How do you find all the zeros of $- 5 x^{2} (x - 9) {(x + 4)}^{3}$ $-5x^2(x-9)(x+4)^3$ ?

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Answer 1 · 2018-06-30T22:52:28

$x = 0, 9$ $x=0, 9$ and $- 4$ $-4$ .

Let's set all terms equal to zero:

$- 5 x^{2} = 0$ $color(blue)(-5x^2=0)$

We need to take the square root of both sides to get

$- 5 x = 0$ $-5x=0$

$x = 0$ $color(blue)(x=0)$

$x - 9 = 0 \Rightarrow x = 9$ $color(purple)(x-9=0=>x=9)$

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

In this case, we need to take the cube root of both sides. We get

$x + 4 = 0 \Rightarrow x = - 4$ $color(red)(x+4=0=>x=-4)$

Therefore, the zeroes are at

$x = 0, 9$ $x=0, 9$ and $- 4$ $-4$ .

Hope this helps!

How do you find all the zeros of −5x2(x−9)(x+4)3-5x^2(x-9)(x+4)^3?