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Monomial: 

i) 5x²y, 

ii) 32 xyz

Binomial: 

i) ax + by, 

ii) 2z – 5

Trinomial : 

i) ax + by + cz, 

ii) p² + q² + r²

Polynomial: 

i) 5x⁴ – 2x² + x – 1, 

ii) 6 + 5x – 4x² + 3y³ – 2z⁴

C) a + 3a + 3

Correct option is (C) \(a^3+3a^2+3a+1\)

\((a+1)^3\) \(=(a+1)^2(a+1)\)

\(=(a^2+2a+1)(a+1)\)

\(=a^3+2a^2+a+a^2+2a+1\)

\(=a^3+3a^2+3a+1\)

Numerical coefficient of,

x³y²z = 1

xy²z³ = 1

–3xy²z³ = -3

5x³y²z = 5

–7x²y²z² = -7

Correct option is  A) 93 × 004

Correct option is (A) 93 x 104

9672 = 93 \(\times\) 104

Correct option is  B) a² – b² 

Correct option is (B) a² - b²

a (a – b) + b (a – b) \(=a^2-ab+ab-b^2\)

= \(a^2-b^2\)

Correct option is  D) 4a²

Correct option is (D) 4a²

Side of given square = 2a.

\(\therefore\) Area of given square \(=(2a)^2=4a^2.\)

Yes, the product of two monomials is a monomial. 

∵ 2x × y = 2xy

Correct option is  D) 6

Correct option is  A) 8

Correct option is A) 7 lm² n²

2.07 × 1.93 

= (2 + 0.07) (2 – 0.07) 

= 2² – (0.07)² 

= 3.9951

Given expression = 4x + x – 2x² + x – 1

= 4x + x + x – 2x² – 1

= – 2x² + 6x – 1

If x = – 1 then, 

the value of – 2x² + 6x – 1 

= – 2 (-1)2 + 6( – 1) – 1

= – 2(1) – 6 – 1

= – 2 – 6 – 1

= – 9

i) If x = 1 ⇒ – x = – (1)= – 1

ii) If x = 1 ⇒ 4x = 4 x 1 = 4

iii) If x= 1 ⇒ – 2x² = – 2(1)² = – 2 × 1= – 2

497 × 505 

= (500 – 3) (500 + 5) 

= 500² + (–3 + 5) × 500 + (–3) (5) [using (x + a) (x + b) = x² + (a + b) x + ab] 

= 250000 + 1000 – 15 

= 250985

a² – b² = (a + b) (a – b) … [from the standard identities]

181² – 19² 

= (181 – 19) (181 + 19) [using a² – b² = (a – b) (a + b)] 

= 162 × 200

= 32400

The speed of car is 55 km/hrs. The distance covered in y hours is 55y.

Because, distance = speed × time

False

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

The expression 13 + 90 is a constant.

13 + 90 = 103

False

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

Like terms in the expression n(n + 1) + 6 (n – 1) are n and 6n.

Consider the given expression, n(n + 1) + 6 (n – 1)

= n² + n + 6n – 6

= n² + 7n – 6

Therefore, like terms are n and 6n

(i), (iii), (iv), (vi) are in standard form.

i) ax² + bx + c

ii) ax + b

iii) 4x³ + 5x² – 6x + 2

iv) 5x⁴ – 3x³ – 2x – 2

v) px³ + qx² + r

True

When we multiply a negative term by a positive term, the resultant will be a negative term, i-e. (-) x (+) = (-).

False

Since, the product of two negative terms is always a positive term, i.e. (-) x (-) = (+).

True

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

(a – b)² + (2ab – 2b²) = a² – b²

= (a – b)² + (2ab – 2b²)

= a² + b² – 2ab + 2ab – 2b²

= a² – b²

Given the term is 4xy 

Numerical coefficient: 

4 Algebraic coefficient: xy

(3/4)x², 5 x², -3x², -(1/4) x²

In the above questions terms having the same literal factors are like terms.

Now add the like terms,

= (3/4)x² + 5x² – 3x² – (1/4) x²

= (3/4)x²– (1/4) x² + 5x² – 3x²

= ((3-1)/4)x² + 2x²

= (2/4)x² + 2x²

= (1/2)x² + 2x²

= ((1 + 4)/2)x²

= (5/2)x²

5a²b, -8a²b, 7a²b

In the above questions terms having the same literal factors are like terms.

Now add the like terms,

= 5a²b – 8a²b + 7a²b

= 12 a²b – 8a²b

= 4a²b

Correct option is B) ax² + bx + c

Sum of x² – y² -1, y² – x² – 1 and 1 – x² – y² = x² – y² – 1 + y² – x² – 1 + 1 – x² – y² 

On combining the like terms,

= x² – x² – x² – y² + y² – y² – 1 – 1 + 1 

= -x² – y² – 1

Now, subtract -(1 + y²) from -x² – y² -1

= -x² – y² -1 – [-(1 + y²)]

= – x² – y² – 1 + 1 + y² 

= -x² – y² + y² – 1 + 1 

= -x²

Given, P = -(x-2),Q = -2(y+ 1)and R = -x + 2y 

Also given, P+Q + R=ax

On putting the values of P,Q and R on LHS, we get

-(x-2)+[-2(y+1)]+(-x+2y) = ax 

=> -x+2 + (-2y-2)-x + 2y = ax

=> -x + 2 – 2y – 2 – x+ 2y = ax

On combining the like terms,

-x – x – 2y + 2y + 2 – 2=ax 

=> -2x = ax

By comparing LHS and RHS, we get a = -2

Correct option is  A) x⁴ – 1

Given, A = 3x² – 4x +1, B = 5x² + 3x – 8 and C = 4x² – 7x + 3

1. (A + B)-C 

= (3x² -4x +1 + 5x² + 3x-8) -(4x² -1x + 3)

On combining the like terms,

= (3x² + 5x² – 4x + 3x + 1 – 8) – (4x² – 1x + 3) 

= (8x²-x-7)-(4x²-7x + 3)

= 8x²-x-7-4x² + 7x-3 

= 8x²-4x² -x + 7x-7-3 

= 4x² + 6x-10

2. B+C-A

= 5x² + 3x-8+4x²-7x + 3-(3x²-4x + 1))

On combining the like terms,

= (5x² + 4x² + 3x-7x-8+ 3)-(3x² – 4x + 1)

= (9x² -4x-5)- (3x² – 4x +1)

= 9x² -4x-5-3x² + 4x-1 

= 9x² – 3x² – 4x + 4x – 5 -1 

= 6x²-6

3. A+B+C

= 3x²-4x+ 1 +5x² + 3x-8+4x²-7x+3 

On combining the like terms,

= 3x² + 5x² + 4x² -4x + 3x-7x +1 – 8 + 3 

= 12x²-8x-4

Given, m = 1, n = -1 and p = 2

So, putting m = 1, n = -1 and p = 2 in the given expressions, we get

(a) m+n+p

= 1-1+2

= 2
(b) m²+n²+P² 

=(1)² + (-1)² + (2)² 

=1+1+ 4

= 6
(c) m³+ n³+p³ 

=(1)³ + (-1)³ + (2)³ 

= 1 -1 + 8 

= 8
(d) mn+np+ pm

= (1)(-1)+(-1)(2)+(2)(1)

= -1-2 + 2 

= -1
(e) m³ +n³ + p³ -3mnp 

= (1)³ + (-1)³ + (2)³ -3(1)(-1)(2) 

= 1-1+8+6

= 14
(f) m²n² +n²p² +p²m² 

= (1)² (-1)² + (-1)² (2)² + (2)² (1)² 

= 1+ 4+ 4

= 9

Correct option is  B) 50

Correct option is (B) 50

\(52^2–48^2\) \(=(52-48)\times(52+48)\)

\(=4\times100=400\)

\(\therefore\) 8x = 400

\(\Rightarrow\) \(x=\frac{400}8=50\)

Correct option is  A) 121 – x² 

Correct option is (A) 121 – x²

(11 + x) (11 – x) \(=11^2-x^2=121-x^2\)

Algebraic expressions in which the variables involved have only non- negative integral exponents are called polynomials. 

For example, by² – y + 2 is a polynomial while a^1/2 + 8a³ – 3 and y^–2 + y are not polynomials.

Required expression is

89a³ -64a²+6a+16-(21a³ -17a²)

= 89a³ -64a²+6a+16 -21a³ + 17a² On combining the like terms,

= 89a³ -21a³ – 64a² +17a² + 6a+16 

= 68a³ -47a² + 6a+16

So,21a³ -17a² is 68a³-47a² + 6a+16 less than 89a³ -64a²+6a+16.

Correct option is  A) Identity

Correct option is (A) Identity

\(a^2-b^2\) = (a + b) (a – b) is true for all a and b.

\(\therefore\) It is an identity.

A) x² + ax – bx – ab

Correct option is (A) x² + ax – bx – ab

(x + a) (x – b) \(=x^2+ax-bx-ab\)

\(=x^2+(a–b)x–ab\)

Correct option is C) 0

Correct option is (C) 0

\(101^2–101^2=0\)

Correct option is  C) 25

Correct option is (C) 25

a = 13, b = 12

Then \(a^2-b^2\) \(=13^2-12^2\) = 169 - 144 = 25

(d) (3p²+5q-9r³+7)

Explanation:

Given (3q+7p²-2r³+4) – (4p²-2q+7r³-3)

According to the rules of subtraction of algebraic equations, we have negative sign will

Becomes positive and so we have to keep the big numerical sign.

Now arrange the variables in rows we get

(3q+7p²-2r³+4) – (4p²-2q+7r³-3) = (3p²+5q-9r³+7)

(\(\frac{-7}{4}\)ab²c - \(\frac{6}{25}\)a²c²) (-50a²b²c²)

= \(\frac{-7}{4}\)ab²c × -50a²b²c² - \(\frac{6}{25}\)a²c² × -50a²b² × c²

= \(\frac{7}{4}\) × 50 × a³ × b⁴ × c³ - \(\frac{6}{25}\) × - 50 × a⁴ × b² × c⁴

= \(\frac{350}{4}\)a³b⁴c³ + 12a⁴b²c⁴

= \(\frac{175}{2}\)a³b⁴c³ + 12a⁴b²c⁴

here the product is given as

[\((-7/4) \) ab²c - \(6/25\) a²c² ] (-50a²b²c²)

= \(225/2\) a³b⁴c³ + 12 a⁴b²c⁴

(b) 1/3²

Explanation: By the law of exponent we know: a^-n = 1/aⁿ.

Hence, 3^-2 = 1/3²

Correct option is  B) 1331

Correct option is (B) 1331

\(11^3=11^2\times11=121\times11=1331\)

\(x^4 + x^2y^2 + y^4\) = x⁴ + 2x²y² + y⁴ - x²y²          (Note : Add and subtract x²y²)

= (x² + y²)² - x²y²

= (x² + y²)² - (xy)²

= (x² + y² + xy)(x² + y² -xy).