Question

If $2x^2+xy-3y^2+x+ay-10=(2x+3y+b)(x-y-2)$ the value of $a$ and $b$ are

• A. $11and5$
• B. $1and-5$
• C. $-1and-5$
• D. $-11and5$
  

Answer 1

Answer: (D)

$-11and5$

Given equation:
$2x^2+xy-3y^2+x+ay-10=(2x+3y+b)(x-y-2)$
$\Rightarrow 2x^2+xy-3y^2+x+ay-10=2x^2+3xy+bx-2xy-3y^2-by-4x-6y-2b$
$\Rightarrow 2x^2+xy-3y^2+x+ay-10=2x^2+xy-3y^2+(b-4)x-(b+6)y-2b$
Comparing the coefficients of $x,y$ and constants will get:
$b-4=1..\left(i\right)$
$-(b+6)=a..\left(ii\right)$
$-2b=-10..\left(iii\right)$
From eq(1), we get
$\Rightarrow b=5$
Now put $b$ in eq(2), we get
$\Rightarrow -(b+6)=a$
$\Rightarrow a=-11$
From eq(3), we get
$\Rightarrow -2b=-10$
$\Rightarrow b=5$
Hence values of $a,b$ are $-11,5$