from which book id you ask/

it is my Holiday Homework

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from which book id you ask/

it is my Holiday Homework

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Factor theorem says that if x - a is a factor of f(x) if and only if f(a) = 0...

F(x) = x^n + a^n

F(-a) = -(1)^n a^n + a^n = 0 for odd n... as -1)^n = -1..

hence x + a is a factor of F(x)

===================

F(x) = x⁴ - 4 / x⁴ = (x⁸ - 4) / x⁴

Let a = √√2 = (2)¹/⁴

then a² = √2 and a⁴ = 2 then a⁸ = 4

F(a) = (a⁸ - 4) / a⁴ = 0 => x - a is a factor of F(x)

F(x) = (x - a) (x⁷ + b x⁶ + c x⁵ + d x⁴ + e x³ + f x² + g x + 4/a) / x⁴

= [x⁸ + (b-a) x⁷ + (c-ab) x⁶ + (d-ac) x⁵ + (e-ad) x⁴ + (f - ae) x³

+ (g - af) x² + (4/a - ag) x - 4 ] / x⁴

= (x⁸ - 4) / x⁴

=> b = a

c = ab = a² = √2

d = ac = a³ = 2³/⁴

e = ad =a⁴ = 2

f = ae = a⁵ = 2⁵/⁴

g = a⁶ = 2³/²

4/a = ag = a⁷ => a⁸ = 4.. which i s true,,

F(x) = (x - a) (x⁷ + a x⁶ + √2 x⁵ + a³ x⁴ + 2 x³ + 2⁵/⁴ x² + 2³/² x + 4/a) / x⁴

= (x - a) (x³ + a x² + √2 x + 2³/⁴ + 2 x⁻¹ + 2⁵/⁴ /x⁻² + 2√2 x⁻³ + 4/a x⁻⁴)

Factorizing :

F(x) = (x⁸ - 4) / x⁴ = (x⁴ - 2) (1 + 2 x⁻⁴)

= (x² - √2) (x² + √2) (1 + 2 x⁻⁴)

= (x - a) (x + a) (x² + √2) (a + 2 x⁻⁴)

where a² = √2

=================================

x+2 is a factor of F(x) = (x + 1)⁷ + (2x + k)³

=> F(-2) = 0

=> -1⁷ + (k-4)³ = 0

=> k-4 = 1

=> k = 5

===================

divisor x^2 + x - 12 = (x + 4) (x - 3)

Let x^3 - 6 x^2 - 15 x + 80

= (x + 4) (x - 3) * (x + a) + [ b (x + 4) + c (x - 3) ]

Thus the remainder after division by divisor is (b + c) x + 4 b - 3 c

Let F(x) = x^3 - 6 x^2 - 15 x + 80 - (b + c) x - 4 b + 3 c

F(-4) = 0 as it should be divisible by x + 4

=> -64 - 96 + 60 + 4 (b+c) - 4 b + 3c = 0

=> c = 100/7

F(3) = 0 as F(x) is divisible by x - 3

=> 27 - 54 - 45 + 80 - 3(b+c) - 4 b + 3 c = 0

=> b = 8/7

So 4 * [ 27 x - 67] /7 must be subtracted from the iven polynomial for the divisor to exatly divide it....

F(x) = x^n + a^n

F(-a) = -(1)^n a^n + a^n = 0 for odd n... as -1)^n = -1..

hence x + a is a factor of F(x)

===================

F(x) = x⁴ - 4 / x⁴ = (x⁸ - 4) / x⁴

Let a = √√2 = (2)¹/⁴

then a² = √2 and a⁴ = 2 then a⁸ = 4

F(a) = (a⁸ - 4) / a⁴ = 0 => x - a is a factor of F(x)

F(x) = (x - a) (x⁷ + b x⁶ + c x⁵ + d x⁴ + e x³ + f x² + g x + 4/a) / x⁴

= [x⁸ + (b-a) x⁷ + (c-ab) x⁶ + (d-ac) x⁵ + (e-ad) x⁴ + (f - ae) x³

+ (g - af) x² + (4/a - ag) x - 4 ] / x⁴

= (x⁸ - 4) / x⁴

=> b = a

c = ab = a² = √2

d = ac = a³ = 2³/⁴

e = ad =a⁴ = 2

f = ae = a⁵ = 2⁵/⁴

g = a⁶ = 2³/²

4/a = ag = a⁷ => a⁸ = 4.. which i s true,,

F(x) = (x - a) (x⁷ + a x⁶ + √2 x⁵ + a³ x⁴ + 2 x³ + 2⁵/⁴ x² + 2³/² x + 4/a) / x⁴

= (x - a) (x³ + a x² + √2 x + 2³/⁴ + 2 x⁻¹ + 2⁵/⁴ /x⁻² + 2√2 x⁻³ + 4/a x⁻⁴)

Factorizing :

F(x) = (x⁸ - 4) / x⁴ = (x⁴ - 2) (1 + 2 x⁻⁴)

= (x² - √2) (x² + √2) (1 + 2 x⁻⁴)

= (x - a) (x + a) (x² + √2) (a + 2 x⁻⁴)

where a² = √2

=================================

x+2 is a factor of F(x) = (x + 1)⁷ + (2x + k)³

=> F(-2) = 0

=> -1⁷ + (k-4)³ = 0

=> k-4 = 1

=> k = 5

===================

divisor x^2 + x - 12 = (x + 4) (x - 3)

Let x^3 - 6 x^2 - 15 x + 80

= (x + 4) (x - 3) * (x + a) + [ b (x + 4) + c (x - 3) ]

Thus the remainder after division by divisor is (b + c) x + 4 b - 3 c

Let F(x) = x^3 - 6 x^2 - 15 x + 80 - (b + c) x - 4 b + 3 c

F(-4) = 0 as it should be divisible by x + 4

=> -64 - 96 + 60 + 4 (b+c) - 4 b + 3c = 0

=> c = 100/7

F(3) = 0 as F(x) is divisible by x - 3

=> 27 - 54 - 45 + 80 - 3(b+c) - 4 b + 3 c = 0

=> b = 8/7

So 4 * [ 27 x - 67] /7 must be subtracted from the iven polynomial for the divisor to exatly divide it....