Respuesta :
Given:
The function is:
[tex]y=3x^3-5x^2-11x-3[/tex]
It is given that -1 is a zero of the given function.
To find:
The other zeroes of the given function.
Solution:
If c is a zero of a polynomial P(x), then (x-c) is a factor of the polynomial.
It is given that -1 is a zero of the given function. So, [tex](x+1)[/tex] is a factor of the given function.
We have,
[tex]y=3x^3-5x^2-11x-3[/tex]
Split the middle terms in such a way so that we get (x+1) as a factor.
[tex]y=3x^3+3x^2-8x^2-8x-3x-3[/tex]
[tex]y=3x^2(x+1)-8x(x+1)-3(x+1)[/tex]
[tex]y=(x+1)(3x^2-8x-3)[/tex]
Again splitting the middle term, we get
[tex]y=(x+1)(3x^2-9x+x-3)[/tex]
[tex]y=(x+1)(3x(x-3)+1(x-3))[/tex]
[tex]y=(x+1)(3x+1)(x-3)[/tex]
For zeroes, [tex]y=0[/tex].
[tex](x+1)(3x+1)(x-3)=0[/tex]
[tex](x+1)=0[/tex] and [tex]3x+1=0[/tex] and [tex]x-3=0[/tex]
[tex]x=-1[/tex] and [tex]x=-\dfrac{1}{3}[/tex] and [tex]x=3[/tex]
Therefore, the other two zeroes of the given function are [tex]-\dfrac{1}{3}[/tex] and [tex]3[/tex].
