Question

Given that x + 2 and x + 4 are the factors of 3x3 + ax2 – 6x – b. determine the values of a and b

Answer 1

Answer:

Using factor theorem,

If x + 2 and x + 4 are factors of 3x³ + ax² - 6x - b, then value of this polynomial must be 0, for x = - 2 or x = - 4.

For x = - 2

→ 3(-2)³ + a(-2)² - 6(-2) - b = 0

→ 3(-8) + a(4) + 12 - b = 0

→ - 24 + 4a + 12 - b = 0

→ 4a - b - 12 = 0

→ b = 4a - 12 --- (1)

For x = - 4

→ 3(-4)³ + a(-4)² - 6(-4) - b = 0

→ 3(-64) + a(16) + 24 - b = 0

→ - 192 + 16a + 24 - b = 0

→ 16a - 168 = b = 4a - 12

→ 16a - 168 = 4a - 12

→ 12a = 156

→ a = 13

Hence,

b = 4a - 12

= 4(13) - 12

= 52 - 12

= 40

Hence a is 13 and b is 40

Answer 2

Solution:-

Using factor theorem,

If x+2 and x+4 are the factors of 3x^3+ ax^2 - 6x - b then the value must be 0 .For x= -2 and x= -4

NoW,

For x = -2,

➡ 3(-2) ^3 + a( 2) ^2 - 6(-2) -b =0

➡ 3(-8) +a(4) -12-b =0

➡ -24 + 4a + 12- b =0

➡ 4a-b-12 =0

➡ b= 4a - 12 -------(1)

For x= - 4,

➡3(-4)^3 + a(4)^2 -6(4)-b =0

➡ 3(-64) + a(16) + 24 - b =0

➡192+16a +24 - b =0

➡16 - 168 = b = 4a - 12

➡ 12a = 156

➡ a= 13

Hence,

b = 4a -12

➡ 4(13) - 12

➡ 52 -12

➡ 40

Hence a = 13 and b = 40.