Question

Q8. The polynomials ax: + 3x2 - 3 and 2° -5x + a when divided by (x-4) leaves the
remainders R and r respectively. Find the value of a if
i) R=r
il) 2R-=0​

Answer 1

Answer:

Step-by-step explanation

First, we will find out the value of x

So,x-4=0

x=4

Then we will put the value 4 on the place of x

P(x) = ax^3+3x^2-3

P(4)=a(4)^3+3(4)^2-3

64a+48-3

64a+45 ------R

P(x) =2x^3-5x+a

P(4)=2(4)^3-5(4)+a

128-20+a

108-a-------r

There are two situation for finding a

First R=r

So, We have values of R1 and R2 . Therefore we will put the values of Randr

64a+45=108+a

64a-a=108+45

63a=6r

a=63/63

a=1

For 1st situation we find out the value of a that is 1

Now 2nd situation,

2R-r=0

So,

2(64a+45)-(108+a)=0

128a+90-108-a=0

127a-18=0

127a=18

a=18/127

So for second situation we find out the value of a that is 18/127

HOPE IT HELPS!