How to factorise $6x^2+5xy+y^2+x+2y-15$

Question

How to factorise $6x^2+5xy+y^2+x+2y-15$?

I've figured out that $$6x^2+5xy+y^2 = (2x+y)(3x+y).$$

But I do not know how I would include $x+2y-15$

$6x^2+5xy+y^2+x+2y-15$.

Answer 1

.Good try thus far. You got the factorization wrong, but it is $2x+y$ times $3x+y$, so the idea to factorize the first three terms was correct.

Write : $$ 6x^2 + 5xy + y^2 + x + 2y - 15 = (2x+y + a)(3x +y + b) $$

and now expand and compare coefficients: $$ 6x^2 + 5xy + y^2 + x + 2y - 15 = 6x^2 + 5xy+ y^2 + by + 2bx + 3ax + ay + ab $$

So what we know by comparing coefficients, is that $2b + 3a = 1$ and $b + a = 2$. This gives after solving that $a = -3$ and $b = 5$, also confirming the fact that $ab =-15$

Hence, the completed factorization is: $$ 6x^2 + 5xy + y^2 + x + 2y - 15 = (2x+y -3)(3x +y + 5) $$

Answer 2

You can always use the quadratic formula on $$ 6x^2+5xy+y^2+x+2y-15=6x^2+(5y+1)x+(y^2+2y-15) $$ The discriminant is $((5y+1)^2-6\cdot4\cdot(y^2+2y-15))$, which turns out to be $(y-19)^2$. Therefore, the roots are $$ x=\frac{-(5y+1)\pm(y-19)}{12} $$

Answer 3

Also we can use the following.

$$6x^2+5xy+y^2+x+2y-15$$ is a quadratic expression of $y$ and we can calculate the $\Delta$,

which must be $(ax+b)^2$ for some reals $a$ and $b$ because otherwise, we have an irreducible polynomial.

Answer 4

In general, it is searched in the form: $$6x^2+5xy+y^2+x+2y-15=(ax+by+c)(dx+ey+f).$$ Plugging $x=0$, it results in: $$y^2+2y-15=(by+c)(ey+f) \Rightarrow$$ $$(y-3)(y+5)=(by+c)(ey+f) \Rightarrow b=e=1,c=-3,f=5.$$ Plugging $y=0$, it results in: $$6x^2+x-15=(ax+c)(dx+f)\Rightarrow$$ $$(2x-3)(3x+5)=(ax+c)(dx+f) \Rightarrow a=2,d=3,c=-3,f=5.$$ The parameters correspond, thus: $$6x^2+5xy+y^2+x+2y-15=(2x+y-3)(3x+y+5).$$