Question

The zeros for f(x)=x^4-4x^3-9x^2+36x are

Answer 1

x^4 - 4x^3 - 9x^2 + 36x = 0
 x(x^3 - 4x^2 - 9x + 36) = 0
x(x^2(x - 4) - 9(x - 4) = 0
x(x + 3)(x - 3)(x - 4) = 0

the zeroes are  -3, 0, 3, 4  Answer

Answer 2

Following are the calculation of the zeros to the given function:

Given:

$\bold{f(x)=x^4-4x^3-9x^2+36x}$

To find:

zeros=?

Solution:

$\bold{f(x)=x^4-4x^3-9x^2+36x}$

Putting the value 3,4,-3 to the above-given function:

When x=3

$\to \bold{f(3)=3^4-4(3^3)-9(3^2)+36(3)}$

            $\bold{=81-4(27)-9(9)+36(3)}\\\\\bold{=81- 108-81+108}\\\\\bold{=0}$

When x=4

$\to \bold{f(4)=4^4-4(4^3)-9(4^2)+36(4)}$

            $\bold{=256-4(64)-9(16)+36(4)}\\\\\bold{=256- 256- 144+144}\\\\\bold{=0}$

When x=-3

$\to \bold{f(-3)=-3^4-4(-3^3)-9(-3^2)+36(-3)}$

            $\bold{=81-4(-27)-9(9)+36(-3)}\\\\\bold{=81+ 108-81-108}\\\\\bold{=0}$

Therefore, the zeros are "3,-3, and 4".

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