Q1: Terms with the same algebraic factors are called ____________ terms.
Ans: Like
Explanation: Like terms have the same variables raised to the same powers (e.g., 3x and 5x).
Q2: A ________________ can take any value and ________________ has a fixed value.
Ans: Variable, constant
Explanation: A variable changes (e.g., x), a constant stays the same (e.g., 5).
Q3: An expression with one or more terms is called _____________
Ans: Algebraic expression
Explanation: An algebraic expression is a combination of variables and constants.
Examples:
2x+3 → algebraic expression
Q4: An expression with one term is called __________________ with two terms is ______________ and with three terms is _______________
Ans: Monomial, binomial, trinomial
Explanation:
1 term = monomial (e.g., 4x)
2 terms = binomial (e.g., x + y)
3 terms = trinomial (e.g., x² + 2x + 3)
Q5: An algebraic expression with equality sign is called ______________
Ans: Equation
Explanation: An equation has an equals (=) sign between two expressions.
Q1: The degree of a constant term is 0
Ans: True
Explanation: A constant term (like 3, −7, or 100) has no variable.
It can be written as: 3 = 3 × x⁰
Q2: The difference between two like terms is a like term.
Ans: True
Explanation: Like terms have the same variables and powers.
For example: 6x and 4x are like terms.
Their difference: 6x − 4x = 2x
Q3: 1 is an algebraic expression
Ans: True
Explanation:
An algebraic expression can include: constants, variables, or both.
The number 1 is a constant, so it's a valid algebraic expression.
Q4: The expression x + y + 5x is a trinomial.
Ans: False
Explanation: Before deciding the number of terms, we must combine like terms.
Here, x and 5x are like terms.
x + y + 5x = 6x + y → Only 2 terms
→ It's a binomial, not a trinomial.
Q5: In like terms, the numerical coefficients should also be the same
Ans: False
Explanation: Like terms need to have the same variables with the same powers, but the coefficients can be different.
For example: 2xy and 7xy are like terms (same variables, different coefficients)
Q1: The volume of a rectangular box where length, breadth, and height are 2a,4b,8crespectively.
Ans: Given: Length of a rectangular box, l=2a
Breadth of rectangular box, b=4b
Height of rectangular box, h=8c
We need to find the volume of the rectangular box with the given dimensions.
We know,
The volume of a cuboid =l×b×h
=2a × 4b × 8c
=64abc
Q2: Carry out the multiplication of the expressions in each of the following pairs.
(i) p − q, 9pq²
(ii) b² − 16, 5b
Ans: (i) (p − q) × 9pq²
We multiply each term of the bracket (p and −q) with 9pq²:
= p × 9pq² − q × 9pq²
= 9p²q² − 9pq³
(ii) (b² − 16) × 5b
Multiply each term in the bracket by 5b:
= b² × 5b − 16 × 5b
= 5b³ − 80b
Q3: Simplify x(2x−1)+5 and find its value at x=−3
Ans:Given: x(2x−1)+5
We need to find the value of the given expression at x=−3
We will substitute x=−3 in the given expression.
Therefore, the expression after simplifying will be
2(−3)2−(−3)+5
=2(9)+3+5
=18+8
=26
Q4: Simplify the expression and evaluate them as directed: 2x(x + 5) - 3(x - 4) + 7 for x = 2
Ans: Simplify 2x(x + 5) - 3(x - 4) + 7:
Ans: x(x − y) + y(y − z) + z(z − x)
First expand each expression:
1. x(x−y)=x2−xy
2. y(y−z)=y2−yz
3. z(z−x)=z2−zx
Add all the expressions:
x2 − xy + y2− yz +z2 − zx
Rearrange like terms:
x2+ y2+ z2- xy -yz -zx
Q6: Multiply: (m² + 3n²) × (2m − n)
Sol: (m² + 3n²) × (2m − n)
= m² × (2m − n) + 3n² × (2m − n)
= 2m³ − m²n + 6mn² − 3n³
Q7: From the sum of 3a−b+9 and −b−9, subtract 3a−b−9
Ans: Given: expressions 3a−b+9, −b−9, 3a−b−9
We need to subtract 3a−b−9
from the sum of 3a−b+9
and −b−9
The sum of the first two terms, −b−9
and 3a−b+9
will be
3a−b+9+(−b−9)=3a−b+9−b−9=3a−2b
Now subtracting 3a−b+9
from 3a−2b
, we get
3a−2b−(3a−b−9)=3a−2b−3a+b+9=−b+9
Q8 Simplify the expression and evaluate them as directed:4y(3y - 2) + 5(y + 3) - 12for y = -1
Ans: Simplify 4y(3y - 2) + 5(y + 3) - 12
Q9:Add 4x(2x + 3) and 5x2 - 7x + 10.
Q10: Simplify (x2−3x+2)(5x−2)−(3x2+4x−5)(2x−1)
Ans:Given: (x2−3x+2) (5x−2) − (3x2+4x−5) (2x−1)
We need to simplify the given expression.
First simplifying, (x2−3x+2) (5x−2),
we will get
(x2−3x+2)(5x−2)
=5x3−15x2+10x−2x2+6x−4
=5x3−17x2+16x−4 ...................(1)
Now simplifying, (3x2+4x−5)(2x−1), we will get
(3x2+4x−5)(2x−1)
=6x3+8x2−10x−3x2−4x+5
=6x3+5x2−14x+5 ..................(2)
Subtract (1)−(2) to get the result
(x2−3x+2)(5x−2)−(3x2+4x−5)(2x−1)
=5x3−17x2+16x−4−[6x3+5x2−14x+5]
=5x3−17x2+16x−4−6x3−5x2+14x−5
=−x3−22x2+30x−9
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