Geometrical meaning of the zeroes of a polynomial.

Rationalising the denominator gives 1/√2 = (1/√2) × (√2/√2) = √2/2. The number √2 is irrational, and dividing it by the non-zero rational number 2 keeps the result irrational. Therefore 1/√2 is irrational. It is nevertheless a real number because √2 is real and non-zero. It cannot be an integer or a natural number, since those are all rational numbers, whereas the expression has just been shown to be irrational. It is also not zero because the numerator is 1 and the denominator √2 is non-zero. Thus the denominator simplification confirms option A rather than any of the numerical alternatives.

The governing concept is factorisation and the distinction between real and non-real zeroes. Factor the polynomial as x^4-1=(x^2-1)(x^2+1)=(x-1)(x+1)(x^2+1). The first two factors provide the known real zeroes x=1 and x=-1. For every real number x, x^2 is non-negative, so x^2+1 is at least 1 and can never be zero. Its formal solutions are x=i and x=-i, which are non-real and must not be counted. Therefore there are no real zeroes beyond the two given ones, so option A is correct. Options B, C, and D result from failing to factor completely or from counting complex roots as real roots.

Answer: C, 2 and −3. A zero of p(x) is an x-value for which p(x)=0. The polynomial is already factorised, so use the zero-product property: a product is zero when at least one factor is zero. Set x−2=0, giving x=2. Set x+3=0, giving x=−3. These are the x-intercepts, or the points where the graph meets the x-axis; they can therefore be marked at 2 and −3 on the number line. Checking gives p(2)=0×5=0 and p(−3)=−5×0=0. A changes −3 to +3, B changes 2 to −2, and D changes both signs incorrectly. The sign warning is important: x−a gives zero a, while x+a gives zero −a.