Question

find the values of "a" and "b" so that the polynomial x3+10x2+ax+b exactly divisible by (x-1) as well as (x-2)

Answer 1

x - 1 = 0

x = 1

p (1) = (1)^3 + 10 (1)^2 + a(1) + b

0 = 1 + 10 + a + b

-11 = a + b ... (1)

x - 2 = 0

x = 2

p (2) = (2)^3 + 10(2)^2 + 2a + b

0= 8 + 40 + 2a + b

-48 = 2a + b

b = - 48 - 2a
from 1
-11 = a + ( -48 - 2a)

-11 + 48 = -a

a = -37

b = -48 - 2a = -48 -(- 74) =. -48+74 = 26

Answer 2

p(x)=x^3+10x^2+ax+b
g(x)=x-1,x-2
       x=1,2
p(1)=1^3+10(1)^2+a+b
0=1+10+a+b
-11=a+b ..........................eq1
p(2)=(2)^3+10(2)^2+2a+b
      =8+40+2a+b               
-48=2a+b   ..............eq2
 eq 1-2
a+b-2a-b=48-11
-a=37
a=-37
put a=-37 in eq 1
-37+b=-11
b=37-11
b=26