# How do you find the zeros of  (-x+1)(x+2)(x-3)^2?

##### 1 Answer
Mar 26, 2016

There are four zero in total: at $x = - 2 , x = 1 , x = 3 \mathmr{and} x = 3$

graph{(-x+1)(x+2)(x-3)^2 [-7.116, 10.05, -12.5, 34.3]}

#### Explanation:

Given: $f \left(x\right) = \left(- x + 1\right) \left(x + 2\right) {\left(x - 3\right)}^{2}$
Required the zeros?
Solution Strategy: Since $f \left(x\right)$ is given by a product of three function $h \left(x\right) , g \left(x\right) , v \left(x\right)$ such $h \left(x\right) = \left(- x + 1\right) , g \left(x\right) = \left(x + 2\right) , v \left(x\right) = {\left(x - 3\right)}^{2}$
$f \left(x\right) = 0$ for any $x$ that sets either $\left(h \left(x\right) \mathmr{and} g \left(x\right) \mathmr{and} v \left(x\right)\right) = 0$ thus
$h \left(x\right) = 0 , x = 1$
$g \left(x\right) = 0 , x = - 2$
$v \left(x\right) = 0 , {x}_{1 , 2} = 3$

There are four zero in total: at $x = - 2 , x = 1 , x = 3 \mathmr{and} x = 3$