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STatiana [176]
3 years ago
5

Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer. How many imaginary r

oots does the polynomial have?
Mathematics
1 answer:
Rainbow [258]3 years ago
8 0

Answer:

<h3>The given polynomial of degree 4 has atleast one imaginary root</h3>

Step-by-step explanation:

Given that " Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer:

<h3>To find how many imaginary roots does the polynomial have :</h3>
  • Since the degree of given polynomial is 4
  • Therefore it must have four roots.
  • Already given that the given polynomial has 1 positive real root and 1 negative real root .
  • Every polynomial with degree greater than 1  has atleast one imaginary root.
<h3>Hence the given polynomial of degree 4 has atleast one imaginary root</h3><h3> </h3>

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<u>Step-by-step explanation:</u>

The first matrix contains the coefficients of the x- and y- values for both equations (top row is the top equation and the bottom row is the bottom equation. The second matrix contains what each equation is equal to.

\begin{array}{c}2x-y\\x-6y\end{array}\qquad \rightarrow \qquad \left[\begin{array}{cc}2&-1\\1&-6\end{array}\right] \\\\\\\begin{array}{c}-6\\13\end{array}\qquad \rightarrow \qquad \left[\begin{array}{c}-6\\13\end{array}\right]

The product will result in the solution for the x- and y-values of the system.

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Help please!
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Step-by-step explanation:

Given

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