Question

x^2-y^2+2y-1 (Show full process.)

Answer 1

: (x-y-1)(x+y+1)

By using a2−b2=(a−b)(a+b)

The only thing which is different from this is you must have to arrange it into this form. So below are the steps.

x2−(y2+2y+1) ==>we take - common

x2−(y+1)2==>(a2+2ab+b2)=(a+b)2

(x−(y+1))(x+(y+1)) =>applying the main rule a2−b2=(a−b)(a+b)

(x−y−1)(x+y+1) =>simplyfying the answer

Answer 2

Answer: (x+y-1)(x-y+1)

Given: x²-y²+2y-1

To find: The factors of the given equations.

Explanation:

$\sf We\:know\:that\:,\\(a^2-b^2)=(a+b)(a-b)\quad ----(1)\\\\Let\:a=x\:and\:b=y-1,\\Substituting\:the\:values\:of\:a\:and\:b\:in\:the\:given\:equation(1),\:we\:get,\\=(x^2-(y-1)^2)\qquad\qquad [Treating\:y-1\:as\:y]\\=(x+y-1)(x-y+1)\\\\\bold Therefore, the \;answer\;is\;\mathbf(x+y-1)(x-y+1).$