Question

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Answer 1

$x^2(4x-3)(5x-1)=0\iff x^2=0\ \vee\ 4x-3=0\ \vee\ 5x-1=0\\\\x^2=0\to\boxed{x=0}\\\\4x-3=0\ \ \ |+3\\4x=3\ \ \ |:4\\\boxed{x=\dfrac{3}{4}}\\\\5x-1=0\ \ \ |+1\\5x=1\ \ \ |:5\\\boxed{x=\dfrac{1}{5}}\\\\Answer:\ x=0\ or\ x=\dfrac{3}{4}\ or\ x=\dfrac{1}{5}$

Answer 2

Hi!

x = 0

or

x = 1/5

or

x = 3/4

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The process:

Step 1: Factor left side of equation.

Step 2: Set factors equal to 0.

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Hope this helped :)