Question

polynomial inequalities
solve each inequality

1. $(x - 4)(x + 1) \leq 0$
2. $(x - 4)(x - 1) \geq 0$
3. $(x - 1)(3x - 4) \geq 0$
4. $(x + 8)(x + 2)(x - 1) > 0$
5. $x^2 + 5x + 4 \leq 0$
6. $x^2 - 14x + 49 \geq 0$
7. $x^2 - 4x - 32 > 0$
8. $x^2 + 16x + 24 > 6x$
9. $(x + 5)(x - 2)(x - 1)(x + 1) < 0$
10. $(x + 8)^2(x - 5)(x + 7)^2 \geq 0$

Explanation:

To solve the polynomial inequality \(x^2 - 14x + 49 \geq 0\) (problem 6), we can follow these steps:

Step 1: Factor the quadratic expression

The quadratic \(x^2 - 14x + 49\) is a perfect square trinomial. Recall that \(a^2 - 2ab + b^2=(a - b)^2\). Here, \(a = x\) and \(b = 7\) (since \(2ab=2\times x\times7 = 14x\)), so we can factor it as:

$$ x^2 - 14x + 49=(x - 7)^2 $$

So the inequality becomes \((x - 7)^2\geq0\).

Step 2: Analyze the square of a real number

For any real number \(y\), \(y^2\geq0\) (by the property of real numbers: the square of a real number is always non - negative). Let \(y=x - 7\). Then \((x - 7)^2\) is the square of the real number \(x - 7\).

No matter what value \(x\) takes (because \(x\) is a real number, \(x-7\) is also a real number), the square of \(x - 7\) will be greater than or equal to \(0\). For example, if \(x=7\), then \((7 - 7)^2=0\); if \(x = 8\), then \((8 - 7)^2=1\gt0\); if \(x=6\), then \((6 - 7)^2=(- 1)^2 = 1\gt0\).

Answer:

The solution to the inequality \(x^2-14x + 49\geq0\) is all real numbers, which can be written in interval notation as \((-\infty,\infty)\) or in set - builder notation as \(\{x|x\in\mathbb{R}\}\)