Chapter 8 - Remainder And Factor Theorems Exercise Ex. 8(B)

Question 1

Using the Factor Theorem, show that:

(i) (x - 2) is a factor of x³ - 2x² - 9x + 18. Hence, factorise the expression x³ - 2x² - 9x + 18 completely.

(ii) (x + 5) is a factor of 2x³ + 5x² - 28x - 15. Hence, factorise the expression 2x³ + 5x² - 28x - 15 completely.

(iii) (3x + 2) is a factor of 3x³ + 2x² - 3x - 2. Hence, factorise the expression 3x³ + 2x² - 3x - 2 completely.

Solution 1

(i) Let f(x) = x³ - 2x² - 9x + 18

x - 2 = 0 x = 2

 

Remainder = f(2)

= (2)³ - 2(2)² - 9(2) + 18

= 8 - 8 - 18 + 18

= 0

Hence, (x - 2) is a factor of f(x).

 

Now, we have:

 x³ - 2x² - 9x + 18 = (x - 2) (x² - 9) = (x - 2) (x + 3) (x - 3)

 

(ii) Let f(x) = 2x³ + 5x² - 28x - 15

x + 5 = 0 x = -5

 

Remainder = f(-5)

= 2(-5)³ + 5(-5)² - 28(-5) - 15

= -250 + 125 + 140 - 15

= -265 + 265

= 0

Hence, (x + 5) is a factor of f(x).

 

Now, we have:

2x³ + 5x² - 28x - 15 = (x + 5) (2x² - 5x - 3)

= (x + 5) [2x² - 6x + x - 3]

= (x + 5) [2x(x - 3) + 1(x - 3)]

= (x + 5) (2x + 1) (x - 3)

 

(iii) Let f(x) = 3x³ + 2x² - 3x - 2

3x + 2 = 0 

 

Hence, (3x + 2) is a factor of f(x).

 

Now, we have:

Question 2

Using the Remainder Theorem, factorise each of the following completely.

(i) 3x³ + 2x² − 19x + 6

(ii) 2x³ + x² - 13x + 6

(iii) 3x³ + 2x² - 23x - 30

(iv) 4x³ + 7x² - 36x - 63

(v) x³ + x² - 4x - 4

Solution 2

(i)

 

(ii) Let f(x) = 2x³ + x² - 13x + 6

For x = 2,

f(x) = f(2) = 2(2)³ + (2)² - 13(2) + 6 = 16 + 4 - 26 + 6 = 0

Hence, (x - 2) is a factor of f(x).

 

 

(iii) f(x) = 3x³ + 2x² - 23x - 30

For x = -2,

f(x) = f(-2) = 3(-2)³ + 2(-2)² - 23(-2) - 30

= -24 + 8 + 46 - 30 = -54 + 54 = 0

Hence, (x + 2) is a factor of f(x).

 

 

(iv) f(x) = 4x³ + 7x² - 36x - 63

For x = 3,

f(x) = f(3) = 4(3)³ + 7(3)² - 36(3) - 63

= 108 + 63 - 108 - 63 = 0

Hence, (x + 3) is a factor of f(x).

 

 

(v) f(x) = x³ + x² - 4x - 4

For x = -1,

f(x) = f(-1) = (-1)³ + (-1)² - 4(-1) - 4

= -1 + 1 + 4 - 4 = 0

Hence, (x + 1) is a factor of f(x).

 

Question 3

Using the Remainder Theorem, factorise the expression 3x³ + 10x² + x - 6. Hence, solve the equation 3x³ + 10x² + x - 6 = 0.

Solution 3

Let f(x) = 3x³ + 10x² + x - 6

For x = -1,

f(x) = f(-1) = 3(-1)³ + 10(-1)² + (-1) - 6 = -3 + 10 - 1 - 6 = 0

Hence, (x + 1) is a factor of f(x).

Now, 3x³ + 10x² + x - 6 = 0

Question 4

Factorise the expression f (x) = 2x³ - 7x² - 3x + 18. Hence, find all possible values of x for which f(x) = 0.

Solution 4

f (x) = 2x³ - 7x² - 3x + 18

For x = 2,

f(x) = f(2) = 2(2)³ - 7(2)² - 3(2) + 18

= 16 - 28 - 6 + 18 = 0

Hence, (x - 2) is a factor of f(x).

Question 5

Given that x - 2 and x + 1 are factors of f(x) = x³ + 3x² + ax + b; calculate the values of a and b. Hence, find all the factors of f(x).

Solution 5

f(x) = x³ + 3x² + ax + b

Since, (x - 2) is a factor of f(x), f(2) = 0

 (2)³ + 3(2)² + a(2) + b = 0

 8 + 12 + 2a + b = 0

 2a + b + 20 = 0 ...(i)

 

Since, (x + 1) is a factor of f(x), f(-1) = 0

 (-1)³ + 3(-1)² + a(-1) + b = 0

 -1 + 3 - a + b = 0

 -a + b + 2 = 0 ...(ii)

 

Subtracting (ii) from (i), we get,

3a + 18 = 0

 a = -6

 

Substituting the value of a in (ii), we get,

b = a - 2 = -6 - 2 = -8

 

f(x) = x³ + 3x² - 6x - 8

Now, for x = -1,

f(x) = f(-1) = (-1)³ + 3(-1)² - 6(-1) - 8 = -1 + 3 + 6 - 8 = 0

Hence, (x + 1) is a factor of f(x).

 

Question 6

The expression 4x³ - bx² + x - c leaves remainders 0 and 30 when divided by x + 1 and 2x - 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.

Solution 6

Let f(x) = 4x³ - bx² + x - c

 

It is given that when f(x) is divided by (x + 1), the remainder is 0.

f(-1) = 0

4(-1)³ - b(-1)² + (-1) - c = 0

-4 - b - 1 - c = 0

b + c + 5 = 0 ...(i)

 

It is given that when f(x) is divided by (2x - 3), the remainder is 30.

 

Multiplying (i) by 4 and subtracting it from (ii), we get,

5b + 40 = 0

b = -8

 

Substituting the value of b in (i), we get,

c = -5 + 8 = 3

 

Therefore, f(x) = 4x³ + 8x² + x - 3

 

Now, for x = -1, we get,

f(x) = f(-1) = 4(-1)³ + 8(-1)² + (-1) - 3 = -4 + 8 - 1 - 3 = 0

 

Hence, (x + 1) is a factor of f(x).

 

Question 7

If x + a is a common factor of expressions f(x) = x² + px + q and g(x) = x² + mx + n; show that: 

Solution 7

f(x) = x² + px + q

It is given that (x + a) is a factor of f(x).

g(x) = x² + mx + n

It is given that (x + a) is a factor of g(x).

From (i) and (ii), we get,

pa - q = ma - n

n - q = a(m - p)

Question 8

The polynomials ax³ + 3x² - 3 and 2x³ - 5x + a, when divided by x - 4, leave the same remainder in each case. Find the value of a.

Solution 8

Let f(x) = ax³ + 3x² - 3

When f(x) is divided by (x - 4), remainder = f(4)

f(4) = a(4)³ + 3(4)² - 3 = 64a + 45

Let g(x) = 2x³ - 5x + a

When g(x) is divided by (x - 4), remainder = g(4)

g(4) = 2(4)³ - 5(4) + a = a + 108

It is given that f(4) = g(4)

64a + 45 = a + 108

63a = 63

a = 1

Question 9

Find the value of 'a', if (x - a) is a factor of x³ - ax² + x + 2.

Solution 9

Let f(x) = x³ - ax² + x + 2

It is given that (x - a) is a factor of f(x).

Remainder = f(a) = 0

a³ - a³ + a + 2 = 0

a + 2 = 0

a = -2

Question 10

Find the number that must be subtracted from the polynomial 3y³ + y² - 22y + 15, so that the resulting polynomial is completely divisible by y + 3.

Solution 10

Let the number to be subtracted from the given polynomial be k.

Let f(y) = 3y³ + y² - 22y + 15 - k

It is given that f(y) is divisible by (y + 3).

Remainder = f(-3) = 0

3(-3)³ + (-3)² - 22(-3) + 15 - k = 0

-81 + 9 + 66 + 15 - k = 0

9 - k = 0

k = 9