Question

If x3+ax2-bx+10 is divisible by x2-3x+2 find the values of a and b

Answer 1

Factorising x² - 3x + 2 we get


[Splitting the middle term]

x² - 2x - x + 2 = 0

=> x (x - 2) -1( x - 2) = 0

=> (x - 1)(x - 2) = 0

This means x - 1 and x - 2 are factors of
x³ + ax² - bx + 10

at x - 1 = 0
we get x = 1

Putting the value we get

(1)³ + a(1)² - b(1) + 10 = 0 (since it is a factor)

=> 1 + a - b + 10 = 0

=> a - b = -10 -1
=> a - b = -11.............(i)

Now at x - 2

x - 2 = 0

=> x = 2

Putting the value we get

(2)³ + a(2)² - b(2) + 10 = 0

=> 8 + 4a - 2b + 10 = 0

=> 4a - 2b + 18 = 0

=> 4a - 2b = -18

=> 2(2a - b) = -18

=> 2a - b =-18/2

=> 2a - b = -9.......... (ii)

Subtracting (i) from (ii) we get

2a - b -(a - b) = -9 - (-11)

=> 2a- b - a + b = -9 + 11

=> a = 2


Now putting the value of a in (i)

2 - b = -11

=> -b = -11 - 2

=> -b = -13

=> b = 13

a = 2 and b = 13


Hope it helps dear friend ☺️✌️

Answer 2

(a,b)≡(2,13)

Consider x2−3x+2.

This can be factorised as (x−1)(x−2).

So, x3+ax2−bx+10 has to be divisible by both (x−1) and (x−2).

By synthetic division, on dividing with (x−1), we get the remainder as a−b+11.

So, we have the first equation: a−b+11=0

And on dividing by (x−2), we get the remainder as 4a−2b+18.

So, the second equation: 2a−b+9=0

So, a+11=2a+9⇒a=2 and b=13.

I hope this helps you. Let us know if there is any confusion!