Find k such that f(x) = x^4-kx^3+kx^2+1 has the factor x+2
(x^4-kx^3+kx^2+1)/(x+2) = x^3 - (k+2)x^2 + (3k+4)x - (6k+8) with remainder 12k+17
So, you want 12k+17 = 0
or
by the Factor Theorem:
f(-2) = (-2)^4 - k(-2)^3 + k(-2)^2 + 1 = 0
16 + 8k + 4k + 1 = 0
12k + 17 = 0 , <------- oobleck's equation
To find the value of k such that \(f(x) = x^4 - kx^3 + kx^2 + 1\) has the factor \(x + 2\), we need to determine if \(x + 2\) is a root of \(f(x)\) and then find the corresponding value of k.
1. To check if \(x + 2\) is a root of \(f(x)\), substitute \(x = -2\) into \(f(x)\):
\(f(-2) = (-2)^4 - k(-2)^3 + k(-2)^2 + 1\)
Simplifying further:
\(f(-2) = 16 + 8k + 4k + 1\)
\(f(-2) = 17 + 12k\)
2. If \(x + 2\) is a factor of \(f(x)\), then \(f(-2) = 0\). Therefore, we need to solve the equation:
\(17 + 12k = 0\)
3. To isolate the value of \(k\), we subtract 17 from both sides of the equation:
\(12k = -17\)
4. Finally, divide both sides of the equation by 12 to solve for \(k\):
\(k = -\frac{17}{12}\)
Therefore, the value of \(k\) such that \(f(x) = x^4 - kx^3 + kx^2 + 1\) has the factor \(x + 2\) is \(k = -\frac{17}{12}\).
To find the value of k such that the polynomial f(x) = x^4 - kx^3 + kx^2 + 1 has the factor x + 2, we can use the factor theorem. According to the factor theorem, if x + c is a factor of a polynomial f(x), then f(-c) = 0.
In this case, we have a factor x + 2, so we substitute -2 for x in the polynomial f(x) and set it equal to 0:
f(-2) = (-2)^4 - k(-2)^3 + k(-2)^2 + 1 = 16 + 8k + 4k + 1 = 0
Simplifying the equation, we get:
16 + 12k + 1 = 0
28k + 17 = 0
k = -17/28
Therefore, the value of k that makes f(x) have the factor x + 2 is k = -17/28.