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P(x)=-3x^4 +6x^3 +27x^2 +54x

Its zeros are x1=___, x2=____, x3= ____, x4=_____, with x1<_x2<_x3<_x4<_

User Jellyfish
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3 votes

Answer:

Explanation:

To find the zeros of the polynomial \(p(x) = -3x^4 + 6x^3 + 27x^2 + 54x\), we need to set it equal to zero and solve for \(x\):

\[ -3x^4 + 6x^3 + 27x^2 + 54x = 0 \]

Factor out the common factor, which is \(3x\):

\[ 3x(-x^3 + 2x^2 + 9x + 18) = 0 \]

Now, we have two factors:

1. \(3x = 0\) (This gives us one zero, \(x_1 = 0\)).

2. \(-x^3 + 2x^2 + 9x + 18 = 0\)

Unfortunately, the cubic equation does not have an easily factored solution. You may need to use numerical methods or a calculator to find approximate solutions.

The zeros \(x_1, x_2, x_3, x_4\) will be the combination of the zero from the linear factor \(3x\) and the solutions to the cubic equation \( -x^3 + 2x^2 + 9x + 18 = 0\).

If you use a tool or a calculator to solve the cubic equation, you can find the numerical values for \(x_2, x_3, x_4\).

User Iridio
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