Find the zeros for: f(x)=x^4+x^3+4x^2 ( show work)

There is an obvious factor of x², so we can reduce this to the problem of finding zeros of a quadratic by factoring that out. Then, we can determine the zeros of the quadratic factor to complete the finding of zeros for the function.

A factor of x² can be factored out, leaving ...

... f(x) = x²(x² +x +4)

The latter factor has only complex zeros. It can be rewritten by completing the square.

... x² +x +4 = (x² +x +1/4) +(3 3/4) = (x +1/2)² +(3 3/4)

So, the real zeros are where x² = 0, at x = 0. The complex zeros are where the above expression is zero, ...

... (x +0.5)² +3.75 = 0

... (x +0.5)² = -3.75

... x + 0.5 = √-3.75 = ±0.5i√15

... x = -0.5(1 ±i√15)

The function zeros are 0 (multiplicity 2) and -0.5±0.5i√15.

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step-by-step explanation:

1,150 give brainliest

step-by-step explanation

2r= 2300

r =1150