Question

plzz solved this question fast

Answer 1

Hey!!!

$let \: f(x) \: be \: {x}^{3} + \: k {x}^{2} \: - 2x + k + 4$
And
$g(x) \: be \: (x + k)$


Therefore, x= -k

Since, g(x) is a factor of f(x)

Therefore, f(-k) =0

Answer 2

Hello!

$\bf{1.)}$ $\sf x^{4} - x \implies \boxed{\red{\bf{x(x^{3} - 1)}}}$

$\bf{2.)}$ Let,

• g(x) = (x + k)
• f(x) = x³ + kx² - 2x + k + 4

$\underline{\bf{g(x)\: is \:a \:factor\: of \:f(x)}}$

Then,

$\sf f(-k) = 0 \\ \\ \implies \sf (-k)^{3} + k(-k)^{2} -2(-k) + k + 4 = 0 \\ \\ \implies \sf -k^{3} + k^{3} + 2k + k + 4 = 0 \\ \\ \implies \sf 3k + 4 = 0 \\ \\ \implies \sf 3k = -4 \implies \boxed{\red{\bf{\sf k = -\dfrac{4}{3}}}}$



Cheers!