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

1 Answer

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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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Alfredo Heathcote
15.5k 3 10 26
answered 9 months ago