# How do you solve (x-5)^2(x-1)(x+3)<=0?

Jun 22, 2016

$x$ is in segment $\left[- 3 , 1\right]$ (including both ends) and $x = 5$.

#### Explanation:

For all $x$ except $x = 5$ the expression ${\left(x - 5\right)}^{2}$ is strictly positive. The value $x = 5$ is a solution since it sets the left hand side to zero.

Therefore, we can
(1) say that $x = 5$ is one of the solutions of this inequality and
(2) for all other $x$ we can reduce both sides of this inequality by positive ${\left(x - 5\right)}^{2}$ without changing the comparison in the inequality getting a simpler inequality to solve:
$\left(x - 1\right) \left(x + 3\right) \le 0$

Next, we divide the set of real numbers into intervals by points that set each multiplier to zero ($x = 1$ and $x = - 3$).
Consider $x$ as moving from $- \infty$ on the left to $+ \infty$ on the right.

(a) While its less than $- 3$, both $x - 1$ and $x + 3$ are negative and, therefore, their product is positive, not good for our inequality.

(b) As soon as $x$ crosses $- 3$, one of these multipliers, $x + 3$, changes sign to positive, while the other, $x - 1$, remains negative. Their product is negative, which is a solution.

(c) finally, as $x$ crosses $1$, both multipliers are positive, their product is positive too, not good for our inequality.

So, remaining points are $\left[- 3 , 1\right]$ and $5$.