The product of a monomial and a binomial is a
A. monomial
B. binomial
C. trinomial
D. none of these
⇒ Monomial has single term while binomial has two terms.
⇒ Let us suppose a monomial = c and a binomial = (a – b)
⇒ product of ‘c’ and (a – b) are as follows:
⇒ c× (a – b)
= c× a – c× b
= (ca – cb) (This is a binomial)
⇒ Thus we can conclude that product of a monomial and a binomial is binomial.
Option (a) does not match to our solution.
Option (b) does match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (b) is correct answer.
In a polynomial, the exponents of the variables are always
A. integers
B. positive integers
C. non-negative integers
D. non-positive integers
⇒ according to definition Polynomial is defined as an expression consisting of variables and coefficients that involves non – negative exponents of variable.
Option (a) does not match to our solution.
Option (b) does not match to our solution.
Option (c) does match to our solution.
Option (d) does not match to our solution.
So, option (c) is correct answer.
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
⇒ (a – b)2 =(a – b)(a – b)
=(a× a – a× b – a× b + b× b)
=(a2 – 2ab + b2)
Option (a) does match to our solution.
Option (b) does not match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (a) is correct answer.
The sum of -7pq and 2pq is
A. -9pq
B. 9 pq
C. 5 pq
D. -5 pq
The addition is as follows:
⇒ -7pq + 2pq
=pq(-7 +2)
=pq(-5)
= (-5pq)
⇒ Result is as follows: (-5pq)
Option (a) does not match to our solution.
Option (b) does not match to our solution.
Option (c) does not match to our solution.
Option (d) does match to our solution.
So, option (d) is correct answer.
If we subtract -3x2y2 from x2y2, then we get
A. -4x2y2
B. -2x2y2
C. 2x2y2
D. 4x2y2
Subtraction is as follows:
⇒ x2y2 – (-3x2y2)
= x2y2 + 3x2y2
= 4x2y2
The result is as follows: 4x2y2
Option (a) does not match to our solution.
Option (b) does not match to our solution.
Option (c) does not match to our solution.
Option (d) does match to our solution.
So, option (d) is correct answer.
Like term as 4m3n2 is
A. 4m2n2
B. -6m3n2
C. 6pm3n2
D. 4m3n
⇒ Like term are the terms which have same variable and powers. The coefficients need not to be matched.
⇒ Here 4m3n2 and – 6m3n2 are like term. Because they have same variables and powers.
Option (a) does not match to our solution.
Option (b) does match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option B. is correct answer.
Which of the following is a binomial?
A. 7 × a + a
B. 6a2 + 7b + 2c
C. 4a × 3b × 2c
D. 6 (a2 + b)
⇒ A binomial is a polynomial which have two terms.
⇒ in option (a)
⇒ given polynomial = (7× a + a)
=7a +a
=8a
⇒ 8a is monomial.
⇒ in option (b)
⇒ Given polynomial = 6a2 + 7b + 2c
⇒ There are three terms in this polynomial. So, this is a trinomial.
⇒ in option (c)
⇒ Given polynomial = 4a× 3b× 2c
=24abc
⇒ There are one term in this polynomial.
∴ This polynomial is a monomial.
⇒ in option (d)
⇒ Given polynomial = 6(a2 + b)
= 6a2 +6b
⇒ there are two terms in this polynomial.
∴ this polynomial is binomial.
Option (a) does not match to our solution.
Option (b) does not match to our solution.
Option (c) does not match to our solution.
Option (d) does match to our solution.
So, option (d) is correct answer.
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
Addition is as follows:
⇒ (a – b + ab)+(b + c – bc)+(c – a – ac)
= (a – b + ab + b + c – bc + c – a – ac)
=(2c + ab – bc – ac)
=(2c + ab – ac – bc)
Option (a) does match to our solution.
Option (b) does not match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (a) is correct answer.
Product of the following monomials 4p, -7q3, -7pq is
A. 196p2q4
B. 196 pq4
C. -196 p2q4
D. 196 p2q2
The multiplication is as follows:
⇒ (4p)× (-7q3)× (-7pq)
= (-28pq3)× (-7pq)
= (196p2 q4)
Option (a) does match to our solution.
Option (b) does not match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (a) is correct answer.
Area of a rectangle with length 4ab and breadth 6b2 is
A. 24a2b2
B. 24 ab3
C. 24ab2
D. 24 ab
⇒ Area of a rectangle = length× width
⇒ Area of a rectangle = 4ab× 6b2
⇒ Area of a rectangle = 24ab3
Option (a) does not match to our solution.
Option (b) does match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (b) is correct answer.
Volume of a rectangular box (cuboid) with length = 2ab, breadth = 3 ac and height = 2ac is
A. 12a3bc2
B. 12a3bc
C. 12a2bc
D. 2ab + 3ac + 2ac
⇒ Volume of rectangular box (cuboid) = length× breadth× height
⇒ Volume of rectangular box (cuboid) = 2ab× 3ac× 2ac
⇒ Volume of rectangular box (cuboid) = 12a3bc2
Option (a) does match to our solution.
Option (b) does not match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (a) is correct answer.
Product of 6a2 – 7b + 5ab and 2ab is
A. 12a3b – 14ab2 + 10ab
B. 12a3b – 14ab2 + 10a2b2
C. 6a2 – 7b + 7 ab
D. 12a2b – 7ab2 + 10ab
The multiplication is follows:
⇒ (6a2 – 7b + 5ab) × 2ab
=(6a2× 2ab – 7b× 2ab + 5ab× 2ab)
=(12a3b – 14ab2 + 10 a2b2)
⇒ Result is as follows: (12a3b – 14ab2 + 10 a2b2)
Option (a) does not match to our solution.
Option (b) does match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (b) is correct answer.
Square of 3x – 4y is
A. 9x2 – 16y2
B. 6x2 – 8y2
C. 9x2 + 16y2 + 24xy
D. 9x2 + 16y2 - 24xy
⇒ To find square of (3x – 4y) using formula –
⇒ (a – b)2=a2 + b2 – 2ab
⇒ here, a = 3x and b= 4y
⇒ (3x – 4y)2= (3x)2 + (4y)2 – 2× (3x)×(4y)
⇒ (3x – 4y)2= 9x2 + 16y2 – 24xy
Option (a) does not match to our solution.
Option (b) does not match to our solution.
Option (c) does not match to our solution.
Option (d) does match to our solution.
So, option (d) is correct answer.
Which of the following are like terms?
A. 5xyz2, -3xy2z
B. -5xyz2, 7xyz2
C. 5xyz2, 5x2yz
D. 5xyz2, x2y2z2
⇒ Like term are the terms which have same variable and powers. The coefficients need not to be matched.
⇒ Here -5xyz2 and 7xyz2 have same variables and same powers.
∴ - 5xyz2 and 7xyz2 are like terms.
Option (a) does not match to our solution.
Option (b) does match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (b) is correct answer.
⇒ As we know that coefficients are the numbers which multiplies the variables.
⇒ In the given term , we can see that ‘y’ is a variable and is a constant which multiplies variable ‘y’.
⇒ In mathematical form –
⇒
Option (a) does not match to our solution.
Option (b) does not match to our solution.
Option (c) does match to our solution.
Option (d) does not match to our solution.
So, option (c) is correct answer.
a2 – b2 is equal to
A. (a – b)2
B. (a – b) (a – b)
C. (a + b) (a – b)
D. (a + b) (a + b)
⇒ Solving option (a)
⇒ (a – b)2 = (a – b)× (a – b)
⇒ =(a× a – a× b – b× a + b× b)
=(a2 – 2ab + b2)
⇒ Solving option (b)
⇒ (a – b)(a – b) = (a× a – a× b – b× a + b× b)
⇒ =(a2 – 2ab + b2)
⇒ Solving equation (c)
⇒ (a + b)(a – b) = (a× a – a× b + b× a – b× b)
⇒ = (a2 – b2)
⇒ Solving option (d)
⇒ (a + b)(a + b) = (a× a + a× b + b× a + b× b)
⇒ = (a2 +2ab + b2)
Option (a) does not match to our solution.
Option (b) does match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (b) is correct answer.
Common factor of 17 abc, 34ab2, 51a2b is
A. 17abc
B. 17 ab
C. 17 ac
D. 17a2b2c
⇒ Factorise the given terms
⇒ 17 abc = 17× a× b× c
⇒ 34 ab = 17× 2× a× b× b
⇒ 51ab = 17× 3× a× a× b
⇒ Common factor = 17× a× b
⇒ Common factor = 17ab
Option (a) does not match to our solution.
Option (b) does match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (b) is correct answer.
Square of 9x – 7xy is
A. 81x2 + 49x2y2
B. 81x2 - 49x2y2
C. 81x2 + 49x2y2- 126x2y
D. 81x2 + 49x2y2- 63x2y
⇒ Applying formula
⇒ (a – b)2 = a2 + b2 – 2ab
⇒ square of (9x – 7xy) means (9x – 7xy)2
⇒ Here, a = 9x and b = 7xy
⇒ (9x – 7xy)2 = (9x) + (7xy) – 2×(9x)×(7xy)
⇒ (9x – 7xy)2 = 81x2 + 49x2y2 – 126x2y
Option (a) does not match to our solution.
Option (b) does not match to our solution.
Option (c) does match to our solution.
Option (d) does not match to our solution.
So, option (c) is correct answer.
Factorised form of 23xy – 46x + 54y – 108 is
A. (23x + 54) (y – 2)
B. (23x + 54y) (y – 2)
C. (23xy + 54y) (-46x – 108)
D. (23x + 54) (y + 2)
lets write factorise form of 23xy – 46x + 54y – 108
⇒ 23xy – 46x + 54y – 108 = 23x(y – 2) + 54(y – 2)
Taking (y-2) as common from the term, we get,
= (y – 2)(23x + 54)
= (23x + 54)(y – 2)
Hence, option (a) is correct
Factorised form of r2 -10r + 21 is
A. (r-1)(r-4)
B. (r-7)(r-3)
C. (r-7)(r+3)
D. (r+7)(r+3)
⇒ Factorise form will be as follows:
⇒ (r2 -10r + 21) = r – (7+3)r + 21
⇒ (r2 -10r + 21) = r – 7r – 3r + 21
⇒ (r2 -10r + 21) = r(r – 7) – 3(r – 7)
⇒ (r2 -10r + 21) = (r – 7)(r – 3)
Option (a) does not match to our solution.
Option (b) does match to our solution.
Option (c) does not match to our solution.
Option (d) does not match to our solution.
So, option (b) is correct answer.
Factorised form of p2 – 17p – 38 is
A. (p-19) (p+2)
B. (p-19) (p-2)
C. (p+19) (p+2)
D. (p+19) (p-2)
We have,
(By splitting the middle term, so that the product of their numerical coefficients is equal constant term)
=p(p-19)+2(p-19)
=(p-19)(p+2)
(Because )
On dividing 57p2qr by 114pq, we get
A. pr
B. pr
C. pr
D. 2 pr
Required value
On dividing p (4p2 – 16) by 4p (p – 2), we get
A. 2p+4
B. 2p-4
C. p+2
D. p-2
We have,
(Because )
The common factor of 3ab and 2cd is
A. 1
B. -1
C. a
D. c
We have monomials 3ab and 2cd
Now,
We see that, there is no common factor between them except 1
(Neither numerical nor letter)
An irreducible factor of 24x2y2 is
A. x2
B. y2
C. x
D. 24x
A factor is said to be irreducible, if it cannot be factorised further
We have, 24x2y2 = 2×2×2×3×x×x×y×y
Hence, an irreducible factor of is
Number of factors of (a + b)2 is
A. 4
B. 3
C. 2
D. 1
We can write as and this cannot be
Factorised further.
Hence, number of factors of is 2
The factorised form of 3x – 24 is
A. 3x × 24
B. 3 (x – 8)
C. 24 (x – 3)
D. 3 (x – 12)
(Taking 3 as common)
The factors of x2 – 4 are
A. (x – 2), (x – 2)
B. (x + 2), (x – 2)
C. (x + 2), (x + 2)
D. (x – 4), (x – 4)
(Because )
Hence, , are factors of
The value of (-27x2y) ÷ (-9xy) is
A. 3xy
B. -3xy
C. -3x
D. 3x
We have,
=3x
The value of (3x3 + 9x2 + 27x) ÷ 3x is
A. 2a + 2b
B. 2a - 2b
C. 2a2 + 2b2
D. 2a2 - 2b2
The value of (3x3 +9x2 + 27x) ÷ 3x is
(a) x2 +9 + 27x (b) 3x3 +3x2 + 27x
(c) 3x3 +9x2 + 9 (d) x2 +3x + 9
(d)
Given: (3x3 +9x2 + 27x) ÷ 3x
= x2 + 3x + 9
The value of (a + b)2 + (a – b)2 is:
(a) 2a + 2b (b) 2a – 2b (c) 2a2 + 2b2 (d) 2a2 – 2b2
C
Given:
(a + b)2 + (a – b)2
⇒ a2 + b2 + 2ab + a2 + b2 – 2ab
⇒ 2a2 + 2b2
The value of (a + b)2 – (a – b)2 is
(a) 4ab (b) – 4ab (c) 2a2 + 2b2 (d) 2a2 – 2b2
(a + b)2 - (a – b)2
⇒ a2 + b2 + 2ab - a2 - b2 + 2ab
⇒ 4ab
Fill in the blanks to make the statements true:
The product of two terms with like signs is a………… term.
Positive
If both like terms are either positive or negative, then the resultant term will always be positive.
Fill in the blanks to make the statements true:
The product of two terms with unlike signs is a …………….. term.
Negative
As the product of a positive term and a negative term is always negative.
Fill in the blanks to make the statements true:
a (b + c) = ax ……….. × ax ……….. .
b,c
We have,
(Using left distributive law => The law relating the operations of multiplication and addition, stated symbolically, a(b+c)=ab+ac; that is, the monomial facto a is distributed, or separately applied, to each term of the binomial factor b+c, resulting in the product ab+ac. From this law it is easy to show that the result of first adding several numbers and then multiplying each separately by the number and then adding the products)
Fill in the blanks to make the statements true:
(a – b) …………….. = a2 – 2ab + b2
(a-b)
We know that,
(Because )
Fill in the blanks to make the statements true:
a2 – b2 = (a + b) …………… .
(a-b)
We have,
(Because )
Alternative method
Let
Fill in the blanks to make the statements true:
(a – b)2 + ……….. = a2 – b2
Let
(Because )
Fill in the blanks to make the statements true:
(a + b)2 - 2ab = ……… + ……… .
We have,
(Because )
Fill in the blanks to make the statements true:
(x + a) (x + b) = x2 + (a + b)x + ……….. .
ab
Fill in the blanks to make the statements true:
The product of two polynomials is a ………. .
polynomial
An expression containing one or more terms with non-zero coefficient, with variables having non-negative exponents, is called a polynomial. When we multiply two polynomials, the coefficients of the terms get multiplied and the power of the variables get added. This gives us the resulting answer as a polynomial, too.
Fill in the blanks to make the statements true:
Common factor of ax2 + bx is ……….. .
x
In both of the above terms, the common factor is .
Fill in the blanks to make the statements true:
Factorised form of 18mn + 10mnp is ……… .
2mn(9 + 5p)
The irreducible factors of 18mn and 10mnp are
18mn = 2×3×3×m×n
10mnp = 2×5×m×n×p
Here, the common factors of the two terms are 2mn. So the factorization of 18mn + 10mnp will be
18mn + 10mnp = (2×3×3×m×n) + (2×5×m×n×p)
= (2mn×9) + (2mn×5p)
= 2mn(9 + 5p)
Fill in the blanks to make the statements true:
Factorised form of 4y2 – 12y + 9 is ……… .
(2y - 3)2
Here, the polynomial has three terms where the first and the last term are perfect squares with a negative sign of the middle term. So it is of the form where a = 2y and b = 3 such that
Now, we know that
Comparing with this, we get,
Fill in the blanks to make the statements true:
38x3y2z÷ 19xy2 is equal to ………. .
2x2z
While dividing two monomials, the coefficient of the dividend is divided by the coefficient of the divisor whereas the variables of the dividend are divided by the variables of the divisor.
Here, dividend =
divisor =
So,
Hence, the answer is
Fill in the blanks to make the statements true:
Volume of a rectangular box with length 2x, breadth 3y and height 4z is ……………. .
24xyz
given:- length of box = 2x
breadth of box = 3y
height of box = 4z
Volume of rectangular box = length × breadth × height
= (2x) × (3y) × (4z)
= 24xyz
Fill in the blanks to make the statements true:
672 – 372 = (67 – 37) × ……………. = ………… .
(67 + 37), 3120
672 – 372 is of the form a2 – b2. Comparing the two we get, a = 67 and b = 37.
We have a2 - b2 = (a + b)(a-b).
So, 672 – 372 = (67 + 37)(67-37)
= (104)(30)
= 104×30 = 3120
Fill in the blanks to make the statements true:
1032 – 1022 = ………….. × (103 – 102) = ……… .
(103 + 102), 205
1032 – 1022 is of the form a2 – b2. Comparing the two we get, a = 103 and b = 102.
We have a2 - b2 = (a + b)(a-b).
So, 1032 – 1022 = (103 + 102)(103-102)
= (205)(1)
= 205×1 = 205
Fill in the blanks to make the statements true:
Area of a rectangular plot with side 4x2 and 3y2 is ……….. .
12x2y2
Let the length of rectangular plot = l = 4x2
The breadth of rectangular plot = b = 3y2
Area of rectangular plot = l×b
= 4x2 × 3y2
= 12x2y2
Fill in the blanks to make the statements true:
Volume of a rectangular box with l = b = h = 2x is …………. .
8x3
Given:- l = b = h = 2x
Volume of rectangular plot = l×b×h
= 2x × 2x × 2x
= 8x3
Fill in the blanks to make the statements true:
The coefficient in – 37abc is …………. .
-37
The numerical factor of a term is called its coefficient. Here, the numerical factor is -37.
Fill in the blanks to make the statements true:
Number of terms in the expression a2 + bc ×d is ……………. .
two
The expression can be written as
A2 + bc×d = a2 + bcd
The multiplication sign does not mean that there are two separate terms. Two terms can be separated only by the operations of addition or subtraction.
Fill in the blanks to make the statements true:
The sum of areas of two squares with sides 4a and 4b is ………………. .
16(a2 + b2)
Given:- side of first square = 4a
⇒ Area of first square = (side)2
= (4a)2
= 16a2
side of second square = 4b
⇒ Area of second square = (side)2
= (4b)2
= 16b2
Sum of areas of the two squares = 16a2 + 16b2
= 16(a2 + b2)
Fill in the blanks to make the statements true:
The common factor method of factorisation for a polynomial is based on ………….. property.
distributive
Fill in the blanks to make the statements true:
The side of the square of area 9y2 is ………. .
3y
Area of a square = (side)2
⇒ 9y2 = (side)2
Taking square root on both sides, we get,
side =
To find the square root of a monomial, we find the square roots of the coefficient and variables separately.
Now, 9y2 = (3y)2
Thus, side of square = 3y
Fill in the blanks to make the statements true:
On simplification = …….. .
We have,
Fill in the blanks to make the statements true:
The factorisation of 2x + 4y is ………. .
2(x + 2y)
2x + 4y = (2×x) + (2×2×y)
= (2×x) + (2×2y)
= 2(x + 2y)
State whether the statements are true (T) or false (F).
(a + b)2 = a2 + b2
Lets recall the formula of (a + b)2.
(a + b)2 = a2 + 2×a×b + b2
⇒ (a + b)2 = a2 + 2ab + b2
But right-hand side of the equation does not match left-hand side of the equation.
That is,
a2 + 2ab + b2 ≠ a2 + b2
⇒ (a + b)2 ≠ a2 + b2
Hence, this statement is false.
State whether the statements are true (T) or false (F).
(a - b)2 = a2 - b2
Expansion of (a – b)2 is,
(a – b)2 = a2 – 2×a×b + b2
⇒ (a – b)2 = a2 – 2ab + b2
And a2 – 2ab + b2 ≠ a2 – b2
⇒ (a – b)2 ≠ a2 – b2
Hence, this statement is false.
State whether the statements are true (T) or false (F).
(a + b) (a – b) = a2 – b2
Let us expand (a + b)(a – b).
Multiply each term with the other term which is in multiplication with it.
(a + b)(a – b) = a2 – ab + ab – b2
⇒ (a + b)(a – b) = a2 +(ab – ab) – b2
⇒ (a + b)(a – b) = a2 + 0 – b2
⇒ (a + b)(a – b) = a2 – b2
Hence, this statement is true.
State whether the statements are true (T) or false (F).
The product of two negative terms is a negative term.
Let us understand this by an example.
Let two negative terms be -a and -b.
Multiplying these two negative terms.
-a × -b = (-1) × (-1) × ab
⇒ -a × -b = +1 × ab [∵, -1 × -1 = 1]
⇒ -a × -b = +ab
The product of -a and -b came out to be ab (positive term).
Whenever two negative terms are multiplied, the product is always positive as the two negatives forms a positive.
Hence, this statement is false.
State whether the statements are true (T) or false (F).
The product of one negative and one positive term is a negative term.
Let us understand this by an example.
Let a negative and a positive term be -a and b respectively.
Then, multiply these two terms.
-a × b = (-1) × (+1) × ab
⇒ -a × b = -1 × ab [∵, -1 × 1 = -1]
⇒ -a × b = -ab
Similarly, take any pair of such numbers, we will always get a negative term as a negative number multiplied by a positive number always gives negative number.
Hence, this statement is true.
State whether the statements are true (T) or false (F).
The coefficient of the term -6x2y2 is -6.
Coefficient is a numerical or constant quantity placed before and multiplying the variable in an algebraic expression.
The algebraic expression in -6x2y2 is x2y2.
And the constant term before x2y2 is -6.
Thus, coefficient of the term -6x2y2 is -6.
Hence, the statement is true.
State whether the statements are true (T) or false (F).
p2q + r2r + r2q is a binomial.
An algebraic expression which consists of two non-zero terms is called a binomial.
Observe in the given question, there are three non-zero terms, namely
p2q, r2r and r2q.
Thus, this is a polynomial having three terms, not a binomial.
Hence, the statement is false.
State whether the statements are true (T) or false (F).
The factors of a2 – 2ab + b2 are (a+b) and (a+b).
Lets factorize a2 – 2ab + b2 by splitting middle term.
a2 – 2ab + b2 = a2 – ab – ab + b2 [∵, -ab – ab = - 2ab & (-ab)(-ab) = a2b2]
⇒ a2 – 2ab + b2 = a(a – b) – b(a – b)
⇒ a2 – 2ab + b2 = (a – b)(a – b) = (a – b)2
So, the factors of a2 – 2ab + b2 are (a – b) and (a – b).
Hence, the statement is false.
State whether the statements are true (T) or false (F).
h is a factor of 2π (h+r).
Factorizing 2π (h + r), we get
2π (h + r) = 2π × (h + r)
⇒ There are three factors of 2π (h + r) namely, 2, π and (h + r).
But h is not any factor of 2π (h + r).
Hence, the statement is false.
State whether the statements are true (T) or false (F).
Some of the factors of are , , n and (n+1).
Factorizing n2/2 + n/2, by simply taking out common variables and constants.
n2/2 + n/2 = n2×1/2 + n×1/2
⇒ n2/2 + n/2 = 1/2 (n×n + n) = 1/2 n (n + 1)
⇒ 1/2, n and (n + 1) are factors of n2/2 + n/2.
Hence, the statement is true.
State whether the statements are true (T) or false (F).
An equation is true for all values of its variables.
An equation is a statement of equality between two quantities or algebraic expressions. Most algebraic equations are TRUE when certain values are substituted for the variable (such as x) and are FALSE for all other values. The values that make equations TRUE are called "solutions". So, an equation is not necessarily true for all values of its variables.
Hence, the statement is false.
State whether the statements are true (T) or false (F).
x2 + (a+b)x + ab = (a+b) (x+ab)
We need to factorize x2 + (a + b)x + ab, by splitting the middle term:
x2 + (a + b)x + ab = x2 + ax + bx + ab [∵, ax + bx = (a + b)x & ax × bx = abx2)
⇒ x2 + (a + b)x + ab = x(x + a) + b(x + a)
⇒ x2 + (a + b)x + ab = (x + a)(x + b)
And (x + a)(x + b) ≠ (a + b)(x + ab)
⇒ x2 + (a + b)x + ab ≠ (a + b)(x + ab)
Hence, the statement is false.
State whether the statements are true (T) or false (F).
Common factor of 11pq2, 121p2q3, 1331p2q is 11p2q2.
Common factors of 11pq2, 121p2q3 and 1331p2q can be found out as:
Common factors (11pq2, 121p2q3, 1331p2q) = Common factors (11×pq, 11×11p2q3, 11×121p2q)
⇒ Common factors (11pq2, 121p2q3, 1331p2q) = 11
[∵, 11 divides all the three term 11pq2, 121p2q3 and 1331p2q; p divides all the three terms 11pq2, 121p2q3 and 1331p2q; and q divides all the three terms 11pq2, 121p2q3 and 1331p2q]
Hence, the statement is false.
State whether the statements are true (T) or false (F).
Common factor of 12a2b2+4ab2-32 is 4.
Let us consider that common factor of 12a2b2 + 4ab2 – 32 = 4
Check: 12a2b2 + 4ab2 – 32 = 4 (3a2b2 + ab2 – 8)
Notice, there are no further common factor in the equation.
Thus, common factor of 12a2b2 + 4ab2 – 32 is 4.
Hence, the statement is true.
State whether the statements are true (T) or false (F).
Factorisation of -3a2+3ab+3ac is 3a(-a–b –c).
Let us factorize -3a2 + 3ab + 3ac.
-3a2 + 3ab + 3ac = 3(-a×a + a×b + a×c)
⇒ -3a2 + 3ab + 3ac = 3a (-a + b + c)
But, 3a (-a + b + c) ≠ 3a (-a – b – c)
Hence, the statement is false.
State whether the statements are true (T) or false (F).
Factorised form of p2+30p+216 is (p+18) (p-12).
Factorizing the equation p2 + 30p + 216, by splitting the middle term of the equation:
p2 + 30p + 216 = p2 + 18p + 12p + 216 [∵, (18p + 12p) = 30p & (18p × 12p) = 216p2)
⇒ p2 + 30p + 216 = p (p + 18) + 12 (p + 18)
⇒ p2 + 30p + 216 = (p + 12) (p + 18)
But, (p + 12)(p + 18) ≠ (p + 18)(p – 12)
Hence, the statement is false.
State whether the statements are true (T) or false (F).
The difference of the squares of two consecutive numbers is their sum.
Let two consecutive numbers be x and (x + 1).
Then square of these numbers are x2 and (x + 1)2.
Difference of squares of these consecutive numbers = (x + 1)2 – x2
= x2 + 1 + 2x – x2 [∵, (a + b)2 = a2 + b2 + 2ab)
= 2x + 1
= x + x + 1
= (x) + (x + 1)
= sum of the two consecutive numbers x and x+1
Thus, difference of squares of two consecutive numbers = sum of the same consecutive numbers.
Hence, the statement is true.
State whether the statements are true (T) or false (F).
abc + bca + cab is a monomial.
An algebraic expression which consists of one non-zero terms is called a monomial.
Observe in the given question, there are three non-zero terms, namely
abc, bca and cab.
Thus, this is a polynomial having three terms, not a monomial.
Hence, the statement is false.
State whether the statements are true (T) or false (F).
On dividing by, the quotient is 9.
Divide p/3 by 3/p, we get
p/3 ÷ 3/p = p/3 × p/3 = p2/9
And p2/9 ≠ 9.
⇒ the quotient is p2/9, not 9.
Hence, the statement is false.
State whether the statements are true (T) or false (F).
The value of p for 512 – 492 = 100p is 2.
Solving for p,
(51)2 – (49)2 = 100p
⇒ (50 + 1)2 – (50 – 1)2 = 100p [∵, 51 = 50 + 1 & 49 = 50 – 1]
⇒ (2500 + 1 + 100) – (2500 + 1 – 100) = 100p [∵, (a + b)2 = a2 + b2 + 2ab and (a – b)2 = a2 + b2 – 2ab)
⇒ 2500 + 1 + 100 – 2500 – 1 + 100 = 100p
⇒ 200 = 100p
⇒ 100p = 200
⇒ p = 200/100 = 2
The value of p is 2.
Hence, the statement is true.
State whether the statements are true (T) or false (F).
(9x – 51) ÷ 9 is x – 51.
Solving (9x – 51) ÷ 9,
(9x – 51) ÷ 9 = (9x – 51)/9
9 divides the term 9x but does not divide 51.
So, the answer comes out to be (9x – 51)/9.
(9x – 51)/9 ≠ x – 51
Hence, the statement is false.
State whether the statements are true (T) or false (F).
The value of (a+1) (a-1) (a2+1) is a4-1
Solving it, we get
(a + 1)(a – 1)(a2 + 1) = {(a + 1)(a – 1)} × (a2 + 1)
⇒ (a + 1)(a – 1)(a2 + 1) = (a2 – 1)(a2 + 1)
[∵, (x + y)(x – y) = x2 – y2; Put x = a and y = 1 ⇒ (a + 1)(a – 1) = a2 – 1]
⇒ (a + 1)(a – 1)(a2 + 1) = a4 – 1
[∵, (x2 + y2)(x2 – y2) = x4 – y4; Put x = a and y = 1 ⇒ (a2 + 1)(a2 – 1) = a4 – 1]
Thus, the value of (a + 1)(a – 1)(a2 + 1) is a4 – 1.
Hence, the statement is true.
Add:
7a2bc, -3abc2, 3a2bc, 2abc2
The addition is as follows:
The result is as follows 10a2bc – abc2
Add:
9ax, + 3by – cz, -5by + ax + 3cz
The addition is as follows:
The result is as follows 10ax-2by + 2cz
Add:
xy2z2 + 3x2y2z–4x2yz2,-9x2y2z + 3xy2z2 + x2yz2
The addition is as follows:
The result is as follows 4xy2z2-6x2y2z-3x2yz2
Add:
5x2 – 3xy + 4y2 – 9, 7y2 + 5xy – 2x2 + 13
: The addition is as follows:
The result is as follows 7x2 + 2xy + 11y2 + 4
Add:
2p4 – 3p3 + p2 – 5p + 7, -3p4-7p3-3p2-p-12
The addition is as follows:
The result is as follows -p4-4p3-2p2-6p-5
Add:
3a(a-b-c), 2b (a-b + c)
To find: 3a(a - b - c) + 2b(a - b + c)
The addition is as follows:
Here,
3a(a - b - c) = 3a2 - 3ab - 3ac
2b(a - b + c) = 2ab - 2b2 + 2bc
Hence the two polynomials to be added are 3a2 - 3ab - 3ac and 2ab - 2b2 + 2bc
3a(a - b - c) + 2b(a - b + c) = 3a2 - 3ab - 3ac + 2ab - 2b2 + 2bc = 3a2 - ab - 3ac + 2bc - 2b2
The result is as follows: 3a2 - ab - 3ac + 2bc - 2b2
Add:
3a (2b + 5c), 3c (2a + 2b)
Here, 3a(2b + 5c) = 6ab + 15ac
3c(2a + 2b) = 6ac + 6bc
Hence the two polynomials to be added are 6ab + 15ac and
6ac + 6bc
The result is as follows 6ab + 21ac + 6bc
Subtract:
5a2b2c2 from -7a2b2c2
The subtraction is as follows:
The result is as follows:- 12a2b2c2
Subtract:
6x2 – 4xy + 5y2 from 8y2 + 6xy – 3x2
The subtraction is as follows:
We can also write 8y2 + 6xy – 3x2 as -3x2 + 6xy + 8y2
Therefore,
The result is as follows: -9x2 + 10xy + 3y2
Subtract:
2a2b2c2 + 4a2b2c - 5a2bc2 from -10a2b2c + 4ab2c2 + 2a2bc2
The subtraction is as follows:
The result is as follows: -14a2b2c + 4ab2c2 + 7a2bc2-2a2b2c2
Subtract:
3t4–4t3 + 2t2–6t + 6 from -4t4 + 8t3-4t2-2t + 11
The subtraction is as follows:
The result is as follows: -7t4 + 12t3-6t2 + 4t + 5
Subtract:
2ab + 5bc–7ac from 5ab–2bc–2ac + 10abc
The subtraction is as follows:
The result is as follows: 3ab -7bc + 5ac + 10abc
Subtract:
7p (3q + 7p) from 8p (2p -7q)
The subtraction is as follows:
Here 8p(2p-7q) = 16p2-56pq and
7p(3q + 7p) = 21pq + 49p2
Therefore,
The result is as follows: -33p2-77pq
Subtract:
-3p2 + 3pq + 3px from 3p (-p-a-r)
The subtraction is as follows:
Here,3p(-p-a-r) = -3p2-3pa-3pr
Therefore,
The result is as follows: 0p2-3pa-3pr-3pq-3px or -3pa-3pr-3pq-3px
Or -3p(a + r + q + x).
Multiply the following:
-7pq2r3, -13p3q2r
The multiplication is as follows:
(-7pq2r3) ×(-13p3q2r)
= (-7×-13×3)×(pq2r3.pq2r)(Here dot implies multiplication)
= (273)×(p2q4r4)
= 273p4q4r4
The product is = 273p4q4r4
Multiply the following:
3x2y2z2, 17xyz
The multiplication is as follows:
(3x2y2z2)×(17xyz)
= (3×17)×(x2 y2z2.xyz)
= 51x3y3z3
The product is = 51x3y3z3
Multiply the following:
15xy2, 17yz2
The multiplication is as follows:
(15xy2)×(17yz2)
= (15×17)×(xy2.yz2)
= 255xy3z2
The product is = 255xy3z2
Multiply the following:
-5a2bc, 11ab, 12abc2
The multiplication is as follows:
(-5a2bc)×(11ab)×(12abc2)
= (-5×11×12)(a2bc.ab.abc2)
= -660a4b3c3
The product is = 255xy3z2
Multiply the following:
-3x2y, (5y - xy)
The multiplication is as follows:
(-3x2y)×(5y-xy)
= (-3x2y.5y) + (3x2y.xy)
= -15x2y2 + 3x3y2
The product is = -15x2y2 + 3x3y2
Multiply the following:
abc, (bc + ca)
The multiplication is as follows:
(vii) 7prq, (p - q + r)
The multiplication is as follows:
(7prq)×(p-q + r)
= (7prq.p-7prq.q + 7prq.r)
= 7p2rq-7prq2 + 7pr2q
The product is = 7p2rq-7prq2 + 7pr2q
Multiply the following:
x2y2z2, (xy – yz + zx)
The multiplication is as follows:
(x2y2z2)(xy-yz + zx)
= (x2y2z2.xy-x2y2z2.yz + x2y2z2.zx)
= x3y3z2-x2y3z3 + x3y2z3
The product is = x3y3z2-x2y3z3 + x3y2z3
Multiply the following:
(p + 6), (q – 7)
The multiplication is as follows:
(p + 6)×(q-7)
= (p.q-7p + 6q-42)
= pq-7p + 6q-42
The product is = pq-7p + 6q-42
Multiply the following:
6mn, 0mn
The multiplication is as follows:
6mn×0mn
= (6×0)mn = 0
The product is = 0
Multiply the following:
a, a5, a6
The multiplication is as follows:
a×a5×a6
= a12
The product is = a12
Multiply the following:
b3, 3b2, 7ab5
The multiplication is as follows:
b3×3b2×7ab5
= 21ab10
The product is = 21ab10
Multiply the following:
-7st, -1, -12st2
The multiplication is as follows:
-7st×-1×-12st2
= -84s2t3
The product is = -84s2t3
Multiply the following:
rs; r3s2
The multiplication is as follows:
= ()
=
The product is =
Multiply the following:
(a2 - b2), (a2 + b2)
The multiplication is as follows:
(a2-b2)×(a2 + b2)
= (a2.a2 + a2.b2-b2.a2-b2.b2)
= a4 + a2b2-a2b2-b4
= a4 + 0-b4
= a4-b4
The product is = a4-b4
Multiply the following:
(ab + c), (ab + c)
The multiplication is as follows:
(ab + c)×(ab + c)
= (ab.ab + ab.c + ab.c + c.c)
= a2b2 + 2abc + c2
The product is = a2b2 + 2abc + c2
Multiply the following:
(pq- 2r), (pq- 2r)
The multiplication is as follows:
(pq-2r)×(pq-2r)
= (pq.pq-2pq.r-2pq.r + 2r.2r)
= p2q2-4pqr + 2r2
The product is = p2q2-4pqr + 2r2
Multiply the following:
The multiplication is as follows:
=
=
= (Here we have calculated the LCM )
=
The product is =
Multiply the following:
, (2p2 – 3q2)
The multiplication is as follows:
=
=
=
=
The product is =
Multiply the following:
(x2 – 5x + 6), (2x + 7)
The multiplication is as follows:
(x2-5x + 6)×(2x + 7)
= (x2.2x + 7x2-5x.2x-5x.7 + 6.2x + 6.7)
= 2x3 + 7x2-10x2-35x + 12x + 42
= 2x3-3x2-23x + 42
The product is = 2x3-3x2-23x + 42
Multiply the following:
(3x2 + 5x - 8), (2x2 – 4x + 3)
The multiplication is as follows:
(3x2 + 5x-8)×(2x2-4x + 3)
= 3x2.2x2-4x.3x2 + 3x2.3 + 5x.2x2-4x.5x + 5x.3-8.2x2 + 8.4x-8.3
= 6x4-12x3 + 9x3 + 10x3-20x2 + 15x-16x2 + 32x-24
= 6x4 + 7x3-36x2 + 47x-24
The product is = 6x4 + 7x3-36x2 + 47x-24
Multiply the following:
(2x – 2y - 3), (x + y + 5)
The multiplication is as follows:
(2x-2y-3)×(x + y + 5)
= 2x.x + 2x.y + 2x.5-2y.x-2y.y-2y.5-3.x-3.y-3.5
= 2x2 + 2xy + 10x-2xy-2y2-10y-3x-3y-15
= 2x2-2y2 + 7x-7y-15
The product is = 2x2-2y2 + 7x-7y-15
Simplify
(3x + 2y)2 + (3x – 2y)2
The given expression is (3x + 2y)2 + (3x – 2y)2
We have,
(3x + 2y)2 = (3x)2 + 2.3x.2y + (2y)2(Here we apply standard identities)
= 9x2 + 12xy + 4y2
and
(3x-2y)2 = (3x)2-2.3x.2y + (2y)2
= 9x2-12xy + 4y2
Therefore,
(3x + 2y)2 + (3x – 2y)2 = 9x2 + 12xy + 4y2 + 9x2-12xy + 4y2
= 18x2 + 8y2
= 2(9x2 + 4y2)
Simplify
(3x + 2y)2 - (3x – 2y)2
The given expression is (3x + 2y)2 - (3x – 2y)2
We have,
(3x + 2y)2 = (3x)2 + 2.3x.2y + (2y)2
= 9x2 + 12xy + 4y2
and
(3x-2y)2 = (3x)2-2.3x.2y + (2y)2
= 9x2-12xy + 4y2
Therefore,
(3x + 2y)2 - (3x – 2y)2 = 9x2 + 12xy + 4y2-9x2 + 12xy-4y2
= 24xy
Simplify
- ab
The given expression is -ab
= -ab
=
Therefore, - ab =
Simplify
+ 2xy
The given expression is
+ 2xy
= + 2xy
= + 2xy
=
Therefore, + 2xy =
Simplify
(1.5p + 1.2q)2 – (1.5p - 1.2q)2
The given expression is
(1.5p + 1.2q)2 – (1.5p - 1.2q)2
We have
(1.5p + 1.2q)2 = (1.5p)2 + 2(1.5p)(1.2q) + (1.2q)2
and
(1.5p - 1.2q)2 = (1.5p)2-2(1.5p)(1.2q) + (1.2q)2
Therefore,
(1.5p + 1.2q)2 – (1.5p - 1.2q)2
= (1.5p)2 + 2(1.5p)(1.2q) + (1.2q)2-(1.5p)2 + 2(1.5p)(1.2q)-(1.2q)2
= 2(1.5p)(1.2q)
Simplify
(2.5m + 1.5q)2 + (2.5m - 1.5q)2
The given expression is
(2.5m + 1.5q)2 + (2.5m - 1.5q)2
We have
(2.5m + 1.5q)2 = (2.5m)2 + 2(2.5m)(1.5q) + (1.5q)2
(2.5m - 1.5q)2 = (2.5m)2-2(2.5m)(1.5q) + (1.5q)2
Therefore,
(2.5m + 1.5q)2 + (2.5m - 1.5q)2
= (2.5m)2 + 2(2.5m)(1.5q) + (1.5q)2 + (2.5m)2-2(2.5m)(1.5q) + (1.5q)2
= 12.50m2 + 4.50q2
Simplify
(x2 - 4) + (x2 + 4) + 16
The given expression is
(x2 - 4) + (x2 + 4) + 16
= 2x2 + 16
Therefore, (x2 - 4) + (x2 + 4) + 16 = 2x2 + 16
Simplify
(ab - c)2 + 2abc
The given expression is
(ab - c)2 + 2abc
= (ab)2-2(ab)(c) + (c)2 + 2abc
= a2b2-2abc + c2 + 2abc
= a2b2 + c2
Therefore, (ab - c)2 + 2abc = a2b2 + c2
Simplify
(a - b) (a2 + b2 + ab) -(a + b) (a2 + b2 - ab)
The given expression is
(a - b) (a2 + b2 + ab) -(a + b) (a2 + b2 - ab)
= (a.a2 + a.b2 + a.ab-b.a2-b.b2-b.ab)-(a.a2 + a.b2-a.ab + b.a2 + b.b2-b.ab)
= a3 + ab2 + a2b-a2b-b3-ab2-a3-ab2 + a2b-a2b-b3 + ab2
= -2b3
Therefore, (a - b) (a2 + b2 + ab) - (a + b) (a2 + b2 - ab) = -2b3
Method 2:
We know that,
(a - b) (a2 + b2 + ab) = a3 - b3 and (a + b) (a2 + b2 - ab) = a3 + b3
Therefore,
(a - b) (a2 + b2 + ab) - (a + b) (a2 + b2 - ab) = a3 - b3 - (a3 + b3 ) = -2b3
Simplify
(b2-49)(b + 7) + 343
The given expression is
(b2-49)(b + 7) + 343
= b2.b + 7b2-49b-49.7 + 343
= b3 + 7b2-49b-343 + 343
= b3 + 7b2-49b
Therefore, (b2-49)(b + 7) + 343 = b3 + 7b2-49b
Simplify
(4.5a + 1.5b)2 + (4.5b + 1.5a)2
The given expression is
(4.5a + 1.5b)2 + (4.5b + 1.5a)2
We have,
(4.5a + 1.5b)2 = (4.5a)2 + 2(4.5a)(1.5b) + (1.5b)2
and
(4.5b + 1.5a)2 = (4.5b)2 + 2(4.5b)(1.5a) + (1.5a)2
Therefore,
(4.5a + 1.5b)2 + (4.5b + 1.5a)2
= (4.5a)2 + 2(4.5a)(1.5b) + (1.5b)2 + (4.5b)2 + 2(4.5b)(1.5a) + (1.5a)2
= (4.5)2(a2 + b2) + 2.2(4.5)(1.5)ab + (1.5)2(a2 + b2)
= (a2 + b2)( (4.5)2 + (1.5)2) + 27ab
= 22.5(a2 + b2) + 27ab
Simplify
(pq-qr)2 + 4pq2r
The given expression is
(pq-qr)2 + 4pq2r
= (pq)2-2(pq)(qr) + (qr)2 + 4pq2r
= p2q2-pq2r + q2r2 + 4pq2r
= p2q2 + 3pq2r + q2r2
Therefore , (pq-qr)2 + 4pq2r = p2q2 + 3pq2r + q2r2
Simplify
(s2t + tq2)2 – (2stq)2
The given expression is
(s2t + tq2)2 – (2stq)2
= (s2t)2 + 2(s2t)(tq2) + 2(s2t)(tq2)-4s2t2q2
= s4t2 + 2s2t2q2 + 2s2t2q2-4s2t2q2
= s4t2
Therefore ,(s2t + tq2)2 – (2stq)2 = s4t2