Question

Given that kx^3+2x^2+2x+3 and kx^3-2x+9 have a common factor, what are the possible values of k?

please solve above problem
​

Answer 1

Required Answer:

GiveN:-

• kx³ + 2x² + 2x + 3 and kx³ - 2x + 9 have a common factor$.$

To FinD:-

• Possible values of k?

Solution:-

Let the first function f(x) and the second function g(x).

• f(x) = kx³ + 2x² + 2x + 3
• g(x) = kx³ - 2x + 9

In order for f(x) and g(x) to have a common factor, there must exist some value a such that:

• f(a) = 0
• g(a) = 0

This can be written as,

• f(a) = ka³ + 2a² + 2a + 3
• g(a) = ka³ - 2a + 9

Equating to 0,

First Equation:

⇒ ka³ + 2a² + 2a + 3 = 0

⇒ ka³ = -2a² - 2a - 3

⇒ k = (-2a² - 2a - 3)/a³

Second Equation:

⇒ ka³ - 2a + 9 = 0

⇒ ka³ = 2a - 9

⇒ k = (2a - 9)/a³

Since these are both values of k, Let's equate them to each other.

• (-2a² - 2a - 3)/a³ = (2a - 9)/a³

Now,

Multiply both sides by a³, to get

⇒ -2a² - 2a - 3 = 2a - 9

Let's solve this:

⇒ -2a² - 4a + 6 = 0

⇒ a² + 2a - 3 = 0

⇒ (a + 3)(a - 1) = 0

⇒ a = -3, 1

We found the possible values for a, which means we can now find the possible values for k. Since

⇒ k = (2a - 9)/a³

Then, if a = -3,

⇒ k = (2(-3) - 9)/(-3)³

⇒ k = (-6 - 9)/(-27)

⇒ k = (-15)/(-27)

⇒ k = 15/27

⇒ k = 5/9

If a = 1, then

⇒ k = (2(1) - 9) / 1³

⇒ k = (2 - 9)/1

⇒ k = -7

Therefore:-

• There are two possible values of k and they are 5/9, -7 (Answer)

Answer 2

Question:-

➡ Given that kx³ + 2x² + 2x + 3 and kx³ - 2x + 9 have a common factor. What are the possible values of k?

Solution:-

Let the first function be f(x) and second function be g(x) .

f(x) = kx³ + 2x² + 2x +3 and

g(x) = kx³ - 2x + 9

There must exist a common value such that,

f(a) = 0 and

g(a) = 0

f(a) = k × a³ + 2 × a² + 2 × a + 3

f(a) = ka³ + 2a² + 2a + 3

Similarly,

g(a) = ka³ - 2a + 9

For the first equation,

➡ ka³ + 2a² + 2a + 3 = 0

➡ ka³ = -2a² - 2a - 3

➡ k = (-2a² - 2a - 3)/a³ ...... (i)

For the second equation,

➡ ka³ - 2a + 9 = 0

➡ ka³ = 3a - 9

➡ k = (2a - 9)/a³ ......(ii)

From (i) and (ii), we get,

➡ (-2a² - 2a - 3)/a³ = (2a - 9)/a³

Cancel a³ from both sides..

➡ -2a² - 2a - 3 = 2a - 9

➡ -2a² - 4a + 6 = 0

➡ 2a² + 4a - 6 = 0

Dividing both sides by 2, we get,

➡ a² + 2a - 3 = 0

Now, let's solve the given quadratic equation.

➡ a² + 2a - 3 = 0

➡ a² - a + 3a - 3 = 0

➡ a(a - 1) + 3(a - 1) = 0

➡ (a + 3)(a - 1) = 0

By zero product rule,

Either a + 3 = 0 or a - 1 = 0

➡ a = -3, 1

We have found the possible values of a.

When a = -3,

k = (2a - 9)/a³

= (2 × -3 - 9)/(-3)³

= (-6-9)/-27

= -15/-27

= 15/27

= 5/9

Again, when a = 1

k = (2a - 9)/a³

= (2 × 1 - 9)/1³

= 2 - 9

= -7

Hence, the possible values of k are 5/9 and -7

Answer:-

➡ The possible values of k are 5/9 and -7 (Answer)