Given that kx^3+2x^2+2x+3 and kx^3-2x+9 have a common factor, what are the possible values of k?
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Answers
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 c^3, 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) / 1^3
⇒ k = (2 - 9)/1
⇒ k = -7
Therefore:-
There are two possible values of k and they are 5/9, -7 (Answer)
Explanation:
If kx^3 + 2x^2 + 2x + 3 and kx^3 - 2x + 9 have an elementary aspect, then their huge difference also has sme elementary aspect. => 2x^2 + 4x - 6 => x^2 + 2x - 3 has similar aspect. x^2 + 2x - 3 = (x+3)(x-a million). positioned x = a million, -3 in any given eqn and equate fee = 0 to get values of ok. kx^3 + 2x^2 + 2x + 3 = ok + 2 + 2 + 3 = 0 => ok = -7 for x = a million. and -27k + 18 -6 + 3 = 0 => ok = 5/9 for x = -3. kx^3 - 2x + 9 = ok - 2 + 9 = 0. => ok = -7 for x = a million, -27k + 6 + 9 = 0 => ok = 5/9 for ok = -3. hence we get 2 values of ok = -7, 5/9.