Find the set of values of x for which p(x) is positive:(x^2+4x+5)(x-1)(x-2)?

x² + 4x + 5

= (x² + 4x + 4) + 1

= (x + 2)² + 1

For all values of x, (x + 2)² ≥ 0

Hence, x² + 4x + 5 = (x + 2)² + 1 ≥ 1

In other words, (x² + 4x + 5) must be positive

p(x) = (x² + 4x + 5)(x -1)(x - 2)

When x < 1: x - 1 < 0 and x - 2 < 0, and thus p(x) > 0

When 1 < x < 2: x - 1 > 0 and x - 2 < 0, and thus p(x) < 0

When x > 2: x - 1 > 0 and x - 2 > 0, and thus p(x) > 0

The answer: x < 1 or x > 2

and can be expressed as: (-∞, 1) or (2, ∞)

The factor x^2 + 4x + 5 will be positive for all values of x, as its roots are not "real." The product p(x) will be positive if (a) both (x-1) and (x-2) are positive or (b) both (x-1) and (x-2) are negative.

So the only non-positive values of p(x) are those where 1 <= x <= 2.

If you mean p(x) = (x² + 4x + 5)(x - 1)(x - 2), the it has two real roots, 1 and 2, and no roots with multiplicity. It is a polynomial, and it changes sign at each root. the leading coefficient is positive, so as x increases, p(x) is positive, then negative, then positive.

p(x) > 0

x ∈ (-∞, 1) ∩ (2, ∞)