Question

15. The polynomials 2x^3 - 7x^2 + ax – 6 and
x^3 - 8x^2+ (2a + 1)x - 16 leave the same
remainder when divided by x - 2. Find the
value of 'a'.​

Answer 1

Answer:

as the remainders are equal, the both can be equalized

$2{x}^{3} - 7 {x}^{2} + ax - 6 = {x}^{3} - 8 {x}^{2} + (2a + 1)x - 16$

now,x-2=0

×=2 substitution this in the equation,

$2( {2})^{3} - 7( {2})^{2} + a(2) - 6 = \\ {2}^{3} - 8 ({2})^{2} + 4a + 2 - 16 \\ 16 - 28 + 2a - 6 = 8 - 32 + 4a - 14 \\ 2a - 4a = 8 - 32 - 14 - 16 + 28 + 6 \\ - 2a = - 20 \\ a = 10$

Answer 2

Answer:

yaa 10 only the answer to this question

Step-by-step explanation:

ok ok