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Expansion of (x + 1/x)² is x² + 1/x² + 2.

(c) (x + a)(x + b)

3x³ + 3x² + x + 1

= 3x² (x + 1) + 1 (x + 1)

= (x + 1) (3x² + 1)

(c) 4x² + 12x + 9

x² + 8x + 16

= x² + 2 × x + 4 + (4)²

= (x + 4)²

2x (2x² + 2x – 9)

= 2x × 2x² + 2x × 2x – 2x × 9

= 4x³ + 4x² – 18x

Factor of 4x² + 8xy + 4y² is (2x + 2y)².

(x + y)² = x² + y² + 2xy

⇒ (20)² = x² + y² + 2(34)

⇒ 400 = x² + y² + 68

⇒ x² + y² = 400 – 68

⇒ x² + y² = 332

(d) 36x² + 12x + 1.

Common factor is (x + y)

The product of expression (2a – 3) (2a + 3) is 4a² – 9.

(a) (2x + 1)(x² + 1)

Common factor of 6x + 18xy is 6x.

(i) (l + m)² − 4lm 

= l² + 2lm + m² − 4lm

= l² − 2lm + m²

= (l − m)²     [(a − b)² = a² − 2ab + b²]

(ii) a⁴ + 2a²b² + b⁴ 

= (a²)² + 2 (a²) (b²) + (b²)²

= (a² + b²)²       [(a + b)² = a² + 2ab + b²]

(0.7)² – (0.3)² 

= (0.7 + 0.3) (0.7 - 0.3)

= 1 x 0.4 = 0.4

16 – 9x² = (4)² – (3x)² = (4 + 3x) (4 – 3x)

1 – 100a² = (1)² – (10a)² = (1 + 10a) (1 – 10a)

4x² – 81y² = (2x)² – (9y)² = (2x + 9y) (2x – 9y)

By taking 1-n as a common factor for the above equation we get,

2m (1-n) + 3(1-n) = (1-n) (2m + 3)

25x³ – x = x(25x² - 1)

= x[(5x)² - (1)²]

= x(5x + 1) (5x - 1)

a⁴ – b⁴ = (a²)² -(b²)² 

= (a² + b²) (a² - b²)

= (a² + b²) (a + b) (a-b)

2a – 4b – xa + 2bx

= 2(a-2b) - x(a-2b)

= (a-2b) (2-x)

(a) 5(2x + 1)(3x + 5) ÷ (2x + 1)

Answer: 5(2x + 1)(3x + 5) ÷ (2x + 1)
= 5(3x + 5) = 15x + 25

(b) 26xy(x + 5) (y – 4) ÷ 13x (y -4)

Answer: 26xy(x + 5) (y – 4) ÷ 13x (y -4)
= 2y(x + 5)(y – 4) ÷(y – 4)
= 2Y (x + 5) = 2xy + 10y

(c) 52pqr(p + q) (q + r)( r + p) ÷ 104 pq(q + r)( r + p)

Answer: 52pqr(p + q) (q + r)( r + p) ÷ 104 pq(q + r)( r + p)
= r (p + q) (q + r) ( r + p) ÷ 2 (q + r) (r + p) 
= r (p + q) ÷ 2

(d) 20 ( y + 4) (y² + 5y + 3) ÷ 5 (y + 4)

Answer: 20 ( y + 4) (y2 + 5y + 3) ÷ 5 (y + 4)
= 4 (y + 4) (y2 + 5y + 3) ÷ (y + 4)
= 4(y2 + 5y + 3)

(e) x ( x + 1) ( x + 2) (x + 3) ÷ x( x + 1)

Answer: x ( x + 1) ( x + 2) (x + 3) ÷ x( x + 1)
= (x + 2)(x + 3)

Question 1: 4(x – 5) = 4x – 5

Answer: 4(x – 5) = 4x – 20

Question 2: x(3x + 2) = 3x² + 2

Answer: = 3x² + 2x

Question 3: 2x + 3y = 5xy

Answer: 2x + 3y
Here; x and y are as different as chalk and cheese and hence cannot be added together.

Question 4: x + 2x + 3x = 5x

Answer: = 6x
It is like adding one apple, two apples and three apples.

Question 5: 5y + 2y + y – 7y = 0

Answer: = y

Question 6: 3x + 2x = 5x²

Answer: = 5x

Question 7: (2x)² + 4(2x) + 7 = 2x² + 8x + 7

Answer: = 4x² + 8x + 7

Question 8: (2x)² + 5x = 4x + 5x = 9x

Answer: 4x² + 5x

Question 9: (3x + 2)² = 3x² + 6x + 4

Answer: Using (a + b)² = a² + 2ab + b²;
= 9x² + 12x + 4

(a) (y² + 7y + 10) ÷ (y + 5)

Answer: (y² + 7y + 10) ÷ (y + 5)
Here, dividend can be factorised by splitting the middle term as follows:
= (y² + 5y + 2y + 10) ÷ (y + 5)
= [y (y + 5) + 2 (y + 5) ] ÷ (y + 5)
= (y + 2) (y + 5) ÷ (y + 5)
= y + 2

(b) (m² – 14m – 32) ÷ (m + 2)

Answer: (m² – 14m – 32) ÷ (m + 2)
Here, dividend can be factorised by splitting the middle term as follows:
= (m² – 16m + 2m – 32) ÷ (m + 2)
= [ m(m – 16) + 2(m – 16) ] ÷ (m + 2)
= (m + 2) (m – 16) ÷ (m +2)
= m – 16

(c) 5p² – 25p + 20) ÷ (p – 1)

Answer: 5p² – 25p + 20) ÷ (p – 1)
Here, dividend can be factorised by splitting the middle term as follows:
= (5p² – 5p – 20p + 20) ÷ (p - 1)
= [5p(p – 1) – 20(p – 1) ] ÷ (p – 1 )
= (5p – 20) (p – 1) ÷ (p – 1)
= 5p – 20

(d) 4yz(z² + 6z – 16) ÷ 2y (z + 8)

Answer: 4yz(z² + 6z – 16) ÷ 2y (z + 8)
= 2z(z² + 8z – 2z – 16) ÷ (z + 8)²= 2z [ z(z + 8) – 2(z + 8) ] ÷ (z + 8)
= 2z (z – 2) (z + 8) ÷(z + 8)
= 2z (z – 2)

(e) 5pq(p² – q²) ÷ 2p (p + q)

Answer: 5pq(p² – q²) ÷ 2p (p + q)
Using (a +b) (a – b) = a² – b²; the equation can be written as follows:
= 5pq (p + q) (p – q) ÷ 2p (p + q)
= 5q(p – q) ÷ 2

(f) 12xy (9x² – 16y²) ÷ 4xy (3x + 4y)

Answer: 12xy (9x² – 16y²) ÷ 4xy (3x + 4y)
This can be solved as previous question;
= 3 (3x + 4y) (3x – 4y) ÷ (3x + 4y)
= 3 (3x – 4y)

(g) 39y³(50y² – 98) ÷ 26y² (5y + 7)

Answer: 39y³(50y² – 98) ÷ 26y² (5y + 7)
= 2 x 39y³(25y² – 49) ÷ 26y² (5y + 7)
= 3y (25y² – 49) ÷ (5y + 7)
= 3y (5y + 7) (5y – 7) ÷ (5y + 7)
= 3y (5y – 7)

(i) ax² + bx

Answer: x(ax + b)

(ii) 7p² + 21q²

Answer: 7(p² + 3q²)

(iii) 2x³ + 2xy² + 2xz²

Answer: 2x³ + 2xy² + 2xz²
= 2x(x²+y²+z²)

(iv) am² + bm² + bn² + an²

Answer: a(m² + n²) + b(m² + n²)
= (a + b)(m² + n²)

(v) (lm + l) + m + 1

Answer: l(m + 1) + 1(m + 1)
= (l + 1)(m + 1)

(vi) y (y + z) + 9 (y + z)

Answer: (y + 9)(y + z)

(vii) 5y² – 20y – 8z + 2yz

Answer: 5y(y + 4) + 2z(y + 4)
= (5 + 2z)(y + 4)

(viii) 10ab + 4a + 5b + 2

Answer: 5b + 10ab + 2 + 4a
= 5b(1 + 2a) + 2(1 + 2a)
= (5b + 2)(1 + 2a)

(ix) 6xy – 4y + 6 – 9x

Answer: 6xy – 4y + – 9x + 6
= 2y (3x – 2) - 3 (3x - 2)
= (2y – 3)(3x – 2)

(i) 4p² – 9q²

Answer: This can be factorised by using the equation; (a + b)(a – b) = a² – b²
Factors = (2p + 3q)(2p – 3q)

(ii) 63a² – 112b²

Answer: 63a² – 112b² = 7(9a² – 16b²)
= 7(3a + 4b)(3a – 4b)

(iii) 49x² – 36

Answer: (7x + 6)(7x – 6)

(iv) 16x⁵ – 144x³

Answer:16x⁵-144x³
= x³(16x²-144)
= x³(4x+12)(4x-12)

(v) (l + m) ² – (l – m) ²

Answer: (l + m + l – m)(l + m – l + m)
2l x 2m = 4lm

(vi) 9x² y² – 16

Answer: (3xy + 4)(3x – 4)

(vii) (x² – 2xy + y²) – z²

Answer: (x² – 2xy + y²) – z²
= (x – y)² – z²
= (x – y + z)(x – y – z)

(viii) 25a² – 4b² + 28bc – 49c²

Answer: 25a² – 4b² + 28bc – 49c²
= (5a)² – (2b)² + 2 x 2b x 7c – (7c)²
= (5a)² – [(2b)² – 2 x 2b x 7c + (7c)²]
= (5a)² – (2b – 7c)²
This can be further factorised by using (a + b)(a – b) = a² – b²
= (5a + 2b – 7c)(5a – 2b + 7c)

(i) a⁴ – b⁴

Answer: a⁴-b⁴ = (a²+b²)(a²-b²)

(ii) p⁴ – 81
=(p²+9)(p²-9)

(iii) x⁴ – (y + z)⁴

Answer: x⁴ – (y + z)⁴
= (x²+(y+z) ²)(x²-(y+z) ²)
= (x²+(y+z)²)[(x+y+z)(x-y-z)]

(iv) x⁴ – (x – z)⁴

Answer: x⁴ – (x – z)⁴
=(x²-(x-z) ²)(x²+(x-z) ²)
=[(x+x-z)(x-x+z)](x²+(x-z) ²]

(v) a⁴ – 2a²b² + b⁴

Answer: a⁴ – 2a²b² + b⁴
This can be factorised by using the identity; (a - b)² = a² – 2ab + b²
Factors = (a² – b²)² = (a² – b²)(a² – b²)

(0.7)² – (0.3)² 

Identity a² – b² = (a + b)(a – b) 

= (0.7 + 0.3)(0.7 – 0.3)

= (1.0)(0.4) 

= 0.4

x⁴ – y⁴ = (x²)² – (y²)² 

Identity a² – b² = (a + b)(a – b) 

= (x² + y²)(x² – y²) 

= (x²+ y²)(x + y)(x – y)

x² - \(\frac{9}{16}\) = x² - (\(\frac{3}{4}\))²

Identity a² - b² = (a + b)(a - b)

= (x + \(\frac{3}{4}\))(x - \(\frac{3}{4}\))

Correct option is  D) 0

Correct option is (D) 0

\((3x–2y)^2–(3x-2y)^2=0\)

4a² – 25 = (2a)² – 5² 

[a² – b² = (a + b)(a - b)] 

= (2a + 5)(2a – 5)

4a² – 9 can be written as (2a) ² – (3)²

By using the formula a² – b² = (a+b) (a-b)

Now solving for the above equation

(2a)² – (3)² = (2a+3) (2a-3)

Correct option is  D) 16x² – 1

Correct option is (D) 16x² – 1

(A) \(x^2\) - 16 \(=x^2-4^2=(x-4)(x+4)\)

\(\therefore\) (4x+1) is not a factor of \(x^2-16\)

(B) 4x+1 is not a factor of 16x - 3 

(C) 4x+1 is not a factor of 10x – 3 

(D) \(16x^2-1=(4x)^2-1^2\)  = (4x-1) (4x+1)

\(\therefore\) 4x - 1 is a factor of \(16x^2-1\)

Correct option is  C) 1

Correct option is (C) 1

\((2a+1)(\frac1{2a-1}-\frac2{4a^2-1})\) \(=(2a+1)(\frac{2a+1-2}{4a^2-1})\)

\(=(2a+1)\frac{(2a-1)}{4a^2-1}\)

\(=\frac{4a^2-1}{4a^2-1}=1\)

B)(x + √7)(x – √7)

Correct option is (B)(x + √7)(x – √7)

\(x^2-7=x^2-(\sqrt7)^2\)

= \((x+\sqrt7)(x–\sqrt7)\)

(x – 4)² = x² – 16

(x – 4)² = x² – 16 = (x)² – (4)²

(a – b)² ≠ a² – b²

∴ (x - 4)² ≠ (x)² – (4)²

∴ The given sentence is wrong.

Correct sentence is (x – 4)² = x² – 8x + 16.

Correct option is  A) a³ + b³

Correct option is (A) a³ + b³

\((a+b)^3–3ab\,(a+b)\) \(=a^3+b^3+3ab(a+b)-3ab(a+b)\)

= \(a^3+b^3\)

A) l² – 16l + 64

Correct option is (A) l² – 16l + 64

\((l–8)^2=l^2-2\times l\times8+8^2\)

= \(l^2–16l+64\)

Correct option is  A) (x – \(\frac{1}{x}\))² 

Correct option is (A) \((x-\frac1x)^2\)

\(x^2-2+\frac1{x^2}\) \(=x^2-2\times x\times\frac1x+(\frac1{x})^2\)

= \((x-\frac1x)^2\)

Correct option is  C) 34

Correct option is (C) 34

x + \(\frac1x\) = 6

\(\Rightarrow\) \((x+\frac1x)^2=6^2\)

\(\Rightarrow\) \(x^2+\frac1{x^2}+2=36\)

\(\Rightarrow\) \(x^2+\frac1{x^2}\) = 36 - 2 = 34

B) (3x + 16)(3x – 16)

Correct option is (B) (3x + 16)(3x – 16)

\(9x^2–256=(3x)^2-16^2\) = (3x-16) (3x+16)

Correct option is  A) 1154

Correct option is (A) 1154

\(x+\frac1x=6\)

\(\Rightarrow\) \(x^2+\frac1{x^2}+2=36\)   (By squaring both sides)

\(\Rightarrow\) \(x^2+\frac1{x^2}=36-2=34\)

\(\Rightarrow\) \((x^2+\frac1{x^2})^2=34^2\)

\(\Rightarrow\) \(x^4+\frac1{x^4}+2=1156\)

\(\Rightarrow\) \(x^4+\frac1{x^4}\) = 1156 - 2 = 1154

Correct option is  C) 0

Correct option is (C) 0

\(81^2–3^8\) \(=(3^4)^2-3^8=3^8-3^8=0\)

A) 36 – p² – q² – 2pq

Correct option is (A) 36 – p² – q² – 2pq

(6 + p + q) (6 – p – q) = (6 + (p + q)) (6 – (p + q))

\(=6^2-(p+q)^2\)

\(=36–(p^2+q^2+2pq)\)

= \(36–p^2–q^2–2pq\)

C) (r + t) (a +b)

Correct option is (C) (r+t) (a+b)

ar+br+at+bt = r(a+b) + t(a+b)

= (a + b) (r + t)

B) (5x-3y)(7a + 5b)

Correct option is (B) (5x-3y) (7a + 5b)

7a (5x – 3y) + 5b (5x – 3y) = (5x-3y) (7a+5b)

A) (a – 3) (a + 3) (a² + 9) 

Correct option is (A) (a – 3) (a + 3) (a² + 9)

\(a^4-81=(a^2)^2-9^2\) \(=(a^2-9)(a^2+9)\)

\(=(a^2-3^2)(a^2+9)\)

= \((a–3)(a+3)(a^2+9)\)

3a² x - bx + 3a² - b

= x (3a² - b) + 1 (3a² - b)

= (x + 1) (3a² - b)