Question

Find the factor: x2 - 2xy + y2 + 3x - 3y + 4 = 0.

Answer 1

Solution:

we know that

${x}^{2} - 2xy + {y}^{2} = ( {x - y)}^{2}$
use this identity to the given polynomial for factorisation

${x}^{2} - 2xy + {y}^{2} + 3x - 3y + 4 = 0 \\ \\ ( {x - y)}^{2} + 3(x - y) + 4 =0$

it is looking like a Quadratic equation in (x-y)

to simplify it,let us substitution
$x - y = k \\ \\ {k}^{2} + 3k + 4 = 0$
it cannot factors,so apply Quadratic formula

$k_{1,2} = \frac{ - 3 ± \sqrt{9 - 16}}{2} \\ \\ k_{1,2} = \frac{ - 3 ± \sqrt{ - 7} }{2} \\ \\ k_{1} = \frac{ - 3 + i \sqrt{7} }{2} \\ \\ k_{2} = \frac{ - 3 - i \sqrt{7} }{2}$

if you are in 11 standard you are familiar with complex roots.

So,factors are

$2(x - y) = - (3 + i \sqrt{7} ) \\ \\ 2(x - y) + 3 + i \sqrt{7} = 0 \\ \\ 2(x - y) + 3 - i \sqrt{7} = 0 \\ \\ [2(x - y) + 3 + i \sqrt{7}][ 2(x - y) + 3 - i \sqrt{7} ] = 0$
are the factors.

if you are in 10 or below 10 th standards than you might not Understand,for that you need to correct the question like

${x}^{2} - 2xy + {y}^{2} + 3x - 3y - 4 = 0$

${x}^{2} - 2xy + {y}^{2} + -3x + 3y + 4 = 0$