Question

The polynomial ax³+3x²+1 and 2x³-5x+a when divided by (x-2) leave the same remainder. The value of a is​

Answer 1

Answer :

By Remainder Theorem,

The zero of the polynomial x-2 is 2

So,

$p(x)=ax^3+3x^2+1$

$p(2) = a(2)^3 + 3(2)^2 + 1\\= 8a +12 +1\\= 8a +13 <---Remainder$

$q(x) = 2x^3 -5x +a$

$q(2)= 2(2)^3 -5(2) +a \\= 16 -10 +a \\= a-4<---Remainder$

Since the remainders are equal (as given in question) we can conclude that :

$8a +13 = a -4$

$8a - a = -4 -13$

$7a = -17$

$a = \frac{-17}{7}$