Question

The polynomials kx^3 +3x^2-8 and3x^3-5x are divided by x +2. If the remainder in each case is the same then find the value of k.​

Answer 1

Step-by-step explanation:

Since x+2 is the divisor

therefore x+2=0

x=-2

putting the value in both the equation

=kx^3+3x^2-8

=k (-2)^3+3 (-2)^2-8

=-8k+12-8

=4-8k

now putting the value of x in the second equation

=3x^3-5x+k

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

=-24+10+k

=k-14

since both the remainders are equal

therefore. k-14=4-8k

=> 9k=18

=> k=(18/9)

=> k=2

Answer 2

Answer:

2

Step-by-step explanation:

since x+2 is the divisor, therefore x+2=0, x=-2

putting the value in both the equations

=kx^3+3x^2-8= k*(-2)^3+3*(-2)^2-8= -8k+12-8=4-8k

now putting the value of x in the second equation

=3x^3-5x+k=3*(-2)^3-5*(-2)+k=-24+10+k=k-14

since both the remainders are equal:

k-14=4-8k => 9k=18 => k=18/9= 2

k=2 is the answer