Question

the polynomial kx^3+3x^2-3 and 2x^3-5x+k leave the same remainder when divide by (x-4) find k​

Answer 1

Answer:

Step-by-step explanation:

let

p(x)=kx³+3x²-3

f(x)=2x³-5x+k

g(x)=x-4

now

p(4)=k×(4)³+3×(4)²-3

     =64k+48-3

     =64k+45

f(4)=2×(4)³-5×4+k

    =128-20+k

    =108+k

A.T.Q.

64k+45=108+k

64k-k=108-45

63k=63

k=1

Answer 2

Answer:

k=1

Step-by-step explanation:

x-4=0,x=4

p(x) = kx^2 +3x^2 -3

p(4) = k(4)^3 + 3 x (4)^2 - 3

= 64k + 3(16) - 3

=64k + 45

p(4) = 2x^3 - 5x +5

= 2(4)^3 - 5(4) +k

= 128 -20 + k

= k +108

so,

=64k + 45 = k + 108

= 64k - k = 108+45

= 63k = 63

= k  = 63/63

k=1