Question

State whether the following statements are true of false. If they are true, give a short justification. If they are false, give a counterexample. For each of the following, P(x) is a polynomial.

(a) If P(x) has only even powers, and P(a) = 0 then x^2 ? a^2 divides P(x).

(b) If P(x) has only odd powers, and P(a) = 0 then x^2 ? a^2 divides P(x).

(c) If P(x) has only even powers, then P(x) has at least one real root.

(d) If P(x) = a7x^7 + a6x^6 + a3x^3 + a0, where ai ? R, ai is not equal to zero, then P(x) has at least one real root.

(e) If P(x) has only even powers, then P(x) has at least one complex root.

(f) If P(x) = a7x^7 + a6x^6 + a3x^3 + a0, where ai ? R, ai is not equal to zero , then P(x) has at least one complex root.

Answer 1

(a). The remainder theorem states that the remainder of the division of a polynomial {\displaystyle f(x)}P(x) by a linear polynomial {\displaystyle x-r}(x-a) is equal to P(a). If P(a) = 0,then (x-a) is a factor of P(x). Further, if P(x) has only even powers and P(a) = 0, then P(-a) = 0 so that (x +a) is also a factor of P(x). Thus, x² – a² divides P(x).

(b).False. (x-a)² is not a factor of x⁵-a⁵.

( c).False. The polynomial P(x) = x⁶ -2x⁴+3x2+5 has all 6 complex roots.

(d). True. If P(x) = a₇ x⁷ + a₆x⁶ + a₃x³ + a₀, where a_i ? R and a₇ is not equal to zero, then P(x) has 7 roots. Further, since complex roots occur in conjugate pairs, hence P(x) has at least one real root.

(e).False. If P(x) = x² – a², where a ? R , then P(x) has 2 real roots a and -a.

(f).False. If P(x) = a₇ x⁷ + a₆x⁶ + a₃x³ + a₀, where a_i ? R and a₇ is not equal to zero, then P(x) has 7 roots. All the 7 roots may be real.

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