Question

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

Answer 1

Hello Mate!

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

Hope it helps☺!✌

Answer 2

Let,

=> x-1 = 0
=> x = 1

Similarly,

=> x = 2

Now, We will put these both value of x in two equations :-.

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

=> P(1) = (1)^3 -10(1)^2+a x 1 + b

=> 0 = 1 - 10 + a+b

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

Again,

=>P(2) = (2)^3 - 10(2)^2 + 2a +b

=> 0 = 8 - 40 + 2a + b

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

Eq (1) - (2)

-9 + a + b = 0

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

-23 + a = 0

=> a = 23

Putting value of 'a' in equation 2.

=> 0 = -32 + 2 x 23 + b

=> -b = -32+46

=> -b = 14

=> b = -14

Again,

Putting value of 'a' in equation 1.

=> 0 = -9 + a + b

=> 0 = -9 + 23 + b

=> -b = 14

=> b = -14