Question

When polynomial 3x square+ax-6 and x cube +2x square -3x+a are divided by x-2 if the remainder remains the same in each case find the value of a polynomials?

Answer 1

Answer:

4.

Step-by-step explanation:

Let the remainder be k.

Since when 3x²+ax-6 when divided by x-2 leaves remainder k,

Let f(x) = 3x²+ax-6

f(2) = k => 3(2)²+2a - 6 = k

=> 2a + 6 = k. -------------------> equation 1

Since x³ +2x² -3x + a when divided by x-2 leaves remainder k,

Let g(x) = x³ +2x² -3x + a

g(2) = k => (2)³ + 2(2)² - 3(2) +  a = k

=> a + 10 = k -------------------> equation 2.

Since remainders are same in each case  the equations are equal.

2a + 6 = a + 10

=> a = 10 - 6

=> a = 4