To solve a polynomial inequality, we factor the polynomial into irreducible factors and find all the realpolynomial. Then we find the intervals determined by the realsign of the polynomial on that interval. Let
$$P(x)=x(x+2)(x-1)$$
Fill in the diagram to find the intervals on which
$P(x) \geq 0$
we see that $P(x) \geq 0$ on the
intervalsand.

To solve a polynomial inequality, we factor the polynomial into irreducible factors and find all the real _zeros_ polynomial. Then we find the intervals determined by the real _zeros and use test points in each interval to find the_ sign of the polynomial on that interval.

If P(x) = x(x+2)(x-1)

And P(x) ≥ 0

We see that P(x) ≥ 0 on the intervals (-2, 0) and (1, ∞).

Step-by-step explanation:

The complete question is attached to this solution

To solve inequality of a polynomial, we first obtain the solutions of the polynomial. The solutions of the polynomial are called the zeros of the polynomial.

If P(x) = x(x+2)(x-1)

The solutions of this polynomial, that is the zeros of this polynomial are 0, -2 and 1.

To now solve the inequality that arises when

P(x) ≥ 0

We redraw the table and examine the intervals

The intervals to be examined as obtained from the zeros include (-∞, -2), (-2, 0), (0, 1) and (1, ∞)

Sign of | x<-2 | -2<x<0 | 0<x<1 | x>1

x               | -ve | -ve | +ve | +ve

(x + 2)       | -ve | +ve | +ve | +ve

(x - 1)         | -ve | -ve | -ve | +ve

x(x+2)(x-1) | -ve | +ve | -ve | +ve

The intervals that satisfy the polynomial inequality P(x) = x(x+2)(x-1) ≥ 0 include

(-2, 0) and (1, ∞)

Hope this Helps!!!

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation: