Question

Which real numbers are zeros of the function? f(x)=2x4−9x3+7x2+6x

Answer 1

Answer:  The real zeroes of the function f(x) are $0,~2,~3,~-\dfrac{1}{2}.$

Step-by-step explanation:  We are given to find the real numbers that are zeroes of the following functions :

$f(x)=2x^4-9x^3+7x^2+6x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

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

$f(x)=0\\\\\Rightarrow 2x^4-9x^3+7x^2+6x=0\\\\\Rightarrow x(2x^3-9x^2+7x+6)=0\\\\\Rightarrow x=0,~~2x^3-9x^2+7x+6=0.$

So, one of the real zeroes of f(x) is x = 0 and the other zeroes are given by $2x^3-9x^2+7x+6=0.$

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

$g(x)=2x^3-9x^2+7x+6.$

At x = 1, we have

$g(x)=2\times1^4-9\times1^3+7\times1^2+6\times1=2-9+7+6=6\neq0.$

At x = 2, we have

$g(x)=2\times1^3-9\times2^2+7\times2+6=8-36+14+6=0.$

So, (x-2) is a factor of g(x).

Therefore, we get

$g(x)\\\\=2x^3-9x^2+7x+6\\\\=2x^2(x-2)-5x(x-2)-3(x-2)\\\\=(x-2)(2x^2-5x-3)\\\\=(x-2)(2x^2-6x+x-3)\\\\=(x-2)(2x+1)(x-3).$

That is, the other two zeroes are given by

$2x+1=0,~~~~~x-3=0\\\\\Rightarrow x=-\dfrac{1}{2},3.$

Since all the four zeroes are real, so all the real zeroes of f(x) are $0,~2,~3,~-\dfrac{1}{2}.$

Thus, the real zeroes of the function f(x) are $0,~2,~3,~-\dfrac{1}{2}.$

Answer 2

-1/2, 0, 2, and 3 are the answers