Question

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

Answer 1

x^3-10x^2+ax+b

x-1=0

x=1

Put value of x in the equation

1^3-10(1)^2+a(1)+b

1-100+a+b

a+b-99 =0---------------------------------(1)

Now putting x value =2

2^3-10(2)^2 +a(2)+b

8-400+2a+b

2a+b-392=0-------------------------------(2)

Using 1 and 2 equation we get

a+b-99=0

2a+b-392

-a+293=0

-a=-293

Then,

a =293

Put this value in (1)equation

293+b-99=0

194+b=0

b=-194

Hence a=293,b=-194

Hope it will help u

Answer 2

Step-by-step explanation:

x - 1 = 0 x - 2 = 0

x = 1 x = 2

P(x) = x³ - 10x² + ax + b

P(1) = 1 - 10 + a + b

=> a + b = 9 _(i)

P(2) = 8 - 40 + 2a + b

=> 2a + b = 32 _(ii)

Solving (i) & (ii) , we get

• a = 23
• b = -14