Factoring Quadratic Trinomials that don't simplify by common terms.

Question

$10x^2 + 13xy - 3y^2$

$(10x^2 + 2xy) + (-15xy - 3y^2)$

$2x(5x + y) + 3y(-3x - y)$

How would this factor out further?

Answer 1

$10x^2+13xy-3y^2$

$=10(x^2+1.3xy-0.3y^2)$

$=10(x^2+2\times x \times 0.65y+0.4225y^2-0.7225y^2)$

$=10(x+0.65y)^2-(\sqrt{7.225}y)^2$

Then you can use the equality $a^2-b^2=(a-b)(a+b)$ to continue:

Next steps (spoilers):

$=\left(\sqrt{10}x+\sqrt{10}\times 0.65y+\dfrac{17\sqrt{10}}{20}y\right)\left(\sqrt{10}x+\sqrt{10}\times 0.65y-\dfrac{17\sqrt{10}}{20}y\right)$

Next step:

$=\sqrt{10}\left(x+ 0.65y+\dfrac{17}{20}y\right)\sqrt{10}\left(x+ 0.65y-\dfrac{17}{20}y\right)$

Finally:

$=10(x+1.5y)(x-0.2y)=(2x+3y)(5x-y)$

Answer 2

These trinomials are homogeneous and so can be reduced to quadratic polynomials in one variable. Here is how this works for the first example. The others are similar.

Let $y=xu$. Then $10x^2 + 13xy - 3y^2=x^2(10+13u-3u^2)=-x^2(3 u + 2) (u - 5)=-(3xu+2x)(xu-5x)=-(3y+2x)(y-5x)=(2x+3y)(5x-y)$.

Answer 3

There was an arithmetic error in your attempt.

$$10x^2 + 13xy - 3y^2 \neq (10x^2 + 2xy) + (-15xy - 3y^2),$$

because $2 + (-15) = -13,$ whereas you needed two coefficients that added to $+13.$

The correct calculation is \begin{align} 10x^2 + 13xy - 3y^2 &= (10x^2 - 2xy) + (15xy - 3y^2) \\ &= 2x(5x - y) + 3y(5x - y) \end{align} and I think you can finish it from there.