Algebraically, find all remaining zeros of f(x) = 8x^3 - 6x^2-36x +27 in simplest form.

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Answer:

$x={-\frac{3\sqrt{2}}{2}, \frac{3}{4}, \frac{3\sqrt{2}}{2}}$

Step-by-step explanation:

In order to find the zeros of a function, we must first set the equation equal to zero, so we get:

$8x^{3}-6x^{2}-36x+27=0$

so we can now solve this by factoring. We can factor this equation by grouping. We start by grouping the equation in pairs of terms, so we get:

$(8x^{3}-6x^{2})+(-36x+27)=0$

and factor each group, so we get:

$2x^{2}(4x-3)-9(4x-3)=0$

and now factor again, so we get:

$(2x^{2}-9)(4x-3)=0$

and now we set each of the factors equal to zero to find the zeros:

$2x^{2}-9=0$

$2x^{2}=9$

we divide both sides into 2 to get:

$x^{2}=\frac{9}{2}$

and take the square root to both sides to get:

$x=\pm\sqrt{\frac{9}{2}}$

which yields:

$x=\pm\frac{3}{\sqrt{2}}$

We rationalize so we get:

$x=\pm\frac{3\sqrt{2}}{2}$

this means that we have two zeros here:

$x=\frac{3\sqrt{2}}{2}$ and $x=-\frac{3\sqrt{2}}{2}$

so we take the other factor and set it equal to zero.

4x-3=0

and solve for x

4x=3

$x=\frac{3}{4}$

and that will be our third zero.

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answered 9 months ago