How do you find the zeroes of the function $g (x) = 2 x^{3} - 5 x^{2} + 4$ $g(x)= 2x^3 - 5x^2 + 4$ ?

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Answer 1 · 2016-06-28T14:43:48

$x = \frac{1 \pm \sqrt{17}}{4} and x = 2$ $x=(1+-sqrt(17))/4 and x=2$

let's consider the known term 4: it is divided by

$\pm 1, \pm 2, \pm 4$ $+-1, +-2, +-4$

By the remainder theorem, let's substitute each of the dividers in x in the function g(x), so

$g (1) = 2 - 5 + 4 \neq 0$ $g(1)=2-5+4!=0$

$g (- 1) = - 2 - 5 + 4 \neq 0$ $g(-1)=-2-5+4!=0$

$g (2) = 16 - 20 + 4 = 0$ $g(2)=16-20+4=0$

So, one of the zeroes is 2

Then we can divide the polynomial g(x) by the bynomial x-2

and obtain:

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

by which you can find two zeroes

$x = \frac{1 \pm \sqrt{17}}{4}$ $x=(1+-sqrt(17))/4$

How do you find the zeroes of the function g(x)=2x3−5x2+4g(x)= 2x^3 - 5x^2 + 4?