Question

if p(x) = x^3+3x-2x^2-6 and q(x) = 3x^2-7x-8 then prove q(-1) is a zero of p (x).​

Answer 1

Step-by-step explanation:

$p(x) = {x}^{3} + 3x - 2 {x}^{2} - 6 \\ \\ q(x) = 3 {x}^{2} - 7x - 8 \\ \\ given \: q( - 1) \: is \: the \: zero \: of \: polynomial \: p(x). \\ \\ \\ q( - 1) = 3 \times { - 1}^{2} - 7 \times - 1 - 8 \\ \\ = 3 + 7 - 8 = 2 \\ \\ p(2) = {2}^{3} + 3 \times 2 - 2 \times {2}^{2} - 6 \\ \\ 8 + 6 - 8 + 6 = 0 \\ \\ hence \: proved$

Answer 2

Step-by-step explanation:

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