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