Question

if the polynomials (2x^3+9x^2+3x-5) and (x^3+x^2-2x+a) leave the same remainder when divided by (x-2) find the value of a. Also,find the remainder in each case.

Answer 1

Answer:

a=45

Step-by-step explanation:

p(x)=2x^3+9x^2+3x-5

f(x)=x^3+x^2-2x+a

g(x) = x -2

x -2 =0

x=2

Putting the value of x in p(x) and f(x),

$2 {(2)}^{3} + 9 {(2)}^{2} + 3 \times 2 - 5 = {2}^{3} + {2}^{2} - 2 \times 2 + a \\ 16 + 36 + 6 - 5 = 8 + 4 - 4 + a \\ 53 = 8 + a \\ a = 45$

HOPE THIS WILL HELP YOU

Answer 2

Given x–2 = 0

SO, X = 2

IN CASE 1&2 .Put the value of X = 2

2x³+9x²+3x-5 = 0. x³+x²-2x+a =0

2(2)³+9(2)²+3(2)-5=0. (2)³+(2)²-2(2)+a =0

16 + 36 + 6 - 5 =0. 8+4-4+a =0

53 = 0. a = -8

Comparmising both Case

we get a = 53-5 = 45 IS THE ANSWER