Question

Find the value of a and b so that x^3-10x^2+ax+b is exactly divisible by x-1 as well as x-2

Answer 1

Hello,


Since, there are two variables , we will get two equations from which we can solve the problem.

Now,
P(x) = x³-10x²+ax+b

The polynomial is divisible by x-1
So ,
By remainder theorem
P(1) =0
So , putting the values :-

P(1) = 1³-10*1²+a+b
Or ,
1-10+a+b = 0
a+b = 9..........(1)

Also ,
Polynomial is divisible by x-2
So ,again by remainder theorem
P(2) =0

So,
2³-10*2²+2a+b =0
8-40 +2a +b =0
2a +b = 32..........(2)


Now ,

Subtracting 1 from 2.
We get

2a +b - (a+b) =32-9
2a + b - a-b = 23
a = 23

Putting the value of a in equation 1 to obtain b
a+b =9

b = 9 - a = 9-23 = (-14)


Hope this will be helping you......

Answer 2

x - 1 and x -2 are factors of x³ - 10x² + ax +b 
So f(x) = x - 1
x - 1 = 0
x = 1
x³ - 10x² + ax + b = 0
(1)³ - 10(1)² + a(1) + b =0
1 - 10 + a + b = 0
a + b =  9 (Equation 1)

f(x) = x - 2
x - 2 = 0
x = 2
x³ - 10x² + ax + b =0
(2)³ - 10(2)² + a(2) + b = 0
8 - 40 + 2a + b = 0
2a + b - 32 = 0
2a + b = 32 (Equation 2)

Subtracting equation 2 from 1
a + b =9
- 2a - b = - 32

- a = - 23
a = 23

23 + b = 9
b = - 14

The values of a and b are 23 and -14

Hope this helps you.