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Oduvanchick [21]
3 years ago
9

Dennis, Emily, and Fernando want to find the zeros of the polynomial p(x)=x3+6x2+9x. Each student worked independently and prese

nted his or her results to the group. Dennis used the Rational Root Theorem to identify all zeros of p(x). According to this theorem he stated that the zeros of p(x) are ±1, ±3, and ±9. Emily used the greatest common factor and perfect-square trinomial methods to factor p(x) as p(x)=x(x+3)2. She applied the definition of zeros and found that the zeros of p(x) are 0 and −3. Fernando used the greatest common factor and perfect-square trinomial methods to factor p(x) as p(x)=x(x+3)2. He applied the definition of zeros and found that the zeros of p(x) are 0 and −3. He then applied the Irrational Root Theorem and stated that since −3 is a zero, 3 must also be a zero. Therefore, the zeros are 0 and ±3. Which statements accurately justify why each student is correct or incorrect? There may be more than one correct answer. Select all correct answers. emily’s response is incorrect because although she correctly factored the polynomial and used the definition of zeros, she did not apply the Irrational Root Theorem. Fernando’s response is correct because he correctly factored the polynomial, correctly used the definition of zeros, and correctly used the Irrational Root Theorem to identify all zeros. Fernando’s response is incorrect because he inappropriately applied the Irrational Root Theorem. Dennis’ response is incorrect. According to the Fundamental Theorem of Algebra, the polynomial p(x) cannot have six roots, or zeros, because it is only of degree 3. Emily’s response is correct because she correctly factored the polynomial, and correctly used the definition of zeros to reach her answer. Dennis’ response is correct because he correctly factored the polynomial, and correctly used the Rational Root Theorem to identify zeros of a polynomial.
Mathematics
1 answer:
alexandr402 [8]3 years ago
3 0

Answer:

  • Fernando’s response is incorrect because he inappropriately applied the Rational Root Theorem.
  • Dennis’ response is incorrect. According to the Fundamental Theorem of Algebra, the polynomial p(x) cannot have six roots, or zeros, because it is only of degree 3.
  • Emily’s response is correct because she correctly factored the polynomial, and correctly used the definition of zeros to reach her answer.

Step-by-step explanation:

The Rational Root Theorem offers a list of possible rational roots. Each needs to be tested to see if it is an actual rational root. Fernando and Dennis made inappropriate assumptions about what the Rational Root Theorem allowed them to conclude.

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8/10=4/5

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Explanation:

When the inequality symbol is replaced by an equal sign, the resulting linear equation is the boundary of the solution space of the inequality. Whether that boundary is included in the solution region or not depends on the inequality symbol.

The boundary line is included if the symbol includes the "or equal to" condition (≤ or ≥). An included boundary line is graphed as a solid line.

When the inequality symbol does not include the "or equal to" condition (< or >), the boundary line is not included in the solution space, and it is graphed as a dashed line.

Once the boundary line is graphed, the half-plane that makes up the solution space is shaded. The shaded half-plane will be to the right or above the boundary line if the inequality can be structured to be of one of these forms:

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Otherwise, the shaded solution space will be below or to the left of the boundary line.

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Just as a system of linear equations may have no solution, so that may be the case for inequalities. If the boundary lines are parallel and the solution spaces do not overlap, then there is no solution.

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The attached graph shows an example of graphed inequalities. The solutions for this system are in the doubly-shaded area to the left of the point where the lines intersect. We have purposely shown both kinds of inequalities (one "or equal to" and one not) with shading both above and below the boundary lines.

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Part of the cake shared = \frac{c}{2}

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