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Find all real values of such that x³+6x²+9x<0. Write the answer in interval notation.

a. (−[infinity],−3)∪(−3,0)
b. (−[infinity],−3)∪(−3,−2)∪(−2,0)
c. (−[infinity],−3)∪(−2,0)
d. (−[infinity],−3)∪(−2,0)

User Lilya
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1 Answer

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Final answer:

The factored form of the inequality x³+6x²+9x<0 is x(x+3)²<0. The solution to the inequality are the intervals where the expression is negative, which is (-∞,-3)∪(-3,0).

Step-by-step explanation:

To find all real values of x such that x³+6x²+9x<0, first factor the left side of the inequality. You can factor out an x, getting x(x²+6x+9)<0. The quadratic x²+6x+9 can be factored further as (x+3)². Therefore, the factored inequality is x(x+3)²<0. The zeros of the expression are x=0 and x=-3 (with a multiplicity of 2).

The inequality is satisfied between the roots where the expression is negative. Since (x+3)² is always non-negative, the sign is determined by the x. Thus, the solution set is the interval when x is less than zero but not including the point where x+3 equals zero since squaring it would result in a non-negative number. Therefore, the real values of x that satisfy the inequality are in the interval (-∞,-3)∪(-3,0).

User Gannaway
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