Question

Factorize `x^2+6xy+81y^2=` ?

Answer 1

`x^2+6xy+81y^2 = (x+(3-6sqrt(2)i)y)(x+(3+6sqrt(2)i)y)`

Explanation:

The task of factoring `x^2+6xy+81y^2` is very similar to that of factoring `x^2+6x+81`.

Putting `a=1`, `b=6` and `c=81`, we find that the discriminant `Delta` is negative:

`Delta = b^2-4ac = 6^2-4(1)(81) = 36 - 324 = -288`

Since `Delta < 0` this quadratic has no factors with Real coefficients. We can factor it using Complex coefficients, for example by completing the square...

`x^2+6xy+81y^2 = x^2+6xy+9y^2+72y^2`

`color(white)(x^2+6xy+81y^2) = x^2+2x(3y)+(3y)^2+(6sqrt(2)y)^2`

`color(white)(x^2+6xy+81y^2) = (x+3y)^2-(6sqrt(2)iy)^2`

`color(white)(x^2+6xy+81y^2) = ((x+3y)-6sqrt(2)iy)((x+3y)+6sqrt(2)iy)`

`color(white)(x^2+6xy+81y^2) = (x+(3-6sqrt(2)i)y)(x+(3+6sqrt(2)i)y)`

Answer 2

No real first degree factorization is possible.

Explanation:

Considering the possibility of

`x^2+6xy+81y^2=(ax+by)(cx+dy)`

If `x^2+6xy+81y^2=0`

would be equivalent to

`(ax+by)(cx+dy)=0` but if we solve

`x^2+6xy+81y^2=0` then

`x = (-3 pm 6 i sqrt[2]) y`

which means that the only solution to `x^2+6xy+81y^2=0` is `x =0, y=0`. So no real factorization for `x^2+6xy+81y^2` of first degree is possible.