Question

if the polynomial p(x)=x3+8x2+17x+ax is divided by (x+2) and (x+1) the remainders are same find the value of a.

Answer 1

Q. If the polynomial p(x) = (x³ + 8x² + 17x + ax) divided by (x+2) and (x+1) the remainders ate same. Find the value of a.

Solution :

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

Answer 2

Given that,

P(x)=x3+8x2+17x+ax

Is divided by (x+2)and(x+1)

And the remainder is same then,

X+2=0

So,x=-2

X+1=0

So,x=-1

A/Q

P(-2)=p(-1)

Now putting the equation,,,

(-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=-10-2a

-10+10=-a+2a

There fore,a=0