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Find all the zeroes of g(x)=4x^4+4x^3-11x^2-12x-3

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LanceLongFei

Answer:

The zeroes of [tex]g(x)=4x^4+4x^3-11x^2-12x-3[/tex] are [tex]-\sqrt{3}, -\frac{1}{2}, \sqrt{3}[/tex], or approximately [tex]-1.732, -0.500, 1.732[/tex].

Step-by-step explanation:

Graphical calculator way (probably intended)

By inputting the function in a graphical calculator we can see that the roots of the graph of [tex]g[/tex] are about [tex]x=-1.732, x = 0.500, x = 1.732[/tex] (make sure you click on the points of interests).

Algebraic way (probably not intended because of obscure factoring)

Notice that [tex]g(x) = 4x^4+4x^3-11x^2-12x-3 = (2x-1)^2(x^2-3).[/tex] So the roots can be calculated with [tex](2x-1)^2 = 0 \vee x^2 - 3 = 0[/tex], yielding [tex]x = -\frac{1}{2} \vee x = \pm\sqrt{3}[/tex].

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