Letp (x) x⁵ - 45x⁴ + 36x³ + 45x² - 36x- 1
Sum of co-efficients = 1 - 45 + 36 + 45 - 36 - 1 = 0
Thus x -1 is a factor of p (x)

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

Let f(x) = ax³ + 4x² + 3x –4 and g(x) = x³ – 4x + a,When f(x) is divided by (x–3), the remainder is f(3).
Now f(3) =a(3)³ + 4(3)² + 3(3) - 4
= 27a + 36 + 9 – 4
f(3) = 27a + 41 (1)
When g(x) is divided by (x–3), the remainder is g(3).
Now g(3) = 3³ – 4(3) + a
= 27 – 12+ a
= 15 + a (2)
Since the remainders are same, (1) = (2)
Given that, f(3) = g(3)
That is 27a + 41 = 15 + a
27a – a = 15 – 41
26a = –26
\(a={-26\over 26}=-1\)
Substituting a = –1,in f(3), we get
f(3) =
= – 27 + 41
f(3) = 14
∴ The remainder is 14.

p(x₁) = 2x³ + ax² + 4x - 12
When it is divided by x - 3,
p(3) = 2(3)³ + a(3)² + 4(3) - 12
= 54 + 9a + 12 - 12
= 54 + 9a .....(R₁)
p(x2) = x³ + x² - 2x + a
When it is divided by x - 3,
p(3) = 3³ + 3² - 2(3) + a
= 27 + 9 -6 +a
= 30 + a  ......(R₂)
The given remainders are same (R₁ = R₂)
\(\therefore\) 54 + 9a = 30 + a
9a - a = 30 - 54
8a = -24
\(\therefore\)  a = - 24/8 = - 3
Consider R₂,
Remainder = 30 - 3
= 27

Length of the rectangle = 3x + 2 units
Breadth of the rectangle = 3x - 2 units
Area of the rectangle = (3x + 2 ) (3x - 2)
= 9x² - 6x + 6x - 4
= 9x²- 4
When x = 20
Area of the rectangle = 9(20)2 - 4
= 9(400) - 4
= 3600 - 4 = 3596 sq.units.

The cost of a chocolate Rs (x +y)
Number of chocolates bought by Amir x + y
Total amount paid by him Rs (x +y) (x +y)
= x²+ xy +xy +y²
= x²+ 2xy+ y²
When x = 10 and y = 5
The total amount paid by him (10)² + 2(10)(5) + (5)²
= 100 + 100 + 25 = Rs 225