Miscellaneous Problems

Find the remainder when 3x³ – x² + 7x + 5 is divided by 3x + 2.

Given that 2x³ – x² – 2x + 3 = (Ax + B)(x – 1)(x + 2) + C(x – 1) + D, find the values of A, B, C and D. Hence or otherwise, deduce the remainder when

2x³ – x² – 2x + 3 is divided by x² + x – 2.

Solve the equation 2x³ + x² – 19x = 6, giving your answers to two decimal places where necessary.

The expression x³ + px² + qx + 6 has the same remainder when divided by

x + 1 and 2 – x. Given that the remainder when the expression is divided by x + 3 is -60, find the value of p and of q.

By remainder theorem,

remainder = f(-2/3)

= 3(-2/3)³ – (-2/3)² + 7(-2/3) + 5

= -1

Let x = 1

2 – 1 – 2 + 3 = D

D = 2

Let x = -2

-16 – 4 + 4 + 3 = -3C + 2

3C = 15

C = 5

Let x = 0

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

-2B = 6

B = -3

Sub any value other than 1, 2 & 0 into x

A = 2

2x³ – x² – 2x + 3 ≡ (Ax + B)(x – 1)(x + 2) + C(x – 1) + D

Since x² + x – 2 = (x – 1)(x + 2), (note the similarity?)

thus, remainder = C(x – 1) + D

= 5(x – 1) + 2

= 5x – 3

2x³ + x² – 19x = 6

let f(x) = 2x³ + x² – 19x – 6

(x – 3) f(3) = 0

(x – 3) is a factor of f(x)

let f(x) = (x – 3)(Ax² + Bx + C)

A = 2

C = 2

let x = 1

2 + 1 – 19 – 6 = -2(2 + B + 2)

-22 = -8 – 2B

B = 7

f(x) = (x – 3)(2x² + 7x + 2)

2x² + 7x + 2 = 0

x =

= -3.19 or -0.31 (3 s.f.)

x = -3.19 or -0.31 or 3

let f(x) = x³ + px² + qx + 6

by remainder theorem, f(-1) = f(2)

-1 + p – q + 6 = 8 + 4p + 2q + 6

p – q – 5 = 4p + 2q + 14

3p + 3q = 9 ——–(1)

by remainder theorem, f(-3) = -60

-27 + 9p – 3q + 6 = -60

3q = 39 + 9p ——–(2)

sub (2) into (1):

3p + 39 + 9p = -9

12p = -48

p = -4

q = 1