Polynomials Test - 1

Let f(x) = x6 - y⁶

g(x) = x + y

x⁶ - y⁶ = (x +y)(x - y)(x⁴ + y⁴ + x²y²)

It is clear that g(x) is a factor of f(x).
So, the remainder is 0.

We have:

p(x) = x³ - ax² + 3x – a

By remainder theorem, we know that p(x) is divided by (x – a), and the remainder is p(a).

Now, p(a) = a³ - aa² + 3a – a

Or, p(a) = a³ - a³ + 2a

Or, p(a) = 2a

x¹⁸ – y¹⁸

= (x⁹)² – (y⁹)²

= (x⁹ – y⁹) (x⁹ + y⁹)

= [(x³)³ – (y³)³] [(x³)³ + (y³)³]

= [(x³ – y³) (x⁶ + x³y³ + y⁶)] [(x³ + y³) (x⁶ - x³y³ + y⁶)]

= [(x – y) (x² + xy + y²) (x⁶ + x³y³ + y⁶)] [(x + y) (x² – xy + y²) (x⁶ - x³y³ + y⁶)]

Let f(x) = (2x³ + 5x² + ax + b)

Here, (x + 3) and (x + 4) are the factors of f(x).
Therefore, by factor theorem:
f(-3) = 0 and f(-4) = 0

(2(-3)³ + 5(-3)² + a(-3) + b) = 0

And (2(-4)³ + 5(-4)² + a(-4) + b) = 0

⇒ (-54 + 45 - 3a + b) = 0 and (-128 + 80 – 4a + b) = 0
(-9 – 3a + b) = 0
b – 3a = 9 ... (i)

and
(-48 + b – 4a) = 0
b - 4a = 48 .... (ii)

Solving further:
b – 3a – b + 4a = 9 – 48
a = -39
Putting (a) in (i), we get:
b – 3(-39) = 9
b + 117 = 9 – 117 = -108

We have:

(x - m)³ + (x - n)³ + (x - o)³ = 3 (x - m) (x - n) (x - o)

Here, x - m = a
x - n = b
x - o = c

Using the identity a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca):

If a + b + c = 0, then a³ + b³ + c³ = 3abc. ,

(x - m) + (x - n) + (x – o) = 0
3x - m - n - o = 0