Question

$p(1) = 0 \\ p( - 2) = 0$
write down p(x) polynomial second degree
​

Answer 1

Answer:

what is the proper question

Answer 2

$let \: p(x) = a {x}^{2} + bx + c \\ now \: according \: to \: question \\ p(1) = 0 \\ therefore \\ p(1) = a \times {1}^{2} + b \times 1 + c \\ a + b + c = 0 \: \: \: \: \: \: \: \: \red{(1)}\\ now \\ p( - 2) = 0 \\ a \times ( - 2) {}^{2} + b \times ( - 2) + c = 0 \\ 4a \: - 2b + c = 0 \: \: \: \: \: \: \: \: \: \ \red{(2)} \\ a + b - 4a + 2b = 0 \\ 3b - 3a = 0 \\ a = b \\ now \: from \: eq(1) \\ a + b + c = 0 \\ 2a + c = 0 \\ c = - 2a \: \: \: \: \: \: \red{(3)} \\ \bold \green{now}\: \: \bold \red{p(x) = a {x}^{2} + b(x) + c} \\ \\ p(x) = a {x}^{2} + a(x) - 2a \\ 0 = {a {x}^{2} + a(x) - 2a } \\ \boxed{x {}^{2} + x - 2 = 0} \: \bold \red{this \: is \: polynomial \: equation} \\ \green{till \: then \: be \: happy}$