Solve the equation in the real number system.
-$$x ^ { 4 } - 3 x ^ { 3 } + 5 x ^ { 2 } - x - 10 = 0$$
A) $\{ - 1,2 \}$
B) $\{ 1,2 \}$
C) $\{ - 2,1 \}$
D) $\{ - 1 , - 2 \}$

Question

Quizplus · Accepted Answer

The given polynomial can be factored as $(x^2 - x - 2)(x^2 - 2x + 5)$. The first factor, $x^2 - x - 2$, can be further factored into $(x - 2)(x + 1)$, giving the real roots $x = 2$ and $x = -1$. The second factor, $x^2 - 2x + 5$, does not have real roots because its discriminant, $b^2 - 4ac = (-2)^2 - 4(1)(5) = 4 - 20 = -16$, is negative. Therefore, the real solutions are $x = -1$ and $x = 2$.

Solve the Equation in the Real Number System x4−3x3+5x2−x−10=0x ^ { 4 } - 3 x ^ { 3 } + 5 x ^ { 2 } - x - 10 = 0x4−3x3+5x2−x−10=0

Solve the Equation in the Real Number System $x ^ { 4 } - 3 x ^ { 3 } + 5 x ^ { 2 } - x - 10 = 0$