Value of P for which (x2 - 10x + P) is exactly divisible by (x - 2) is
Answers
Aim :-
- To find the value of P in the polynomial x² - 10x + P
Given :-
- p(x) = x² - 10x + P is exactly divisible by g(x) = (x-2)
Using factor theorem,
g(x) = x - 2
=≥ x - 2 = 0
=≥ x = 2
Substituting x for 2,
p(2) = (2)² - 10(2) + P
=≥ 4 - 20 + P
=≥ -16 + P
In order to satisfy the theorem, P should take a value such that the sum is 0.
Hence P = 16
Verification :-
Let us substitute and verify the answer.
p(x) = x² - 10x + 16
=≥ x² - 8x - 2x + 16
=≥ x(x - 8) - 2(x - 8)
=≥ (x-8)(x-2)
Hence verified.
We get (x-2) as one of the factors of the quadratic equation. Therefore, we can conclude that the value of P is 16.
[The value of x can also be 8 as (x-8) is also a factor of the polynomial]
The roots of the polynomial are :- 8,2
Given : (x^2 - 10x + P) is exactly divisible by (x-2)
To find : Value of p
Solution:
Long division method :
x - 8
(x-2) _| (x^2 - 10x + P) | _
x² - 2x
________
-8x + p
-8x + 16
______
p - 16
Remainder must be zero hence
p - 16 = 0
=> p = 16
Method 2 :
divisible by x - 2 hence x - 2 is a factor
x - 2 = 0 => x = 2
Substituting x = 2 in expression we must get 0
Hence 2² - 10(2) + p = 0
=> 4 - 20 + p = 0
=> p = 16
Hence Value of p is 16
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