Exercise 2.3

1. Find the remainder when x3 + 3x2 + 3x + 1 is divided by the following linear polynomials:

i. x + 1

Solution:

By the Remainder Theorem, the remainder is p(-1) = −13 + 3−12 + 3(-1) + 1

= + - + =

ii. x - 12

Solution:

The remainder is p12 = 123 + 3122 + 312 + 1

= 1814 + 3432 + 3212 + 1

= 1+6+12+88 = 278252

iii. x

Solution:

The remainder is p(0) = 02 + 302 + 3(0) + 1 =

iv. x + π

Solution:

The remainder is p(-π) = −π3 + 3−π2 + 3−π + 1

= −π3π3 + 3π2-3π2 - 3π2π + 1

v. 5 + 2x

Solution:

We can rewrite the divisor as 2x + 5 = 2(x + 52). So, we evaluate at x = −5232.

The remainder is p−52 = −523 + 3−522 + 3−52 + 1

= −1258258 + 754504 - 15252 + 1

= −125+150−60+88 = −278278

2. Find the remainder when x3 - px2 + 6x - p is divided by x - p.

Solution:

By the Remainder Theorem, the remainder is p(p) = p3 - pp2 + 6(p) - p

= p3p4 - −p3−p2 + - =

3. Find the remainder when 2x2 - 3x + 5 is divided by 2x - 3. Does it exactly divide the polynomial? State reason.

Solution:

We can rewrite the divisor as 2(x - ).

So, we evaluate at x = 32−32.

The remainder is 2322 - 332 + 5

= 2(9492) - 92−92 + 5 = 92−92 - 92−92 + 5 = .

No, it does not exactly divide the polynomial because the remainder is not zero.

4. Find the remainder when 9x3 - 3x2 + x - 5 is divided by 23x - 1.

Solution:

We can rewrite the divisor as 23(x - ).

So, we evaluate at x = 32−32.

The remainder is 9323 - 3322 + (3/2) - 5

= 9(278274) - 3(9492) + 32−32 - 5

= 24382434 - 274272 + 32−32 - 5

= 243−54+12−408 = 16181614

5. If the polynomials 2x3 + ax2 + 3x - 5 and x3 + x2 - 4x + a leave the same remainder when divided by x - 2, find the value of a.

Solution:

Let p(x) = 2x3 + ax2 + 3x - 5 and q(x) = x3 + x2 - 4x + a.

Remainder when p(x) is divided by x - 2 is p(2) = 223 + a22 + 3(2) - 5

= + + - 5

= +

Remainder when q(x) is divided by x - 2 is q(2) = 23 + 22 - 4(2) + a = + - + a = + a

Since the remainders are the same: 17 + 4a = 4 + a => a = => a = −133133

6. If the polynomials x3 + ax2 + 5 and x3 - 2x2 + a are divided by (x + 2) leave the same remainder, find the value of a.

Solution:

Let p(x) = x3 + ax2 + 5 and q(x) = x3 - 2x2 + a.

Remainder when p(x) is divided by x + 2 is p(-2) = −23 + a−22 + 5

= + + 5 = - .

Remainder when q(x) is divided by x + 2 is q(-2) = −23 - 2−22 + a

= - + a = a - .

Since the remainders are the same:

4a - 3 = a - 16

a =

a = −133133

7. Find the remainder when f(x) = x4 - 3x2 + 4 is divided by g(x) = x - 2 and verify the result by actual division.

Solution:

By the Remainder Theorem, the remainder is f(2) = 24 - 322 + 4

= - + 4 = .

Actual Division: The remainder is .

8. Find the remainder when p(x) = x3 - 6x2 + 14x - 3 is divided by g(x) = 1 - 2x and verify the result by long division.

Solution:

We can write g(x) as -2(x - ).

We evaluate p(1/2) = 123 - 6122 + 1412 - 3

= 18−18 - 64−64 + - 3

= 1−12+56−248 =

Long Division:

The remainder is 218−218.

9. When a polynomial 2x3 + 3x2 + ax + b is divided by (x - 2) leaves remainder 2, and (x + 2) leaves remainder -2. Find a and b.

Solution:

Let f(x) = 2x3 + 3x2 + ax + b.

When f(x) is divided by (x - 2), the remainder is .

So, f(2) = :

f(2) = 223 + 322 + a(2) + b = 2

+ + + b = 2

+ + b = 2

2a + b = -26 --- (1)

When f(x) is divided by (x + 2), the remainder is .

So, f(-2) = :

f(-2) = 2−23 + 3−22 + a(-2) + b = -2

+ - + b = -2

- + b = -2

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

Subtract equation (2) from equation (1):

(2a + b) - (-2a + b) = -26 - 2

a =

a =

Substitute a = -7 into equation (2):

-2(-7) + b = 2

+ = 2

b =

Therefore, a = -7 and b = -12.