p(x) = x² – 5x – 6, find the values of p(1), p(2), p(3), p(0), p(–1), p(–2), p(–3).
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Solution :
p(x) = x² - 5x - 6
i ) p(1) = 1² - 5(1) - 6
= 1 - 5 - 6
= 1 - 11
= -10
ii ) p(2) = 2²-5(2) - 6
= 4 -10 - 6
= -12
iii ) p(3) = 3² - 5(3) - 6
= 9 - 15 - 6
= -12
iv ) p(0) = 0² - 5(0) - 6 = -6
v ) p(-1) = (-1)² - 5(-1) - 6
= 1 + 5 - 6
= 0
v ) p(-2) = (-2)² - 5(-2) - 6
= 4 + 10 - 6
= 8
vi ) p(-3) = (-3)² - 5(-3) - 6
= 9 + 15 - 6
= 18
•••••
p(x) = x² - 5x - 6
i ) p(1) = 1² - 5(1) - 6
= 1 - 5 - 6
= 1 - 11
= -10
ii ) p(2) = 2²-5(2) - 6
= 4 -10 - 6
= -12
iii ) p(3) = 3² - 5(3) - 6
= 9 - 15 - 6
= -12
iv ) p(0) = 0² - 5(0) - 6 = -6
v ) p(-1) = (-1)² - 5(-1) - 6
= 1 + 5 - 6
= 0
v ) p(-2) = (-2)² - 5(-2) - 6
= 4 + 10 - 6
= 8
vi ) p(-3) = (-3)² - 5(-3) - 6
= 9 + 15 - 6
= 18
•••••
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