Question

Find numbers a and k so that x-2 is a factor of f(x)=x^4-2ax^3+ax^2- x+k and f(-1)=3

Answer 1

Answer:

$a = \frac{16}{15} = 1\frac{1}{15}$

$k = -5\frac{1}{5}$

Step-by-step explanation:

$f(x) = x^4 - 2ax^3 + ax^2 - x + k$

$x - 2$ is a factor:

∴ $(2)^4 - 2a(2)^3 + a(2)^2 - (2) + k = 0$

$=>16 - 16a + 4a + 2 + k = 0$

$=> -12a + k = - 18$

$=> 12a - k = 18$

∴ $k = 12a - 18$ . . . . . . . . . . . . . . (i)

Also, given that, $f(-1) = 3$

∴ $(-1)^4 - 2a(-1)^3 + a(-1)^2 - (-1) + k = 0$

$=> 1 + 2a + a + 1 + k = 0$

$=> 3a + 12a - 18 = -2$  [using equation (i)]

$=> 15a = -2 + 18 = 16$

∴ $a = \frac{16}{15} = 1\frac{1}{15}$

∴ $k = 12a - 18 = 12. \frac{16}{15} - 18 = \frac{64}{5} - 18 = \frac{64 - 90}{5} = -\frac{26}{5} = -5\frac{1}{5}$