I need help please rndnddmdndnxxnxx

9514 1404 393

  a) f(-2.5) = 62.6875; f(-1.5) = -13.0625; (-2.5, -1.5) contains a real root

  b) there are 1 each of positive and negative real roots; 2 complex roots

  c) -2 and 1/3 are the rational real roots

  d) x ∈ {-2, 1/3, -1/3 +i(√2)/3, -1/3 -i(√2)/3}

Step-by-step explanation:

a) f(-2.5) = 62.6875; f(-1.5) = -13.0625. The function changes sign between -1.5 and -2.5, so there is a real root in that interval.

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b) The given coefficients have one sign change, so there is one positive real root. Changing the signs of odd-degree terms makes the signs + - + - -, so there are three sign changes. There are 1 or 3 negative real roots, and possibly 2 complex roots. The constant term is non-zero, so zero is not a root. Here's your table:

  positive real roots: 1

  negative real roots: 1 or 3

  complex roots: 2 or 0

  zero roots: none

The graph is attached. It shows 2 real roots, one positive and one negative, so there are 2 complex roots. (There are 4 roots in total, so if only 2 are real, the other 2 are complex.)

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c) Possible rational roots are of the form ...

  ±{1, 2}/{1, 3, 9}

Listed, they are ...

  ±{1/9, 2/9, 1/3, 2/3, 1, 2}

The actual real roots are -2 and 1/3.

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d) The graph shows the real roots to be -2 and 1/3. The original leading coefficient of 3 is the product of the leading coefficients of the factors. Writing the factor (x -1/3) as (3x -1) lets us reduce the leading coefficient of the quadratic term from 9 to 3. The result of dividing factors (x+2) and (3x-1) from the original function is shown in the attachment. It is a parabola with vertex (-1/3, 2/3), so the factorization is ...

  f(x) = (x +2)(3x -1)(3(x +1/3)² +2/3)

The roots of the quadratic term are ...

  3(x +1/3)² = -2/3

  x +1/3 = ±√(-2/9)

  x = (-1 ±i√2)/3

So, the entire list of roots is ...

  x ∈ {-2, 1/3, -1/3 +i(√2)/3, -1/3 -i(√2)/3}

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Additional comment

The roots of a quadratic are easily found by knowing its leading coefficient and its vertex. If factoring known real roots from a higher-degree polynomial results in a quadratic, then the remaining complex roots are easily found. A graphing calculator is a wonderful tool for this process.