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First we have to find the sum of 13x – 4y + 7z and – 6z + 6x + 3y

Therefore, sum of (13x – 4y + 7z) and (–6z + 6x + 3y)

= (13x – 4y + 7z) + (–6z + 6x + 3y)

= (13x – 4y + 7z – 6z + 6x + 3y)

= (13x + 6x – 4y + 3y + 7z – 6z)

= (19x – y + z)

Now we have to find the sum of (6x – 4y – 4z) and (2x + 4y – 7)

= (6x – 4y – 4z) + (2x + 4y – 7)

= (6x – 4y – 4z + 2x + 4y – 7)

= (6x + 2x – 4z – 7)

= (8x – 4z – 7)

Now, required expression = (8x – 4z – 7) – (19x – y + z)

= 8x – 4z – 7 – 19x + y – z

= 8x – 19x + y – 4z – z – 7

= –11x + y – 5z – 7

Correct option is  A) 25ab

18ab + (-5ab) + 12ab = 30ab - 5ab = 25ab

Given 6x³ −7x² + 5x − 3 and 4 − 5x + 6x² − 8x³

= (4 – 5x + 6x² – 8x³) – (6x³ – 7x² + 5x – 3)

= 4 – 5x + 6x² – 8x³ – 6x³ + 7x² – 5x + 3

= – 8x³– 6x³ + 7x² + 6x²– 5x – 5x + 3 + 4

= – 14x³ + 13x² – 10x + 7

\(\frac{7}{5}(\frac{3}{5}x^3y^3+\frac{2}{5}x^3y)\)

= \(\frac{21}{25}x^3y^3+\frac{14}{25}x^3y\)

250 × 5 (x²yz + \(\frac{xy\times y}{10}\))

= 250 (5x²yz + \(\frac{x\times y\times y}{2}\))

= 250 × 5x²yz + 125xy²

(7a + 9b) (7a – 9b)

= (7a)² – (9b)²

= 49a² – 81b²

-3 × 4 – a² × b²

= -12 × a² × b²

= -12a²b²

\(\frac{1}{2}\times \frac{2}{3}\times x\times x^{2}\times y\times y\times z^{2}\)

= \(\frac{1}{3}\times x^{3}\times y^{2}\times z^{2}\)

= \(\frac{1}{3}x^{3}y^{2}z^{2}\)

Correct option is A) -2x² y

4x²y - 6x²y = (4 - 6)x²y = -2x²y

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

(-5) × (-5) × x × x² × y × y × z

= 15 × x³ × y² × z

= 15x³y²z

(i) (x + 1) (x + 2)
= x² + (1 + 2)x + 1 × 2 (Using given identity)
= x² + 3x + 2

(ii) (3x + 5) (3x + 1)
= (3x)² + (5 + 1) 3x + 5 x 1 (Using given identity)
= 9x² + 18x + 5

(iii) (4x – 5) (4x – 1)
= {4x + (- 5)} {4x + {- 1)}
= (4x)² + {(- 5) + (- 1)} 4x + (- 5) (- 1) (Using given identity)
= 16x² + (- 6) 4x + (5)
= 16x² – 24x + 5

(iv) (3a + 5) (3a – 8)
= (3a + 5) {3a + (- 8)}
= (3a)² + {5 + (- 8)} 3a + (5) (- 8) (Using given identity)
= 9a² + (- 3) 3a – 40
= 9a² – 9a – 40

(v) {xyz – 1) (xyz – 2)
= {xyz + {- 1)} {xyz + {- 2)}
= {xyz)² + {(- 1) + (- 2)} xyz + (- 1) (- 2)
= x²y²z² – 3xyz + 2

Let A = 2l² – 3lm + 5m² and 

B = 3l² – 4lm + 6m² 

A – B = A + (- B) 

Additive inverse of B is 

– B = – (3t² – 4lm + 6m² ) 

= – 3l² + 4lm – 6m² 

∴ A – B = A + (- B) 

= (2l² – 3lm + 5m² ) + (- 3l² + 4lm – 6m² )

= 2l² – 3lm + 5m² – 3l² + 4lm – 6m² 

= 2l² – 3l² – 3lm + 4lm + 5m² – 6m²

= (2 – 3)l² + (- 3 + 4)lm + (5 – 6)m² 

= (- 1) l² + 1 lm + (- 1)m² 

∴ A – B = – l² + lm – m²

(a) 196 p²q⁴

= 4p × (– 7q³) × (–7pq)

= (4 × (-7) × (-7)) × p × q³ × pq

= 196p²q⁴

(d) – 5pq

The given two monomials are like terms.

Then sum of -7pq and 2pg = – 7pq + 2pq

= (-7 + 2) pq

= -5pq

Correct option is B) {7p, – 2p, 3p}

Like terms have the the same variables to the same power. It is cleared that only {7p, -2p, 3p} are like terms in all given options.

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)

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

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)

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

(i) \(7x^{2}yz-5xy\)

This equation consists of two terms that are:

\(7x^{2}yz\) and \(-5xy\)

The coefficient of \(7x^{2}yz\) is 7

The coefficient of – 5xy is –5

(ii) \(x^{2}+x+1\)

This equation consists of three terms that are:

\(x^{2},x,1\)

The coefficient of \(x^{2}\) is 1

The coefficient of x is 1

The coefficient of 1 is 1

(iii) \(3x^{2}y^{2}-5x^{2}y^{2}z^{2}+z^{2}\)

This equation consists of three terms that are:

\(3x^{2}y,-5x^{2}y^{2}z^{2}\) and \(z^{2}\)

The coefficient of \(3x^{2}y\) is 3

The coefficient of \(-5x^{2}y^{2}z^{2}\) is -5

The coefficient of \(z^{2}\) is 1

(iv) 9-ab+bc-ca

 (v)  \(\frac{a}{2}+\frac{b}{2}-ab\) 

(vi) 0.2x-0.3xy+0.5y

201 x 202

= (200 + 1) x (200 + 2)

= (200)² + (1 + 2) (200) + 1 x 2 (Using given identity)

= 40000 + 600 + 2

= 40602

(a + 3b) and (x + 5)

= a (x + 5) + 3b (x + 5)

= a × x + a × 5 + 3b × x + 3b × 5

= ax + 5a + 3bx + 3 × 5 × b

= ax + 5a + 3 bx + 15 b

(i) (2a + 7) (2a – 7)

= (2a)² – (1)²

= 2²a² – 7²

= 4a² – 49

(ii) (p² + q²) (p² – q²)

= (p²)² – (q²)²

= p⁴ – q⁴

(1.2)² – (0.8)²

= (1.2 + 0.8) (1.2 – 0.8) (Using identity III)

= (2) (0.4)

= 0.8

True

(x + a) (x + b) = x² + (a + b) x + ab is a identity.

(2lm + 3m²) and (3lm – 5m²)

= 2lm (3lm – 5m²) + 3m² (3lm – 5m²)

= 2lm × 3lm – 2lm × 5m² + 3m² × 3lm – 3m² × 5m²

= (2 × 3) l²m² – (2 × 5) lm³ + (3 × 3) lm³ – (3 × 5) m⁴

= 6l2m² – 10lm³ + 9lm³ – 15m⁴

= 6l2m² – lm³ – 15m⁴

(1.5p – 0.5q) and (1.5p + 0.5q)

= 1.5p (1.5p + 0.5q) – 0.5q (1.5p + 0.5q)

= 1.5p × 1.5p + 1.5p × 0.5q – 0.5q × 1.5p – 0.5q × 0.5q

= 1.5 × p × 1.5 × p + 1.5 × p × 0.5 × q – 0.5 × q × 1.5 × p – 0.5 × q × 0.5 × q

= 1.5 × 1.5 × p × p + 1.5 × 0.5 × p × q – 0.5 × 1.5 × q × p – 0.5 × 0.5 × q × q

= 2.25 × p² + 0.75 × pq – 0.75 × qp – 0.25 × q²

= 2.25p² + 0.75pq – 0.75pq – 0.25q²

= 2.25p² – 0.25q²

(c) 27

64a³ + 48a²b + 12a²b + b³

⇒ (4a + b)³   [∴ (a + b)² = a^{3 +} b³ + 3a²b + 3ab²]

∴ Reqd. value = (4-1)³ = 3³ = 27.

The multiplication is as follows:

(-7pq²r³) ×(-13p3q²r)

= (-7×-13×3)×(pq²r³.pq²r)(Here dot implies multiplication)

= (273)×(p²q⁴r⁴)

= 273p⁴q⁴r⁴

The product is = 273p⁴q⁴r⁴

(i) Given x³ -2x²y + 3xy²– y³, 2x³– 5xy² + 3x²y – 4y³

Collecting positive and negative like terms together, we get

= x³ +2x³ – 2x²y + 3x²y + 3xy² – 5xy² – y³– 4y³

= 3x³ + x²y – 2xy² – 5y³

(ii) Given a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴, – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

= a⁴ – 2a³b + 3ab³ + 4a²b² + 3b⁴ – 2a⁴ – 5ab³ + 7a³b – 6a²b² + b⁴

Collecting positive and negative like terms together, we get

= a⁴ – 2a⁴– 2a³b + 7a³b + 3ab³ – 5ab³ + 4a²b² – 6a²b² + 3b⁴ + b⁴

= – a⁴ + 5a³b – 2ab³ – 2a²b² + 4b⁴

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

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

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

We have, the sum of first n natural numbers

= n²/2 + n/2

Factorasation of given expression = 1/2 (n² + n) = 1/2n(n  + 1)

(i) x+y

This expression contains two terms x and y

So, it is called ‘Binomial’

(ii) 1000

It contains one term 1000

So, it is called monomial

(iii) \(x+x^{2}+x^{3}+x^{4}\)

It contains four terms

So, it is not a monomial, binomial and trinomial

(iv) 7+a+5b

It contains three terms

So, it is called trinomial

(v) \(2b-3b^{2}\)

It contains two terms

So, it is called binomial

(vi) \(2y-3y^{2}+4y^{3}\)

It contains three terms

So, it is called trinomial

(vii) 5x-4y+3x

8x – 4y

It contains two terms

So, it is called binomial

(viii) \(4a-15a^{2}\)

It contains two terms

So, it is called binomial

(ix) xy+yz+zt+tx

It contains four terms

So, it is not a monomial, binomial and trinomial

(x) pqr

It contains one term

So, it is called monomial

(xi) \(p^{2}q+pq^{2}\)

It contains two terms

So, it is called binomial

(xii) 2p+2q

It contains two terms

So, it is called monomial

(i) Given a²

a² is a monomial expression because it contains only one term

(ii) Given a² − b²

a² − b² is a binomial expression because it contains two terms

(iii) Given x³ + y³ + z³

x³ + y³ + z³ is a trinomial because it contains three terms

(iv) Given x³ + y³ + z³ + 3xyz

x³ + y³ + z³ + 3xyz is a quadrinomial expression because it contains four terms

(v) Given 7 + 5

7 + 5 is a monomial expression because it contains only one term

(vi) Given a b c + 1

a b c + 1 is a binomial expression because it contains two terms

(vii) Given 3x – 2 + 5

3x – 2 + 5 is a binomial expression because it contains two terms

(viii) Given 2x – 3y + 4

2x – 3y + 4 is a trinomial because it contains three terms

(ix) Given x y + y z + z x

x y + y z + z x is a trinomial because it contains three terms

(x) Given ax³ + bx² + cx + d

ax³ + bx² + cx + d is a quadrinomial expression because it contains four terms

(i) Since x + y contains two terms. Therefore it is binomial.
(ii) Since 1000 contains one terms. Therefore it is monomial.
(iii) Since x + x² + x³ + x⁴ contains four terms. Therefore it is a polynomial and it does not fit in above three categories.
(iv) Since 7 + y + 5x contains three terms. Therefore it is trinomial.
(v) Since 2y -3y² contains two terms. Therefore it is binomial.
(vi) Since 2y -3y² + 4y³ contains three terms. Therefore it is trinomial.
(vii) Since 5x - 4y + 3xy contains three terms. Therefore it is trinomial.
(viii) Since 4x -15z² contains two terms. Therefore it is binomial.
(ix) Since ab+bc +cd + da contains four terms. Therefore it is a polynomial and it does not fit in above three categories.
(x) Since pqr contains one terms. Therefore it is monomial.
(xi) Since p²q + pq contains two terms. Therefore it is binomial.
(xii) Since 2 p + 2q contains two terms. Therefore it is binomial.

x⁴ – 256 = (x²)² – (16)² 

= (x² + 16) (x² – 16) (using a² – b² = (a + b) (a – b))

= (x² + 16) (x² – 4²) 

= (x² + 16) (x + 4) (x – 4) (using a² – b² = (a + b) (a – b))

y²/9 - 9

= (y/3)² - (3)²

= (y/3 + 3)(y/3 - 3) (Since a² – b² = (a + b) (a – b))

(2x + 5) and (3x – 7)

= 2x(3x – 7) + 5(3x – 7)
= 2x × 3x – 2x × 7 + 5 × 3x – 5 × 7
= 2 × x × 3 × x – 2 × x × 7 + 5 × 3 × x – 35
= 2 × 3 × x × x – 2 × 7 × x + 15 × x – 35
= 6 × x² – 14 × x + 15x – 35
= 6x² – 14x + 15x – 35
= 6x² + x – 35

Identity (I) is

(a + b)² = a² + 2ab + b²

We put – b in place of b

{a + (- b)}² = a² + 2a (- b) + (- b)² = (a – b)² = a² – 2ab + b²

Which is identity (II). Yes, we get identity (II).

Let a = 5x – y, b = y – 4z, c= 4z – 5x

a + b + c = 5x - y + y - 4z + 4z - 5x = 0

∴ a³ + b³ + c³ = 3abc

⇒ (5x - y)³ + (y - 4z)³ + (4z - 5x)³ + (4z - 5x)³ = 3(5x - y)(y - 4z)(4z - 5x)

i. Monomial 

ii. Binomial 

iii. Trinomial 

iv. Polynomial 

v. Binomial 

vi. Monomial

vii. Monomial 

viii. Trinomial

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

Factorised form of 18 mn + 10 mnp is 2mn (9 + 5p)

= (2 × 9 × m × n) + (2 × 5 × m × n × p)

= 2mn (9 + 5p)

False

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)

(b) 3 (x – 8)

The factorised form of 3x – 24 is,

Take out 3 as common,

= 3 (x – 8)

a³ – 4a² + 12 – 3a 

= a² (a – 4) – 3a + 12 

= a² (a – 4) – 3 (a – 4) 

= (a – 4) (a² – 3)