How do you solve ${(x + 4)}^{3} {(x + 2)}^{2} = 0$ $(x+4)^3(x+2)^2=0$ ?

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Answer 1 · 2017-06-14T13:48:16

$x = - 4, - 2$ $x = - 4, - 2$

We have: ${(x + 4)}^{3} {(x + 2)}^{2} = 0$ $(x + 4)^(3) (x + 2)^(2) = 0$

Using the null factor law:

$\Rightarrow {(x + 4)}^{3} = 0$ $Rightarrow (x + 4)^(3) = 0$

$\Rightarrow x + 4 = 0$ $Rightarrow x + 4 = 0$

$therefore x = - 4$

$or$

$Rightarrow (x + 2)^(2) = 0$

$Rightarrow x + 2 = 0$

$therefore x = - 2$

Therefore, the solutions to the equation are $x = - 4$ and $x = - 2$ .

How do you solve (x+4)^3(x+2)^2=0(x+4)3(x+2)2=0(x+4)^3(x+2)^2=0?