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

 ANS: Let p(x) = x³ - 10x² +ax +b Using factor theorem we find that if p(x) is divided by x-1 the remainder, i.e. p(1) = 0

Therefore, p(1) = (1)³ - 10(1)² + a(1) +b = 0

=> 1 - 10 +a +b =0

=> a+ b = 9=> b = 9-a .........equation (1) 
In the same way if p(x) is divided by x-2 the remainder, i.e. p(2) = 0 Therefore, p(2) = (2)³ - 10(2)² + a(2) +b = 0

=> 8 - 40 +2a +b =0=> -32+2a+b=0...........equation (2)

Substituting the value of equation(1) in equation(2) we get, -32 + 2a+9 -a = 0

=> a = 23

Substituting the value of a in equation (1) we get b= 9- 23

=> b = -14

Hence, a=23 and b=-14

 ANS: Let p(x) = x ³ + 10x ² +ax +b Using factor theorem we find that if p(x) is divided by x-1 the remainder, i.e. p(1) = 0

Therefore, p(1) = (1)³ + 10(1)² + a(1) +b = 0

=> 1 + 10 +a +b =0

=> a+ b = 11=> b = 11-a .........equation (1) 
In the same way if p(x) is divided by x-2 the remainder, i.e. p(2) = 0 Therefore, p(2) = (2)³ + 10(2)² + a(2) +b = 0

=> 8 + 40 +2a +b =0=> 48+2a+b=0...........equation (2)

Substituting the value of equation(1) in equation(2) we get, 48 + 2a + 11 - a = 0

=> a = -59

Substituting the value of a in equation (1) we get b= 9- (-59)

=> b = 68

Hence, a=-59 and b=68

 sorry again