Question

2x^2+3xy+y^2+5x+2y-3=0

Answer 1

You didn't really ask a question, but that's a pretty interesting conic.

This is the rare conic that factors on the left:

$2 x^2 + 3 x y + y^2 + 5 x + 2 y - 3 = (x + y + 3) (2 x + y - 1) = 0$

So this is really two lines,

$x + y + 3 = 0 \textrm{ and } 2x + y - 1 = 0$

For fun, let's calculate the meet of those two lines:

$y = -x -3$

$2x + -x -3 - 1 = 0$

$x = 4$

$y = -4 -3 = -7$

This conic is two lines which meet at (4,-7)

Answer 2

The factors of the equation are 4 and -7.

Given

Equation; $\rm 2x^2+3xy+y^2+5x+2y-3=0$

What is the factor theorem?

Factor theorem is used when factoring the polynomials completely.

Therefore,

The factors of the polynomial are;

$\rm 2x^2+3xy+y^2+5x+2y-3=0\\\\2x^2+3xy+5x+y^2+2y-3=0\\\\ 2x^2+xy-x+2xy+y^2-y+6x+3y-3=0\\\\ (x+y+3)(2x+y-1)=0$

On solving the equations

$\rm x+y=-3\\\\2x+y=1$

On subtracting both the equation

$\rm x+y-2x-y=-3-1\\\\-x=-4\\\\x=4$

Substitute the value of x in the equation first

$\rm x+y=-3\\\\4+y=-3\\\\y=-3-4\\\\y=-7$

Hence, the factors of the equation are 4 and -7.

To know more about the Factor theorem click the link given below.

https://brainly.com/question/12959513