Question

If the polynomials 2x³+ax²+3x-5 and x³+x²-4x+a leave the same remainder when divided by x-2, find the value of a.

Answer 1

Given : If the polynomials 2x³ + ax² + 3x - 5 and x³ + x² - 4x + a  when divided by (x - 2) leave the same remainder.

 

Let p (x) = 2x³ + ax² + 3x - 5 and q (x) = x³ + x² - 4x + a    be the given polynomials. The remainders when p(x) and q(x) are divided by (x - 2) are p (2) and q (2) .

By the given condition, we have :

p(2) = q(2)

⇒ 2 (2)³ + a (2)² + 3 (2) – 5 = (2)³ + (2)² – 4 (2) + a

⇒ 2 × 8 + a × 4 + 6 - 5 = 8 + 4 - 8 + a

⇒ 16 + 4a + 1 =  4 + a

⇒ 17 + 4a =  4 + a

⇒ 4a - a = 4 - 17

⇒ 3a  = - 13

⇒ a = - 13/3

 Hence, the value of 'a' is - 13/3.

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Answer 2

Answer:

x-2 = 0

=> x = 2

Remainder when 2x³+ax²+3x-5 is divided by x-2:

p(2) = 2(2)³+a(2)²+3(2)-5

=> 16 + 4a + 6 - 5

=> 17 + 4a

Remainder when x³+x²-4x+a is divided by x-2:

p'(2) = (2)³+(2)²-4(2)+a

=> 8 + 4 - 8 + a

=> 4 + a

Remainders are equal. Thus:

=> 17 + 4a = 4 + a

=> 13 = -3a

=> a = -13/3

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