Find the x-intercepts of the polynomial function. State whether the graph crosses the x-axis, or touches the x-axis and turns around, at each intercept.
-$$x ^ { 5 } - 27 x ^ { 3 } + 50 x = 0$$
A) 0 , crosses the $x$-axis;
5 , crosses the $x$-axis;
$- 5$, crosses the $x$-axis;
$\sqrt { 2 }$, crosses the $x$-axis;
$- \sqrt { 2 }$, crosses the $x$-axis
B) 0 , touches the $x$-axis and turns around;
5 , crosses the $x$-axis;
$- 5$, crosses the $x$-axis;
$\sqrt { 2 }$, crosses the $x$-axis;
$- \sqrt { 2 }$, crosses the $x$-axis
C) 0 , crosses the $x$-axis;
25 , touches the $x$-axis and turns around;
2 , touches the $x$-axis and turns around
D) 0 , touches the $x$-axis and turns around;
25 , touches the $x$-axis and turns around;
2 , touches the $x$-axis and turns around

Question

Quizplus · Accepted Answer

First, factor the polynomial: $x(x^4 - 27x^2 + 50) = 0$. The roots can be found by setting each factor equal to zero. The quadratic factor can be factored further or solved using the quadratic formula, revealing roots at $x = 0$, $x = \pm \sqrt{2}$, and $x = \pm 5$. Since the original polynomial is of odd degree and each factor is to the first power, the graph crosses the x-axis at each intercept.

Find the X-Intercepts of the Polynomial Function x5−27x3+50x=0x ^ { 5 } - 27 x ^ { 3 } + 50 x = 0x5−27x3+50x=0

Find the X-Intercepts of the Polynomial Function $x ^ { 5 } - 27 x ^ { 3 } + 50 x = 0$