DIVIDING POLYNOMIALS WITH MISSING TERMS

Problem 1 :

If y = 1 is a zero of the polynomial

q(y) = 4y³ + ky² - y - 1

then find the value of k.

Solution :

Since y = 1 is one of the zero, it will satisfy the polynomial q(y).

q(y) = 4y³ + ky² - y - 1

q(1) = 4(1)³ + k(1)² - 1 - 1

0 = 4 + k - 2

0 = 2 + k

k = -2

So, the value of k is -2.

Problem 2 :

For what value of m is x³ - 2mx² + 16 is divisible by x + 2.

Solution :

Since the given polynomial x³ - 2mx² + 16 is divisible by x + 2, then p(-2) = 0

p(x) = x³ - 2mx² + 16

p(-2) = (-2)³ - 2m(-2)² + 16

0 = -8 - 8m + 16

0 = 8 - 8m

8m = 8

m = 8/8

m = 1

Problem 3 :

The polynomials x³ + 2x² - 5ax - 7 and x³ + ax² - 12x + 6 when divided by x + 1 and x - 2 respectively, leave remainders R₁ and R₂ respectively. Find the value of a in each of the following cases :

i) R₁ = R₂

ii) R₁ + R₂ = 0

iii) 2R₁ + R₂ = 0

Solution :

Let p(x) = x³ + 2x² - 5ax - 7 and q(x) = x³ + ax² - 12x + 6

x + 1 = 0, then x = -1

Finding remainder p(x) = x³ + 2x² - 5ax - 7 by x + 1

p(-1) = (-1)³ + 2(-1)² - 5a(-1) - 7

R₁ = -1 + 2 + 5a - 7

R₁ = -8 + 2 + 5a

R₁ = -6 + 5a

x - 2 = 0, then x = 2

Finding remainder q(x) = x³ + ax² - 12x + 6 by x + 2

q(2) = 2³ + a(2)² - 12(2) + 6

R₂ = 8 + 4a - 24 + 6

R₂ = 14 - 24 + 4a

R₂ = - 10 + 4a

i) When R₁ = R₂

-6 + 5a = -10 + 4a

5a - 4a = -10 + 6

a = -4

ii) R₁ + R₂ = 0

-6 + 5a - 10 + 4a = 0

9a - 16 = 0

a = 16/9

iii) 2R₁ + R₂ = 0

2(-6+ 5a) + (-10 + 4a) = 0

-12 + 10a - 10 + 4a = 0

14a - 22 = 0

a = 22/14

a = 11/7

Problem 4 :

When a polynomial

p(x) = x⁴ - 2x³ + 3x² - ax + b

is divisible by x - 1 and x + 1, the remainders are 5 and 19 respectively. Find the remainder when p(x) is divided by x - 2.

Solution :

p(x) = x⁴ - 2x³ + 3x² - ax + b 

When the polynomial p(x) is divided by x - 1, the remainder is 5.

x - 1 = 0

x = 1

p(1) = 1⁴ - 2(1)³ + 3(1)² - a(1) + b 

5 = 1 - 2 + 3 - a + b

5 = 2 - a + b

-a + b = 5 - 2

-a + b = 3 -----(1)

When the polynomial p(x) is divided by x + 1, the remainder is 19.

x + 1 = 0

x = -1

p(-1) = (-1)⁴ - 2(-1)³ + 3(-1)² - a(-1) + b 

19 = 1 + 2 + 3 + a + b

19 = 6 + a + b

a + b = 19 - 6

a + b = 13 ----(2)

(1) + (2)

2b = 3 + 13

2b = 16

b = 8

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

-a + 8 = 3

-a = 3 - 8

-a = -5

a = 5

So, the values of a and b are 5 and 8 respectively.

Applying the values of a and b, we get

p(x) = x⁴ - 2x³ + 3x² - 5x + 8

Dividing p(x) by x - 2, we get

x - 2 = 0

x = 2

p(2) = 2⁴ - 2(2)³ + 3(2)² - 5(2) + 8

p(2) = 16 - 16 + 12 - 10 + 8

p(2) = 10

Problem 5 :

If x - 3 and x - 1/3 are both factors of ax² + 5x + b, show that a = b.

Solution :

Let p(x) = ax² + 5x + b

When x - 3 = 0, then x = 3

p(3) = a(3)² + 5(3) + b

p(3) = 9a +15 + b

0 = 9a +15 + b

9a + b = -15  ------(1)

When x - 1/3 = 0, then x = 1/3

p(1/3) = a(1/3)² + 5(1/3) + b

p(1/3) = a/9 +5/3 + b

0 = a/9 +5/3 + b

a/9 + b = -5/3 ------(2)

(1) - (2)

9a - a/9 = -15 + (5/3)

a = -3/2

By applying the value of a in (1), we get

9(-3/2) + b = -15

-27/2 + b = -15

b = -15 - (-27/2)

b = -3/2