find the factor x2–2xy+y2+3x–3y–4

Question

find the factor
x2–2xy+y2+3x–3y–4

Answer 1

x²-2xy + y² + 3x - 3y -4

 

If you recall the formula for (a-b)² = a²-2ab + b²

This fits the first part of your equation if a=x and b=y

 

x²-2xy+y² = (x-y)²

 

So we have:

(x-y)² + 3x- 3y - 4

 

You can factor a 3 out of the 3x-3y giving 3(x-y)

Giving:

 

(x-y)² + 3(x-y) -4

 

What we have is actually a quadratic equation of the form

ax² + bx + c   where   x = (x-y)

 

To look at this in an easier format pick a variable to

represent x-y ...   Let  z = x-y    so you have:

 

z² +3z -4

This can be factored.  The factors of -4 that add to 3 are (4)(-1) so

 

z² + 3z - 4

= (z+4)(z-1)

 

But z = x-y   so we have:

 

(x-y+4)(x-y-1)

 

 

Answer 2

 

Hi Faraz,

 

x² - 2xy + y² + 3x - 3y - 4

 

     = (x² - 2xy + y²) + 3(x - y) - 4

 

     = (x - y)² + 3(x - y) - 4

 

     = A² + 3A - 4                       (let A = x - y)

 

     = (A - 1)(A + 4)

 

     = (x - y - 1)(x - y + 4)         (put x - y back to A)