Question

using remainder theorem, find the remainder when 3x^4-4x^3-1 by x-1​

Answer 1

Answer:

16

Step-by-step explanation:

p(x)=3x^4-4x^3-1 g(x)= x-1 =0=x=1

so by substituting the value of x in p(x)

p(1)=3*1^4-4*1^3-1

p(1)=3^4-4^3-1

p(1)= 81-64-1

p(1)=81-65

p(1)=16 ans

Answer 2

Answer:

Step-by-step explanation:

$p(x) =3x^{4} - 4x^{3} -1g(x) = x-1\\since \\g(x) \\is \\a \\factor \\of \\p(x)\\therefore g(x) = 0\\=> x-1 = 0\\x = 1\\\\p(1) =3(1)^{4} - 4(1)^{3} -1\\ =(3)^{4} - (4)^{3} -1\\ = 81 - 64 - 1\\ = 81 - 65\\ = 16$

therefore the remainder is 16