WHAT SHOULD BE ADDED TO THE POLYNOMIAL SO IT IS COMPLETELY DIVISIBLE

Problem 1 :

What must be subtracted from

p(x) = 8x⁴ + 14x³ - 2x² +7x -8

so that the resulting polynomial is exactly divisible by

g(x) = 4x² + 3x -2?

Solution :

If the polynomial is divisible by another polynomial, then the remainder will be 0.

Since the remainder is 14x - 10, to get 0 as a remainder we have to subtract 14x - 10 from the given polynomial.

Problem 2 :

What must be added to

f(x) = 4x⁴ + 2x³ -2x² + x - 1

so that the resulting polynomial is divisible by

g(x) = x² +2x - 3?

To make the remainder as 0, we need 61x - 65.

So, 61x - 65 to be added to the polynomial.

Problem 3 :

If the polynomial

x⁴ + 2x³ + 8x² + 12x + 18

is divided by another polynomial x²+5, the remainder comes

out to be px + q. Find the values of ' p' and ' q'?

Solution :

Remainder = 2x + 3

Comparing with the given remainder px + q

p = 2 and q = 3

Problem 4 :

Given P(x) = x³ + px² + qx - 2 when P(x) is divided by x - 3 and x + 2, the remainders are 4 and -16. Find the values of p and q.

P(x) = x³ + px² + qx - 2

x - 3 = 0

x = 3

Applying x = 3, we get the remainder 4

P(3) = 3³ + p(3)² + q(3) - 2

4 = 27 + 9p + 3q - 2

4 = 25 + 9p + 3q

9p + 3q = 4 - 25

9p + 3q = -21

Dividing by 3, we get

3p + q = -7 --------(1)

x + 2 = 0

x = -2

Applying x = -2, we get the remainder -16

P(-2) = (-2)³ + p(-2)² + q(-2) - 2

-16 = -8 + 4p - 2q - 2

-16 = -10 + 4p - 2q

-16 + 10 = 4p - 2q

4p - 2q = -6

Dividing by 2, we get

2p - q = -3 --------(2)

(1) + (2)

3p + 2p = -7 - 3

5p = -10

p = -2

Applying the value of p, we get

3(-2) + q = -7

-6 + q = -7

q = -7 + 6

q = -1

So, the values of p and q are -2 and -1 respectively.

Problem 5 :

When the polynomial x³ - 3x + 6 and x³ - 2x² + x + 4 are divided by x - m, the remainders are equal. Find the possible values of m.

Solution :

x - m = 0

x = m

P(x) = x³ - 3x + 6

Q(x) =  x³ - 2x² + x + 4

P(m) = m³ - 3m + 6 ----(1)

Q(m) =  m³ - 2m² + m + 4 -----(2)

(1) = (2)

m³ - 3m + 6 =  m³ - 2m² + m + 4

2m²- 3m - m + 6 - 4 = 0

2m²- 4m + 2 = 0

m²- 2m + 1 = 0

(m - 1)(m - 1) = 0

m = 1 and 1

So, the possible value of m is 1.

Problem 6 :

When a polynomial is divided by x + 1 the remainder is 5 and when it is divided by x- 4, the remainder is 15. Find the remainder when P(x) is divided by (x + 1)(x - 4)

Solution :

From the given information, it is clear that the required polynomial must be a quadratic.

x + 1 = 0 and x - 4 = 0

x = -1 and x = 4

P(-1) = 5 and P(4) = 15

Using division algorithm, 

P(x) = (x + 1)(x - 4) Q(x) + a x + b

When x = -1

5 = (-1 + 1)(-1 - 4) Q(-1) + a(-1) + b

5 = -a + b

-a + b = 5 -----(1)

When x = 4

15 = (4 + 1)(4 - 4) Q(4) + a(4) + b

15 = 4a + b

4a + b = 15 -----(2)

(1) - (2)

-a - 4a = 5 - 15

-5a = -10

a = 2

Applying the value of a in (1), we get

-2 + b = 5

b = 5 + 2

b = 7

The remainder is a x + b, then 2x + 7.

Problem 7 :

The polynomials are ax⁴ + 2x³ + bx + 3 leaves a remainder of 15x + 19 when divided by x² - x - 2. Find the value of a and b.

Solution :

Factoring x² - x - 2,

= (x - 2) (x + 1)

x + 1 = 0 and x - 2 = 0

x = -1 and x = 2

P(x) = ax⁴ + 2x³ + bx + 3

Applying x = -1, we get

P(-1) = a(-1)⁴ + 2(-1)³ + b(-1) + 3

15x + 9 = a - 2 - b + 3

15(-1) + 9 = a - b + 1

a - b + 1 = -15 + 9

a - b = -6 - 1

a - b = -7 --------(1)

Applying x = 2, we get

P(2) = a(2)⁴ + 2(2)³ + b(2) + 3

15(2) + 9 = 16a + 16 + 2b + 3

30 + 9 = 16a + 2b + 19

16 a + 2b = 39 - 19

16a + 2b = 20

8a + b = 10 ------------(2)

(1) + (2)

a + 8a = -7 + 10

9a = 3

a = 1/3

Applying the value of a, we get

(1/3) - b = -7

1/3 + 7 = b

b = (1 + 21)/3

b = 22/3

Problem 8 :

If x + 1 is a factor of P(x) = 4x³ + 14x² + kx - 3, determine the value of k.

Solution :

P(x) = 4x³ + 14x² + kx - 3,