Question

If remainder is same when polynomial p(x) = x^3 + 8x^2+17x+ax is divided by (x+2) and (x+1) . Find the value of "a".

Answer 1

p(x)=x³ + 8.x² +17.x + a.x

So,p(-2)= -8 + 32 - 34 - 2.a

p(-2)= -2.a - 10

And,p(-1)= -1 + 8 - 17 -a

p(-1) = -a -10

Remainders are same....so,

-2.a - 10= -a - 10

-a=0

a=0

Answer 2

Answer:

Given,

p(x) = (x³ + 8x² + 17x + ax)

First,

we will take value of x

Here two factors are given : (x+2) and (x+1)

(x+2) = 0

x = -2

(x+1) = 0

x = -1

Now,

According to the question,

[ p(x) = (x³ + 8x² + 17x + ax) divided by (x+2) and (x+1) the remainders ate same. ]

So,

p(-2) = p(-1)

(-2)³+ 8(-2)² + 17(-2) + a(-2) = (-1)³ + 8(-1)² + 17(-1) + a(-1)

=> -8 + 32 - 34 - 2a = -1 + 8 - 17 - a

=> -10 -2a = - 10 - a

=> -2a + a = -10 + 10

=> -a = 0

=> a = 0

Hence, the value of 'a' = 0

Hope it helps you....