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a² + 3a = (- 3)² + 3(- 3) (when a = – 3) 

= (- 3 × – 3) – 9

(Error is 3(- 3) = 0) 

= 9 – 9 = 0

3 – a = 3 – (- 3) (when a = – 3) 

= 3 + 3 = 6 

(Error is – (- 3) = – 3)

(P + 2) (P – 2) (P² + 4) 

Identity (a + b)(a – b) = a² – b² 

= (P² – 2²)(P²+ 4) 

= (P² – 4)(P² + 4) 

= (P²)² – 4² 

= P⁴ – 16

a³ – a² – 3 

= (- 3)³ – (- 3)² – 3 (when a = – 3) 

= (- 3 × – 3 × – 3) – (- 3 × – 3) – 3 

= – 27 – (9) – 3 

(Error is (- 3)³ = – 9 and (- 3)² = 6) 

= – 27 – 9 – 3 = – 39

Given expression is 5x² + 3y + 7

Numerical terms = 7 

Algebraic terms = 5x² , 3y

(2a + 3)(2a – 3) (4a² + 9) 

Identity (a + b)(a-b) = a² – b² 

= [((2a)² – 3²)(4a²+9)² 

= (4a² – 9)(4a² + 9) 

= (4a²)² – 9² 

= 16a⁴ – 81

Given,

(9x² + 24x + 16)

x = 12

So, we can also write it as;

= (3x)² + 2(3x)(4) + (4)²

→ By the formula (a + b)² we get,

= (3x + 4)² 

= [3 (12) + 4]² 

= [36 + 4]² 

= [40]² = 1600

Hence the value of the expression is 1600 when x =12.

(x – 3)(x + 3) (x² + 9) 

Identity(a + b)(a-b) = a² – b² 

= (x² – 3²)(x² + 9) 

= (x² – 9)(x² + 9) 

= (x²)²– 9² 

= x⁴ – 81

(i) Given expression is 5x² + 3y + 7

Number of terms = 3

(ii) Given expression is 5x²y + 3

Number of terms = 2

(iii) Given expression is 3x²y

Number of terms = 1

(iv) Given expression is 5x – 7

Number of terms = 2

(v) Given expression is 7x³ – 2x

Number of terms = 2

24x² (1 – 2x)

= 24x² – 48x³

According to question,

When x = 3

= 24x² – 48x³

= 24 (3)² – 48 (3)³

= 24 (9) – 48 (27)

= 216 – 1296

= -1080

Correct option is B) 0

x² - 1 = (-1)² - 1 = 1 - 1 = 0

102 × 98 = (100+ 2) (100 – 2) 

Identity (a + b)(a – b) = a² – b² 

Here a = 100, b= 2 

(100 + 2)(100 – 2) = 100² – 2² 

=10000 – 4 

102 × 98 = 9996

8.5 × 9.5 = (9 – 0.5) (9 + 0.5) 

Here a = 9, b 

= 0.5 (9 – 0.5)(9 + 0.5) 

= (9²) – (0.5)² 

= 81 – 0.25 

= 80.75

(i) Given x + y – 3z + y    

x + y – 3z + y = x + y – (3z – y)

(ii) Given 3x – 2y – 5z – 4

3x – 2y – 5z – 4 = 3x – 2y – (5z + 4)

(iii) Given 3a – 2b + 4c – 5

3a – 2b + 4c – 5 = 3a – 2b – (–4c + 5)

(iv) Given 7a + 3b + 2c + 4

7a + 3b + 2c + 4 = 7a + 3b – (–2c – 4)

(v) Given 2a² – b² – 3ab + 6

2a² – b² – 3ab + 6 = 2a² – b² – (3ab – 6)

(vi) Given a² + b² – c² + ab – 3ac

a² + b² – c² + ab – 3ac = a² + b² – c² – (- ab + 3ac)

(x + a) (x + b) = x² + (a + b) x + ab

= (x + a) (x + b)

= x × (x + b) + a × (x + b)

= x² + xb + xa + ab

= x² + x (b + a) + ab

Required expression is

y⁴ -12y² +y+14-(17y³+34y² -51y+68)

= y⁴ -12y² +y+14-17y³-34y² +51y-68 On combining the like terms,

= y⁴-12y²-34y² + y+51y+14-68-17y³ 

= y⁴-46y² + 52y-17y³ -54 

= y⁴ -17y³ -46y² + 52y-54

So, y⁴ -12y² +y+14 is y⁴-17y³-46y² + 52y-54 greater than 17y³+34y² -51y+68.

Polynomial

As the product of two polynomials is again a polynomial.

Correct option is  C) 13

Correct option is (C) 13

\(85^2–84^2\) \(=(85-84)\times(85+84)\)

\(=1\times169=169\)

\(\therefore\) 13x = 169

\(\Rightarrow\) \(x=\frac{169}{13}=13\)

Required expression is

93p² -55p+4 —(13p³ -5p² + 17p-90)

= 93p² -55p+4 -13p³+5p² – 17p + 90

On combining the like terms,

= 93p² + 5p²-55p-17p+4+90-13p³ 

= 98p²-72p+94-13p³ 

=-13p³ + 98p²-72p+ 94

So, 93p² -55p+4 is -13p³ + 9p² -72p+ 94 exceed from 13p³ -5p² + 17p-90.

In order to find the solution, we will subtract 99x³ -33x² -13x-41 from 0.

Required expression is

0- (99x³ -33x² -13x-41) 

= 0- 99x³ +33x² +13x+41

= – 99x³ +33x² +13x+41

So, If we add -99x³ +33x² +13x+41 to 99x³ -33x² -13x-41, then the sum is zero.

(b) (6x²+7x-3)

Explanation:

Given (2x+3) (3x-1)

By solving in horizontal method we get

(2x+3) (3x-1)= 2x (3x-1) + 3 (3x-1)

(2x+3) (3x-1)= (6x²+7x-3)

 (8/5)x + (11/7)y + (9/4)xy, (-3/2)x – (5/3)y – (9/5)xy

Required sum,

[(8/5)x + (11/7)y + (9/4)xy] + [(-3/2)x – (5/3)y – (9/5)xy]

Collecting like terms,

= [(8/5)x – (3/2)x] + [(11/7)y – (5/3)y] + [(9/4)xy – (9/5)xy]

= (1/10)x – (2/21)y + (9/20)xy

(3/2)x³ – (1/4)x² + (5/3), (-5/4)x³ + (3/5)x² – x + (1/5), -x² + (3/8)x – (8/15)

Required sum,

= (3/2)x³ – (1/4)x² + (5/3), (-5/4)x³ + (3/5)x² – x + (1/5), -x² + (3/8)x – (8/15)

Collecting like terms,

= [(3/2)x³ – (5/4)x³] + [- (1/4)x² + (3/5)x² -x²] + [– x + (3/8)x] + [(5/3) + (1/5) – (8/15)]

= (1/4)x³ – (13/20)x² – (5/8)x + (4/3)

(2/3)a – (4/5)v + (3/5)c, -(3/4)a – (5/2)b + (2/3)c, (5/2)a + (7/4)b – (5/6)c

Required sum,

= [(2/3)a – (4/5)b + (3/5)c] + [-(3/4)a – (5/2)b + (2/3)c] + [(5/2)a + (7/4)b – (5/6)c]

Collecting like terms,

= (2/3)a -(3/4)a + (5/2)a – (4/5)b – (5/2)b + (7/4)b + (3/5)c + (2/3)c – (5/6)c

= [(2/3) – (3/4) + (5/2)]a + [(-4/5) – (5/2) + (7/4)]b + [(3/5) + (2/3) – (5/6)]c

= [(8 – 9 + 30)/12)a] + [(-16 -15 +30)/20)b] + [(18 + 20 – 25)/ 30)c]

= (29/12)a +- (31/20)b + (13/30)c

(c) (x²+8x+16)

Explanation:

Given (x+4) (x+4)=(x+4)²

By expanding the given expression by using (a + b)² = a²+2ab+b² we get

(x+4)² = x²+2(x) (4)+4² = x²+8x+16

(d) (x²-12x+36)

Explanation:

Given (x-6) (x-6)=(x-6)²

By expanding the given expression by using (a – b)² = a²-2ab+b² we get

(x-6)² = x²-2(x) (6)+6² = x²-12x+36

(3/5)x, (2/3)x, (-4/5)x

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

= (3/5)x + (2/3)x + (-4/5)x

LCM of 5, 3, and 5 is 15

= (9x + 10x – 12x)/ 15

= (19x – 12x)/ 15

= (7/15)x

(c) -1/3

-y/3 can also be written as y × (-1/3)

So, Coefficient of y is -1/3

5x, 7x, -6x

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

Now add the like terms,

= 5x + 7x + (-6x) … [∵ + × – = -]

Add terms having same sign first,

= 5x + 7x – 6x

= 12x – 6x

= 6x

Given 6ax-2by+3cz, 6by-11ax-cz and 10cz-2ax-3by

To add the given expression we have arrange them column wise is given below:

6ax-2by+3cz

-11ax+6by-cz

-2ax-3by+10cz

(b) – 5xyz², 7xyz²

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.

Given 2x³-9x²+8, 3x²-6x-5, 7x³-10x+1 and 3+2x-5x²-4x³

To add the given expression we have arrange them column wise is given below:

2x³-9x²+8

7x³-10x+1

3x²– 6x- 5

-4x³-5x²+2x+3

Positive

If both the like terms are either positive or negative, then the resultant term will always be positive.

(d) 9x² + 16y² – 24xy

As per the condition in the question, (3x – 4y)²

The standard identity = (a – b)² = a² – 2ab + b²

Where, a = 3x, b = 4y

Then,

(3x – 4y)² = (3x)² – (2 × 3x × 4y) + (4y)²

= 9x² – 24xy + 16y²

Sum or difference of two like terms is a like term.

Let us consider the tow like terms = 2y and 3y

Sum of two like terms = 2y + 3y

= 5 y

Difference of two like terms = 2y – 3y

= -y

Negative

As the product of a positive term and a negative term is always negative.

(i) Given x = -2, y = -1, z = 3

Consider (x/y) + (y/z) + (z/x)

On substituting the given values we get,

= (-2/-1) + (-1/3) + (3/-2)

The LCM of 3 and 2 is 6

= (12 – 2 – 9)/6

= (1/6)

(ii) Given x = -2, y = -1, z = 3

Consider x² + y² + z² – xy – yz – zx

On substituting the given values we get,

= (-2)² + (-1)² + 3² – (-2) (-1) – (-1) (3) – (3) (-2)

= 4 + 1 + 9 – 2 + 3 + 6

= 23 – 2

= 21

In the formula, area of circle = πr², the numerical constant of the expression πr² is π.

(b) 12a³b – 14ab² + 10a²b²

Now we have find product of trinomial and monomial,

= (6a² – 7b + 5ab) × 2ab

= (2ab × 6a²) – (2ab × 7b) + (2ab × 5ab)

= 12a³b – 14ab² + 10a²b²

b,c

We have , a(b+c)=a x b + a x c [using left distributive law]

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.

(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

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

Correct option is  C) 2 + m + n

(i) Given x²y − xy² + 7xy − 3

The constant term in the given algebraic expressions is -3.

(ii) Given a³ − 3a² + 7a + 5

The constant term in the given algebraic expressions is 5.

Correct option is C) x – 4a

Binomial expression is a mathematical expression of two terms connected by a plus sign or minus sign.

It is clear that 19 x 4 and 7xyz are example o monomial, 3x + y - z is example of trinomial and x - 4a is an example of binomial.

Correct option is D) x - 11/3

Numerical expression contains numbers and at least on e operation. it does not contain any variable . It is clear that 1 + 2 - 9, 9 + (6 - 4) and -3 - 5 are examples of numerical expression.

(a – b) (a – b) = (a – b)²= a² – 2ab + b²

(a – b) (a – b) = a × (a – b) – b × (a – b)

= a² – ab – ba + b²

= a² – 2ab + b²

-37

The constant term (with their sign) involved in term of an algebraic expression is called the numerical coefficient of that term.

(i) Given xy

 The numerical coefficient in the term xy is 1.

(ii) Given -6yz

The numerical coefficient in the term – 6yz is – 6.

(iii) Given 7abc

The numerical coefficient in the term 7abc is 7.

(iv) Given -2x³y²z

The numerical coefficient in the term −2x³y²z is -2.