How do you find all the zeros of $y = ( x -3)^4 (2x+4) (3x-2)$ ?

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Answer 1 · 2018-06-22T13:57:03

To find the zeroes, set each binomial equal to 0.

Once we set them up like this, ...

$(x-3)^4=0$
$2x+4=0$
$3x-2=0$

…we can simplify them.

$root(4)((x-3)^4)=root(4)(0)$
$x-3=0$
$x=3$

$2x=-4$
$x=-4/2=-2$

$3x=2$
$x=2/3$

Zeroes: $x=3, -2, 2/3$

How do you find all the zeros of y = ( x -3)^4 (2x+4) (3x-2)y = ( x -3)^4 (2x+4) (3x-2)?