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)?

Question

Answer

Answer

Solution:

Given: The polynomial (x3 - 10x2 + ax + b) is exactly divisible by (x - 1) and (x - 2).

Step 1: Find the Remainder When Dividing by (x - 1)

When a polynomial is divided by x - a, the remainder is the value of the polynomial when x = a. So, to find the remainder when dividing (x3 - 10x2 + ax + b) by (x - 1), we substitute x = 1:

R(1) = (13 - 10a + b)

Therefore, (x - 1) divides (x3 - 10x2 + ax + b) if and only if R(1) = 0.

Step 2: Find the Remainder When Dividing by (x - 2)

Similarly, to find the remainder when dividing (x3 - 10x2 + ax + b) by (x - 2), we substitute x = 2:

R(2) = (2a + 2b - 28)

Therefore, (x - 2) divides (x3 - 10x2 + ax + b) if and only if R(2) = 0.

Step 3: Set Up and Solve Equations

To satisfy the conditions of the problem, we need to find values of a and b such that:

R(1) = 0

R(2) = 0

Substituting the expressions for R(1) and R(2) from steps 1 and 2, we get:

13 - 10a + b = 0

2a + 2b - 28 = 0

Solving these two equations simultaneously, we get:

a = 9

b = -4

Therefore, the values of a and b that satisfy the conditions of the problem are a = 9 and b = -4.

Answer

The first x =1 so the put of 1st f (x)  = x3-10×2 +ax +b so f(1) = 1 ka cube -20+ a×1+b =1-20+a+b =19+a+b =19-a=b the first and 2nd x=2 hoga Toh Ussko bhi aaise put kr dena toh likhne put the value of (11)-(1) we have so a jitna likhlega Ussko b wala put kr dena ho jayega slove

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