Question

Find the values of ‘a’ & ‘b’ when x3 – 6x2 + ax + b is exactly divisible by x-1 and x-2

Answer 1

Question:-

• Find the values of a and b when x³-6x²+ax+b=0 is exactly divisible by x-1 and x-2

Solution:-

We have two value of x

i) x - 1= 0

x = 1

ii) x - 2= 0

x= 2

Now put the value of x on given equation:-x³-6x²+ax+b=0

Put x=1 on given equation

=> (1)³ - 6×(1)² + a × 1 +b=0

=> 1 - 6 + a + b = 0

=> a + b = 5 .............(i) eq

Now put the x= 2 on given equation

=> (2)³ - 6 × (2)² + a × 2 + b = 0

=> 8 - 6 × 4 + 2a + b = 0

=> 8 - 24 + 2a + b = 0

=> 2a + b = 16 ..............(ii) eq

Now we have 2 unknown variables and two equation ,

use substitution method

a + b = 5 ...........(i) eq

2a + b = 16 ........... (ii)

Take (i)st equation find value of a , we get

a + b = 5

a = 5 - b

Now put the value of a on second equation

2a + b = 16

2(5 - b ) + b = 16

10 - 2b + b = 16

10 - b = 16

- b = 16 - 10

b = - 6

Put the value of b on a, we get

a = 5 - ( -6)

a = 5 + 6

a= 11

Value of a = 11 and b = -6

Answer 2

Step-by-step explanation:

Answer:

Here one root is x = 1 so (x - 1) is a factor. i.e

{x}^{3} - {x}^{2} - 3 {x}^{2} + 3x + 2x - 2 =x

3

−x

2

−3x

2

+3x+2x−2=

{x}^{2} (x - 1) - 3x(x - 1) + 2(x - 1) =x

2

(x−1)−3x(x−1)+2(x−1)=

(x - 1)( {x}^{2} - 3x + 2) =(x−1)(x

2

−3x+2)=

(x - 1)( {x}^{2} - 2x - x + 2) =(x−1)(x

2

−2x−x+2)=

(x - 1)(x(x - 2) - 1(x - 2)) =(x−1)(x(x−2)−1(x−2))=

(x - 1)(x - 2)(x - 1) = (x - 1) {}^{2} (x - 2)(x−1)(x−2)(x−1)=(x−1)

2

(x−2)