Respuesta :
Answer: The real zeroes of the function f(x) are [tex]0,~2,~3,~-\dfrac{1}{2}.[/tex]
Step-by-step explanation: We are given to find the real numbers that are zeroes of the following functions :
[tex]f(x)=2x^4-9x^3+7x^2+6x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
We know that
the zeroes of a polynomial function y = p(x) are given by p(x) = 0.
So, the zeroes of the given polynomial (i) are
[tex]f(x)=0\\\\\Rightarrow 2x^4-9x^3+7x^2+6x=0\\\\\Rightarrow x(2x^3-9x^2+7x+6)=0\\\\\Rightarrow x=0,~~2x^3-9x^2+7x+6=0.[/tex]
So, one of the real zeroes of f(x) is x = 0 and the other zeroes are given by [tex] 2x^3-9x^2+7x+6=0.[/tex]
Factor theorem : If at x = a, the value of a function p(x) is zero, then (x-a) is a factor of p(x).
Let us consider that
[tex]g(x)=2x^3-9x^2+7x+6.[/tex]
At x = 1, we have
[tex]g(x)=2\times1^4-9\times1^3+7\times1^2+6\times1=2-9+7+6=6\neq0.[/tex]
At x = 2, we have
[tex]g(x)=2\times1^3-9\times2^2+7\times2+6=8-36+14+6=0.[/tex]
So, (x-2) is a factor of g(x).
Therefore, we get
[tex]g(x)\\\\=2x^3-9x^2+7x+6\\\\=2x^2(x-2)-5x(x-2)-3(x-2)\\\\=(x-2)(2x^2-5x-3)\\\\=(x-2)(2x^2-6x+x-3)\\\\=(x-2)(2x+1)(x-3).[/tex]
That is, the other two zeroes are given by
[tex]2x+1=0,~~~~~x-3=0\\\\\Rightarrow x=-\dfrac{1}{2},3.[/tex]
Since all the four zeroes are real, so all the real zeroes of f(x) are [tex]0,~2,~3,~-\dfrac{1}{2}.[/tex]
Thus, the real zeroes of the function f(x) are [tex]0,~2,~3,~-\dfrac{1}{2}.[/tex]
