Question

Choose the correct option.
at the GCD of x^2(p-1) and x^2-px+(p-1) is x-1. then the value of p is​

Answer 1

Step-by-step explanation:

ब्रो प्लीज सर्च ऑन गूगल नॉट हियर

Answer 2

Given:

the GCD of x^2(p-1) and  is x-1.

To Find:

Value of p is?

Solution:

it is given that- (x -1) is the GCD of $x^2(p-1)$ and $x^2-px+(p-1)$.

therefore, on putting value of x= 1 , we get

⇒

$1^{2}(p-1)=0\\2(p-1)=0\\p=1$

and,

$(1)^2-p(1)-(p-1)=0\\1-p-p+1=0\\2-2p=0\\p=1$

therefore, value of p is 1.