Find the solutions of the inequality by drawing appropriate graphs. State the solution rounded to two decimals. (Enter your answer using interval notation.) 25x^3 + 60x^2 > −21x − 2

Answer 1

Hello Jonny,

To find the solution of 25x^3 + 60x^2 > -21x - 2, begin by adding 21x and 2 on both sides so that,

25x^3 + 60x^2 + 21x + 2 > 0

Next, we factor the polynomial into irreducible factors by using the rational zero theorem and synthetic division as follows:

According to the rational zero theorem, the possible rational zeros of 25x^3 + 60x^2 + 21x +2 are the factors of 2 (the constant) divided by the factors of 25 (the leading coefficient). Namely, {+/- 1, +/- 2} over {+/- 1, +/- 5, +/- 25} which simplify to +/- 1, +/- 1/5, +/- 1/25, +/- 2, +/- 2/5, +/- 2/25.

Use synthetic division to test which are actual zeros:

Testing: - 1/5 25 60 21 2

-5 -11 -2

25 55 10 0 <----- Remainder is 0, so - 1/5 is a zero.

Hence, [x - (-1/5)] = (x+1/5) is a factor and our inequality may be written as

(x + 1/5)(25x^2 + 55x +10) > 0

(x + 1/5)(5)(5x^2 + 11x + 2) > 0 factor out the greatest common factor from the quadratic factor

5(x + 1/5)(5x + 1)(x + 2) > 0 factor 5x^2 + 11x + 2

(5x + 1)(5x + 1)(x+2) > 0 distribute the 5

Hence x = -1/5 is a factor of multiplicity 2 and x = -2 is a factor of multiplicity 1.

So, the intervals determined by the zeros of the polynomial are (-∞, -2), (-2, -1/5), (-1/5, ∞)

Make a diagram to indicate the sign of each factor on each interval as follows:

-2 -1/5

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Sign of (5x + 1)^2 + ¦ + ¦ +

Sign of (x + 2) - ¦ + ¦ +

Sign of (5x + 1)^2 (x+2) - ¦ + ¦ +

Since we are interested in the intervals were the polynomial is greater than zero (i.e. positive), the solution set is (-2, -1/5) ∪ (-1/5, ∞).

Answer 2

25x³ + 60x² + 21x + 2 = 0; 25x³ + 50x² + 10x² + 20x + x + 2 = 0; (x + 2)(25x² + 10x + 1) = 0; (x + 2)(5x + 1)² = 0; x = - 2 or x = - 1/5.

So, 25x³ + 60x² + 21x + 2 > 0 if we have intervals for x: (- 2, - 1/5)U(-1/5, ∞)