Question

If p(x) = x³ + x² – 9x – 9, find p(0), p(3), p(–3) and p(–1). What do you conclude about the zero of p(x)? Is 0 a zero of p(x)?

Answer 1

p(x) = x³ + x² – 9x – 9 .....(1)

Putting x = 0 in (1), we get

p(0) = 0³ + 0² – 9 × 0 – 9 = 0 + 0 – 0 – 9 = –9 ≠ 0

Thus, x = 0 is not a zero of p(x).

Putting x = 3 in (1), we get

p(3) = 3³ + 3² – 9 × 3 – 9 = 27 + 9 – 27 – 9 = 0

Thus, x = 3 is a zero of p(x).

Putting x = –3 in (1), we get

p(–3) = (–3)³ + (–3)² – 9 × (–3) – 9 = –27 + 9 + 27 – 9 = 0

Thus, x = –3 is a zero of p(x).

Putting x = –1 in (1), we get

p(–1) = (–1)³ + (–1)² – 9 × (–1) – 9 = –1 + 1 + 9 – 9 = 0

Thus, x = –1 is a zero of p(x).