Question

If the polynomials(2x3+ax2+3x-5)and (x3+x2-7x-6) leave the same remainder when divided by (x-2) find the value of a

Answer 1

ATQ, the polynomial (2x³ + ax² + 3x - 5)and (x³ + x² - 7x - 6) leave the same remainder when divided by (x - 2)

➡ p(2) = (2)³ + (2)² - 7(2) - 6

= 8 + 4 - 14 - 6

= 12 - 20

= -8

therefore p(2) = 2(2)³ + a(2)² + 3(2) - 5 = -8 (since remainder of both polynomial is same when divided by (x - 2))

➡ 16 + 4a + 6 - 5 = -8

➡ 17 + 4a = -8

➡ 4a = -8 - 17

➡ 4a = -25

➡ a = -25/4

➡ a = -6.25

Answer 2

Answer:

-25/4

Step-by-step explanation:

let p(x) = 2x³+ax²+3x-5

  & q(x) = x³+x²-7x-6

given, p(x) & q(x) leaves the same remainder when divided by (x-2)

⇒ p(2) = q(2)

⇒  2(2)³+a(2)²+3(2)-5  =  (2)³+(2)²-7(2)-6

⇒  16 + 4a + 6 - 5  = 8 + 4 - 14 - 6

⇒ 17 + 4a = -8

⇒ 4a = -8 - 17

⇒ a = -25/4