if p(x)=x³+x²-9x-0, then find p(0), p(3), p(-3), p(-1)
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Answered by
2
P(0)=9
P(3)=9
P(-3)=9
P(-1)=9
P(3)=9
P(-3)=9
P(-1)=9
Answered by
1
p(0)=0
p(3)=3^3+3^2-9(3)-0= 27+9-27-0 = 9
p(-3)= (-3)^3+(-3)^2-9(-3)-0 = -27+9+27-0 = 9
p(-1) = (-1)^3+(-1)^2-9(-1)-0 = -1+1+9-0 = 9
p(3)=3^3+3^2-9(3)-0= 27+9-27-0 = 9
p(-3)= (-3)^3+(-3)^2-9(-3)-0 = -27+9+27-0 = 9
p(-1) = (-1)^3+(-1)^2-9(-1)-0 = -1+1+9-0 = 9
paramjeetkaursaggi:
divide p(x) by g(x) p(x)=3x4-4x³-3x-1 and g(x)=x-2
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