6x²+ 17x + 12 = 6x² + 9x + 8x + 12
= 3x (2x + 3) + 4 (2x + 3)
= (2x + 3)(3x + 4)

(4a+3b) (4a-3b) = (4a)² - (3b)² = 16a² - 9b²  [We have (a+b)(a-b) = a² - b²] Put [a = 4a, b = 3b]

(4m-3m)² = (4m)² - 2(4m) (3m) + (3m)² = 16m² - 24mn + 9n²

Let the breadth of the rectangle be "b"
Length of the rectangle = x² + 4
Area of the rectangle = x⁴ + 9x² + 20

Length x Breadth = x⁴ + 9x² + 20
(x⁴ + 4)x b = x⁴ + 9x² + 20
b = \(\frac{x^4+9x^2+20}{x^2+4}\)
= x²+ 5
Breadth of the rectangle = x²+ 5

Let the side AC be x.

Perimeter of a ∆ABC = 8m²-9m+7
AB+BC+AC = 8m²-9m+7
m² - m + 6 + 3m² - 2m - 4 + x = 8m²-9m+7
4m² - 3m + 2 + x = 8m²-9m+7
x = 8m²-9m+7-(4m²-3m+2)
= 8m²-9m+7-4m²+3m-2
= 4m²-6m+5
The third side AC = 4m²-6m+5

37a + 29b - 45 = 0
29a + 37b - 21 = 0

\(\cfrac { a }{ 29(-21)-37(-45) } =\cfrac { b }{ (-45)(29)-(-21)37 } '=\cfrac { 1 }{ 37(37)-29(29) } \)
\(\cfrac { a }{ -609+1665 } =\cfrac { b }{ -1305+777 } =\cfrac { 1 }{ 1369-841 } \)
\(\cfrac { a }{ 1056 } =\cfrac { b }{ -528 } =\cfrac { 1 }{ 528 } \)
\(\cfrac { a }{ 1056 } =\cfrac { b }{ -528 } \Rightarrow a\cfrac { 1056 }{ 528 } =2\)
\(\cfrac { b }{ -528 } =\cfrac { 1 }{ 528 } \Rightarrow b=\cfrac { -528 }{ 528 } =-1\)

2x- 3y = 7
(s + 2) x - (2s + 1)y == 3 (2s - 1)
\(\cfrac { 2 }{ s+2 } =\cfrac { 3 }{ 2s+1 } =\cfrac { 7 }{ 3(2s-1) } \)
\(\cfrac { 2 }{ s+2 } =\cfrac { 3 }{ 2s+1 } \)
2(2s + 1 ) = 3 (s + 2)
4s + 2 = 3s + 6
4s - 3s = 6- 2
s = 4

5x - y = 5 -------------(1)
3x +y = 11 ----------(2)
(1) \(\Rightarrow\) y = 5x-5
(2) \(\Rightarrow\) 3x + 5x - 5 = 11
8x = 11+5
8x = 16
\(x=\cfrac { 16 }{ 8 } =2\)
(1) \(\Rightarrow\) 5 (2) - y = 5 \(\Rightarrow\) 10 - 5 =y \(\Rightarrow\) y = 5
\(\therefore\) Solution is x = 2, y = 5

x⁴ - 9x² = x² (x² - 9) = x² (x² - 32) = x² (x - 3)(x + 3)

25m^-2 - 16n² = (5m)² - (4n)²             [ \(\because \) a²-b² = (a - b)(a + b)]
= (5m - 4n) (5m+ 4n)

Let p (x) 2x³ - x² - 12x - 9
Sum of the co-efficients = 2 - 1- 12- 9 = -20 \(\neq \) 0
Hence x-1 is not a factor
Sum of co-efficients of even powers with constant = -1 - 9 = -10
Sum of co-efficients of odd powers = 2 - 12= -10
Hence x + 1 is a factor of x.
Now we use synthetic division to find the other factors.

Then p (x) = (x + 1)(2x² - 3x - 9)
Now 2x² - 3x - 9 = 2x² - 6x + 3x - 9 = 2x (x - 3) + 3 (x - 3)
= (x - 3)(2x + 3)
Hence 2x³ - x² - 12x - 9 (x + 1) (x - 3) (2x + 3)

P (x) = 5x³ + 7x² + 3x + 2
d(x) = 3x + 2
Standard form of p (x) = 5x³ + 7x² + 3x + 2
and d (x) = 3x + 2

5x³ + 7x² + 3x + 2 = \(\left( x+\cfrac { 2 }{ 3 } \right) \left( 5{ x }^{ 2 }-{ \cfrac { 11 }{ 3 } x+\cfrac { 5 }{ 9 } } \right) +\cfrac { 44 }{ 27 } =\left( \cfrac { 3x+2 }{ 3 } \right) \left( 5x2-\cfrac { 11 }{ 3 } x+\cfrac { 5 }{ 9 } \right) +\cfrac { 44 }{ 27 } \)
= \(\left( \cfrac { 3x+2 }{ 3 } \right) 3\left( \cfrac { 5 }{ 3 } { x }^{ 2 }-\cfrac { 11 }{ 9 } x+\cfrac { 5 }{ 27 } \right) +\cfrac { 44 }{ 27 } \)
= \(\left( 3x+2 \right) \left( \cfrac { 5 }{ 3 } { x }^{ 2 }-\cfrac { 11 }{ 9 } x+\cfrac { 5 }{ 27 } \right) +\cfrac { 44 }{ 27 } \)
Hence the quotient \(\cfrac { 5 }{ 3 } { x }^{ 2 }-\cfrac { 11 }{ 9 } x+\cfrac { 5 }{ 21 } \) and remainder  IS \(\cfrac { 44 }{ 27 } \)

P (x) = 4x³ + 6x² + 7x + 2
d(x) = x - 2
Standard form of p (x) 4x³ + 6x² + 7x + 2 and
d(x) x - 2

4x² + 6x² + 7x + 2 = (x - 2)(4x³ + 14x + 35) + 72
Hence the quotient is 4x³ + 14x + 35 and remainder is 72

p(x) 5x³ - 9x² + 10x + 2
d (x) = x + 2
Standard form ofp (x) 5x² - 9x² + 10x + 2 and
d (x) = x + 2

5x³ - 9x² + 10x + 2 = (x + 2) (5x² - 19x + 48) - 94
Hence the quotient is 5x² - 19x + 48 and remainder is - 94