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Practice each question, then open the answer or worked solution.
The product of a monomial and a binomial is a
(a) monomial (b) binomial
(c) trinomial (d) none of these
Solution:-
(b) binomial
Let monomial = 2x, binomial = x + y
Then, product of a monomial and a binomial = (2x) × (x + y)
= 2x2 + 2xy
In a polynomial, the exponents of the variables are always
(a) integers (b) positive integers
(c) non-negative integers (d) non-positive integers
Solution:-
(b) positive integers
Which of the following is correct?
(a) (a – b)2 = a2 + 2ab – b2 (b) (a – b)2 = a2 – 2ab + b2
(c) (a – b)2 = a2 – b2 (d) (a + b)2 = a2 + 2ab – b2
Solution:-
(b) (a – b)2 = a2 – 2ab + b2
We have, = (a – b) × (a – b)
= a × (a – b) – b × (a – b)
= a2 – ab – ba + b2
= a2 – 2ab + b2
The sum of –7pq and 2pq is
(a) –9pq (b) 9pq (c) 5pq (d) – 5pq
Solution:-
(d) – 5pq
The given two monomials are like terms.
Then sum of -7pq and 2pg = – 7pq + 2pq
= (-7 + 2) pq
= -5pq
If we subtract –3x2y2 from x2y2, then we get
(a) – 4x2y2 (b) – 2x2y2 (c) 2x2y2 (d) 4x2y2
Solution:-
(d) 4x2y2
We have,
The given two monomials are like terms.
Subtract –3x2y2 from x2y2 = x2y2 – (- 3x2y2)
= x2y2 + 3x2y2
= x2y2 (1 + 3)
= 4x2y2
Like term as 4m3n2 is
(a) 4m2n2 (b) – 6m3n2 (c) 6pm3n2 (d) 4m3n
Solution:-
(b) – 6m3n2
Like terms are formed from the same variables, and the powers of these variables are also the same. But coefficients of like terms need not be the same.
Which of the following is a binomial?
(a) 7 × a + a (b) 6a2 + 7b + 2c
(c) 4a × 3b × 2c (d) 6 (a2 + b)
Solution:-
(d) 6 (a2 + b)
Expressions that contain exactly two terms are called binomials.
= 6 (a2 + b)
= 6a2 + b
Sum of a – b + ab, b + c – bc and c – a – ac is
(a) 2c + ab – ac – bc (b) 2c – ab – ac – bc
(c) 2c + ab + ac + bc (d) 2c – ab + ac + bc
Solution:-
(a) 2c + ab – ac – bc
We have,
= (a – b + ab) + (b + c – bc) + (c – a – ac)
= a – b + ab + b + c – bc + c – a – ac
Now, grouping like terms
= (a – a) + (-b + b) + (c + c) + ab – bc – ac
= 2c + ab – bc – ac
Product of the following monomials 4p, – 7q3, –7pq is
(a) 196 p2q4 (b) 196 pq4 (c) – 196 p2q4 (d) 196 p2q3
Solution:-
(a) 196 p2q4
= 4p × (– 7q3) × (–7pq)
= (4 × (-7) × (-7)) × p × q3 × pq
= 196p2q4
Area of a rectangle with length 4ab and breadth 6b2 is
(a) 24a2b2 (b) 24ab3 (c) 24ab2 (d) 24ab
Solution:-
(b) 24ab3
We know that the area of a rectangle = length × breadth
Given, length = 4ab, breadth = 6b2
= 4ab × 6b2
= 24ab3