Question

Factorise:
-7x2 – 3y2 + 22xy – 44x + 12y - 12
​

Answer 1

Answer:

Hi buddy hope this will be helpful to you

Step-by-step explanation:

-7x2-3y2+22xy-44x+12y-12

-1(7x2-3y2)+22x(y-2)+12(y-1)

Answer 2

Answer:

The factorization of the given expression is

$-7x^{2}-3y^{2}+22xy-44x+12y-12=(7x-y+2)(3y-x-6)$

Step-by-step explanation:

The given second degree equation to be factorized is

$-7x^{2}-3y^{2}+22xy-44x+12y-12$

We can write this equation as a product of two linear factors as mentioned below

$-7x^{2}-3y^{2}+22xy-44x+12y-12=(a_{1}x+b_{1}y+1)(a_{2}x+b_{2}y+c)$

Multiplying both the factors the expression becomes as follows

$(a_{1}x+b_{1}y+1)(a_{2}x+b_{2}y+c)=a_{1}a_{2}x^{2}+a_{1}b_{2}xy+a_{1}cx+b_{1}a_{2}xy+b_{1}b_{2}y^{2}+b_{1}cy+a_{2}x+b_{2}y+c\\=a_{1}a_{2}x^{2}+b_{1}b_{2}y^{2}+(a_{1}b_{2}+b_{1}a_{2})xy+(a_{1}c+a_{2})x+(b_{1}c+b_{2})y+c$

Comparing the expression obtained above with the given expression,we get

$c=-12\\a_{1}a_{2}=-7= > a_{2}=\frac{-7}{a_{1}} \\b_{1}b_{2}=-3= > b_{2}=\frac{-3}{b_{1}} \\a_{1}b_{2}+a_{2}b_{1}=22-- > (1)\\-12a_{1}+a_{2}=-44--- > (2)\\-12b_{1}+b_{2}=12---- > (3)$

Solving equation (2),we get

$-12a_{1}+a_{2}=-44= > -12a_{1}+\frac{-7}{a_{1}} =-44\\= > 12a_{1}^{2}-44a_{1}+7=0\\= > a_{1}=\frac{7}{2} ,\frac{1}{6} === > a_{2}=-2,-42(\therefore\frac{-7}{a_{1}}=a_{2})\\we\:choose\:a_{1}=\frac{7}{2}$

Similarly,solving equation(3),we get

$-12b_{1}+b_{2}=12= > -12b_{1}+\frac{-3}{b_{1}} =12\\= > 12b_{1}^{2}+12b_{1}+3=0\\= > b_{1}=\frac{-1}{2} === > b_{2}=6(\therefore\frac{-3}{b_{1}}=b_{2})$

Substituting these values in the above calculated expression,we get

$(a_{1}x+b_{1}y+1)(a_{2}x+b_{2}y+c)=(\frac{7}{2}x-\frac{1}{2}y+1)(-2x+6y-12)\\=(\frac{7x-y+2}{2} )2(-x+3y-6)=(7x-y+2)(3y-x-6)$

Therefore,the factorization of the given expression is

$-7x^{2}-3y^{2}+22xy-44x+12y-12=(7x-y+2)(3y-x-6)$

#SPJ2