What are the possible number of positive real, negative real, and complex zeros of $y = (2 - x) (3 - 2 x)$ $y= (2-x)(3-2x)$ ?

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What are the possible number of positive real, negative real, and complex zeros of $y = (2 - x) (3 - 2 x)$ $y= (2-x)(3-2x)$ ?

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Answer 1 · 2015-07-30T13:17:22

I found two:
$x = 2$ $x=2$
$x = \frac{3}{2}$ $x=3/2$

Explanation:

If you set $y = 0$ $y=0$ you get:
$(2 - x) (3 - 2 x) = 0$ $(2-x)(3-2x)=0$
this is tru when:
$2 - x = 0$ $2-x=0$ or $x = 2$ $x=2$
and
$3 - 2 x = 0$ $3-2x=0$ or $x = \frac{3}{2}$ $x=3/2$
these are the zeros of your function.

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What are the possible number of positive real, negative real, and complex zeros of $y = (2 - x) (3 - 2 x)$ $y= (2-x)(3-2x)$ ?

1 Answer

Explanation:

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What are the possible number of positive real, negative real, and complex zeros of y= (2-x)(3-2x)y=(2−x)(3−2x)y= (2-x)(3-2x)?

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Explanation:

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What are the possible number of positive real, negative real, and complex zeros of $y = (2 - x) (3 - 2 x)$ $y= (2-x)(3-2x)$ ?