Polynomial inequalities involve finding the intervals on the real number line where a polynomial function is positive, negative, or zero. To solve them, one first identifies the critical points by setting the polynomial equal to zero and solving for the roots. These roots partition the number line into distinct intervals. By selecting a test value from each interval and substituting it into the inequality, the sign of the polynomial can be determined for the entire interval. This method, combined with an understanding of root multiplicity and the leading coefficient's sign, allows for the accurate construction of the solution set, typically expressed in interval notation.
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اختر إجابة واحدة لكل سؤال. عند الاختيار ستظهر النتيجة فورًا: الأخضر صحيح، والأحمر خطأ، وسيظهر تفسير الإجابة مباشرة إن كان متوفرًا. وبعد آخر سؤال ستظهر الدرجة النهائية تلقائيًا.
The roots of the polynomial are x = 2 and x = -3. Since the parabola opens upward (leading coefficient is positive) and we are looking for values greater than or equal to zero, the solution includes the regions outside the roots, including the roots themselves.
The roots are -2, 2, 6. Testing the intervals: for x < -2 the expression is negative; for -2 < x < 2 it is positive; for 2 < x < 6 it is negative; and for x > 6 it is positive. Since we need the expression to be less than zero, the solution is \((-\infty, -2) \cup (2, 6)\).
Factoring the polynomial as \(x(x - 4) \ge 0\) gives the roots 0 and 4. As a parabola opening upward, the expression is greater than or equal to zero in the intervals outside the roots: \((-\infty, 0]\) and \([4, \infty)\).
The roots are 3 and -6. Because the leading coefficient is negative (-x2), the parabola opens downward, meaning the expression is non-negative between the roots: [-6, 3].
The polynomial can be factored as \(-(x - 6)(2x + 1)(x + 3) \le 0\). The critical points are \(-3, -\frac{1}{2}, 6\). Testing intervals with the negative leading coefficient shows that the expression is less than or equal to zero in the intervals \([-3, -\frac{1}{2}]\) and \([6, \infty)\).
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