Question

find the values of of p(1),p(-1),p(2),p(-2),p(0) in p(x)= x2-7x-4​

Answer 1

Answer:

$p(x) = {x}^{2} - 7x - 4 \\ p(1) = {1}^{2} - 7(1) - 4 \\ = = > 1 - 3 = - 2 \\ p( - 1) = {( - 1)}^{2} - 7( - 1) - 4 \\ = = > 1 + 7 - 4 = 1 \\ p(2) = {2}^{2} - 7(2) - 4 \\ = = > 4 - 14 - 4 = - 14 \\ p( - 2) = {( - 2)}^{2} - 7( - 2) - 4 \\ = = > 4 + 14 - 4 = 14 \\ p(0) = {0}^{2} - 7(0) - 4 \\ = = > 0 - 0 - 4 \\ = - 4$