Problem Study (Polynomial)

    When a polynomial f(x) is divided by (x – 1) and (x + 5),
    the remainders are -6 and 6 respectively. Let r(x be the
    remainder when f(x) is divided by x² + 4x – 5. Find the  
    value of r(-2).   

Solution 

    By the problem,

    f(x) = p(x) (x – 1) – 6 …………(1)

    f(x) = q(x) (x + 5) + 6 …………(2)

    f(x) = Q(x) (x² + 4x – 5) + r(x) …………(3) 

    (1) × (x + 5) ⇒ (x + 5) f(x) = p(x) (x – 1) (x + 5) – 6x – 30   

    (2) × (x – 1)  ⇒  (x – 1) f(x) = q(x) (x – 1) (x + 5) + 6x – 6 

    Subtracting the two equations,
    6 f(x) = [p(x) – q(x)] (x – 1) (x + 5) – 12x – 24
    Hence ,
    f(x) =  p ( x ) – q ( x ) 6 (x – 1) (x + 5) – 2x – 4 ——-(4)
    Comparing equations (3) and (4), we have
    r(x) = – 2x – 4
    Therefore, r (-2) = -2 (-2) – 4 = 0.

    Credit : Sayar U Pyi Kyaw

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

Factor Theorem အသံုးခ် ပုစာၦမ်ား

Example 2

Given that x² + x – 6 is a factor of 2x⁴ + x³ – ax + bx + a + b – 1, find the value of a and b.

x² + x – 6 = (x + 3)(x + 2)

Let f(x) = 2x⁴ + x³ – ax² + bx + a + b – 1

f(-3) = 2(-3)⁴ + (-3)³ – a(-3)² – 3b + a + b – 1 = 0

134 – 8a – 2b = 0

4a + b = 67 ——–(1)

f(2) = 2(2)⁴ + 2³ – a(2)² + 2b + a + b – 1 = 0

39 – 3a + 3b = 0

a – b = 13 ——–(2)

(1) + (2) : 5a = 80

a = 16

when a = 16, b = 3

Example 3

Use the factor theorem to find the value of k for which (a + 2b) where a does not = 0 and b does not = 0, is a factor of a⁴ + 32b⁴ + a³b(k + 3).

Solution

Let f(a) = a⁴ + 32b⁴ + a³b(k + 3)

f(-2b) = (-2b)⁴ + 32b⁴ + (-2b)³b(k + 3) = 0

48b⁴ – 8b⁴(k+ 3) = 0

8b⁴[6 – (k + 3)] = 0

8b⁴(3 – k) = 0

Since b does not = 0, 3 – k = 0

k = 3

Example 4

Determine the value of k for which x + 2 is a factor of (x + 1)⁷ + (2x + k)³.

Solution

Let f(x) = (x + 1) + (2x + k)

f(-2) = (-2 + 1) + (-4 + k) = 0

(k – 4) – 1 = 0

(k – 4) = 1

k – 4 = 1

k = 5

Example 5

Given that f(x) = x³ – x² – ax – b, where a and b are constants, has the factor x – 3 but has a remainder of 13x – 11 when divided by x + 4. Calculate the values of a and b.

Hence (i) factorise f(x) completely. (ii) Solve the equation 27x³ – 3ax = ax² + b.

Solution

f(3) = 0

3³ – 3² – 3a – b = 0

3a + b = 18 ——–(1)

f(-4) = (-4)³ – (-4)² – a(-4) – b = 13(-4) – 11

4a – b = 17 ——–(2)

sub (1) into (2): a = 5, b = 3

(i) Factorising.

f(x) = (x – 3)(x² + 2x + 1)

= (x – 3)(x + 1)²

(ii) 27x³ – 9x² – 3ax – b = 0

(3x)³ – (3x)² – a(3x) – b = 0

Let y = 3x

y³ – y² – ay – b = 0

(y – 3)(y + 1)² = 0 [from (1)]

y = 3, y = -1

x = 1, x = -1/3

Example 6

Show that the expression x³ + (k-2) x² + (k-7)x – 4 has a factor x+1 for all values of k .If the expression also has a factor x+2, find the value of k and the third factor.

Solution

Let f(x) = x³ + (k-2) x² + (k-7)x – 4

f(-1) =(-1)³ + (k-2)(-1)² + (k-7)(-1) – 4 = -1+ k – 2 – k + 7 – 4 = 0

Therefore (x+1) is a factor of f(x).

x + 2 is a factor of f(x). ——-(given)

Therefore f(-2)=0

(-2)³ + (k-2)(-2)² + (k-7)(-2) – 4 = 0

-8 + 4k -8 – 2k +14 – 4 = 0

2k = 6

k = 3

Therefore f(x) = x³ + x² – 4x – 4

Therefore (x + 1) (x + 2) = x² + 3x + 2 is also a factor of f(x).

Therefore the third factor of f(x) is x – 2.

Example 7

Given that kx³ + 2x² + 2x + 3 is a factor of kx³ – 2x + 9, have a common factor, what are the possible values of k?

Solution

Let f(x) = >kx³ + 2x² + 2x + 3 and g(x) = kx³ – 2x + 9.

Let (x – c) be a common factor of both f(x) and g(x).Therefore …

f(c) = 0

kc³ + 2c² + 2c + 3 = 0 ———(1)

g(c) = 0

kc³ – </span2c + 9= 0 ———(2)

Subtracting equation(2) from (1),we have

2c² + 4c – 6 = 0

c² + 2c – 3 = 0

(c + 3) (c – 1) =0

c = -3 or c = 1

Substituting c = -3 or c = 1 in equation (2), we get

k = 5/9 when c = -3 and

k = -7 when c = 1

The Factor Theorem

6 = 3 × 2

6÷2 => remainder = 0

6÷3 => remainder = 0

အၾကြင္း 0 ရေအာင္စားႏိုင္ေသာ စားကိန္းကို တည္ကိန္း၏ factor ဟုေခၚသည္။

The Factor Theorem:

Let f(x) be a polynomial. Then (x-k) is a factor of f(x) if and only if f(k) = 0.

Example 1

Find what value p must have in order that x – p may be a factor of
4x³ – (3p + 2)x² – (p² – 1)x + 3.

Let f(x) = 4x³ – (3p + 2)x² – (p² – 1)x + 3.
x – p is a factor of f(x) only if
f(p) = 0
4p³ – (3p + 2)p² – (p² – 1)p + 3 = 0
2p² – p – 3 = 0
(p + 1) (2p – 3) = 0
p = -1 or p = 3/2