Question

find the cofficient of x^2 and x in the product of (x-3) (x+7) (x-4)​

Answer 1

Answer:

$product \: of \\ \\( x - 3)(x + 7)(x - 4) \\ \\ = (x ^{2} + 7x - 3x - 21)(x - 4) \\ \\ = (x ^{2} + 4x - 21)(x - 4) \\ \\ = x ^{3} + 4x ^{2} - 21x - 4x ^{2} - 16x + 84 \\ \\ = x ^{3} - 21x - 16x + 84 \\ \\ ( + 4x ^{2} \: and \: - 4x^{2} \: cancelled) \\ \\ = x^{3} - 37x + 84 \\ \\ so \: co \: efficent \: of \: x = 37 \\ \\ x^{2} = 0 \\ \\ x ^{3} = 1$

hope it helps uH ☺