Question

for what value of k,[-4] is a zero of the polynomial x²-x[2k+2],answer it with steps with common method i will mark you as brainliest and send thanks ..

Answer 1

Answer:

9

Step-by-step explanation:

p(x) = x² - x - [ 2k + 2 ]

p(x) = x² - x - [ 2k + 2 ]

p(x) = x² - x - [ 2k + 2 ] p(-4) = [-4]² - [-4] - [ 2k + 2 ]

p(x) = x² - x - [ 2k + 2 ] p(-4) = [-4]² - [-4] - [ 2k + 2 ] = 16 + 4 - 2k - 2 = 0

p(x) = x² - x - [ 2k + 2 ] p(-4) = [-4]² - [-4] - [ 2k + 2 ] = 16 + 4 - 2k - 2 = 0 = 20 - 2k - 2 = 0

p(x) = x² - x - [ 2k + 2 ] p(-4) = [-4]² - [-4] - [ 2k + 2 ] = 16 + 4 - 2k - 2 = 0 = 20 - 2k - 2 = 0 18 - 2k = 0

p(x) = x² - x - [ 2k + 2 ] p(-4) = [-4]² - [-4] - [ 2k + 2 ] = 16 + 4 - 2k - 2 = 0 = 20 - 2k - 2 = 0 18 - 2k = 018 = 2k

p(x) = x² - x - [ 2k + 2 ] p(-4) = [-4]² - [-4] - [ 2k + 2 ] = 16 + 4 - 2k - 2 = 0 = 20 - 2k - 2 = 0 18 - 2k = 018 = 2kk = 18/2

p(x) = x² - x - [ 2k + 2 ] p(-4) = [-4]² - [-4] - [ 2k + 2 ] = 16 + 4 - 2k - 2 = 0 = 20 - 2k - 2 = 0 18 - 2k = 018 = 2kk = 18/2k = 9

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