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

Answer:

Factor theorum states that p( x ) = 0

x = 1, 2

p( x ) = x^3 -10x^2+ax+ b

0 = 1 - 10 + a + b

0 = - 9 + a + b ------(1)

0 = 2^3 - 10(2)^2 + a(2) + b

0 = 8 - 40 + 2a + b

0 = - 32 + 2a + b ------(2)

(1) - (2)

-9 + a + b

-( - 32 + 2a - b )

23 - a = 0

a = 23

keeping value of a in eq. 1

0 = - 9 + 23 + b

-14 = b