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

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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, ...

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

…we can simplify them.

$\sqrt[4]{{(x - 3)}^{4}} = \sqrt[4]{0}$ $root(4)((x-3)^4)=root(4)(0)$
$x - 3 = 0$ $x-3=0$
$x = 3$ $x=3$

$2 x = - 4$ $2x=-4$
$x = - \frac{4}{2} = - 2$ $x=-4/2=-2$

$3 x = 2$ $3x=2$
$x = \frac{2}{3}$ $x=2/3$

Zeroes: $x = 3, - 2, \frac{2}{3}$ $x=3, -2, 2/3$

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