question 52. Find the product : \(\left(a - \frac{b}{2} - 1\right)\left(a + \frac{b}{2} + 1\right)\)
Answer: Rearrange the expression by grouping terms:
\(= \left[ a - \left(\frac{b}{2} + 1\right) \right] \left[ a + \left(\frac{b}{2} + 1\right) \right]\)
Let \(x = a\) and \(y = \left(\frac{b}{2} + 1\right)\). The expression becomes \((x - y)(x + y)\).
Using identity \((x - y)(x + y) = x^2 - y^2\):
\(= (a)^2 - \left(\frac{b}{2} + 1\right)^2\)
Expand \(\left(\frac{b}{2} + 1\right)^2\) using \((u + v)^2 = u^2 + 2uv + v^2\):
\(= a^2 - \left[ \left(\frac{b}{2}\right)^2 + 2\left(\frac{b}{2}\right)(1) + (1)^2 \right]\)
\(= a^2 - \left[ \frac{b^2}{4} + b + 1 \right]\)
Distribute the negative sign:
\(= a^2 - \frac{b^2}{4} - b - 1\)
In simple words: Group terms cleverly to treat the expression as a difference of two squares.

Exam Tip: Regrouping terms into a form matching \((x - y)(x + y)\) avoids long, tedious multiplication.