Question

Apply The Remainder Theorem, Fundamental Theorem, Rational Root Theorem, Descartes Rule, and Factor Theorem to find the remainder, all rational roots, all possible roots, and actual roots of the given function. Make Sure to graph the function and state the turning points (Relative maximum and relative minimum) and end behavior (Graph is rising or falling on the right or left side, etc.)
f(x) = 3x^3 − x^2 + 8x + 12
Please answer all the questions fully asap!!

Answer 1

9514 1404 393

Answer:

  possible rational roots: ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12}

  actual roots: -1, (2 ±4i√2)/3

  no turning points; no local extrema

  end behavior is same-sign as x-value end-behavior

Step-by-step explanation:

The Fundamental Theorem tells us this 3rd-degree polynomial will have 3 roots.

The Rational Root Theorem tells us any rational roots will be of the form ...

  ±{factor of 12}/{factor of 3} = ±{1, 2, 3, 4, 6, 12}/{1, 3}

  = ±{1/3, 2/3, 1, 4/3, 2, 3, 4, 6, 12} . . . possible rational roots

Descartes' Rule of Signs tells us the two sign changes mean there will be 0 or 2 positive real roots. Changing signs on the odd-degree terms makes the sign-change count go to 1, so we know there is one negative real root.

The y-intercept is 12. The sum of all coefficients is 22, so f(1) > f(0) and there are no positive real roots in the interval [0, 1]. Synthetic division by x-1 shows the remainder is 22 (which we knew) and all the quotient coefficients are all positive. This means x=0 is an upper bound on the real roots.

The sum of odd-degree coefficients is 3+8=11, equal to the sum of even-degree coefficients, -1+12=11. This means that -1 is a real root. Synthetic division by x+1 shows the remainder is zero (which we knew) and the quotient coefficients alternate signs. This means x=-1 is a lower bound on real roots. The quotient of 3x^2 -4x +12 is a quadratic factor of f(x):

  f(x) = (x +1)(3x^2 -4x +12)

The complex roots of the quadratic can be found using the quadratic formula:

  x = (-(-4) ±√((-4)^2 -4(3)(12)))/(2(3)) = (4 ± √-128)/6

  x = (2 ± 4i√2)/3 . . . . complex roots

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The graph in the third attachment (red) shows there are no turning points, hence no relative extrema. The end behavior, as for any odd-degree polynomial with a positive leading coefficient, is down to the left and up to the right.