Question

find the sum of 3x^2-2x+9 and the quotient of xyz(x+y)(y+z)(z+x) divided by xy^2+xyz​

Answer 1

Answer:

Step-by-step explanation:

Take 2nd term xyz(x+y) (y+z) (z+x) /(xy^2+xyz) =xyz(x+y) (y+z) (z+x) /xy(y+z)

On solving, =z(x+y) (z+x) =z(xz+yz+x^2+xy) =xz^2+yz^2+x^2z+xyz

Add3x^2-2x+9 and xz^2+yz^2+x^2z+xyz, we get

=x^2(3+z) +x(z^2-2+yz) +yz^2+9

Answer 2

Given:

$3x^2-2x+9$  and

$\frac{xyz(x+y)(y+z)(z+x)}{xy^2+xyz}$

To find:

Sum of $3x^2-2x+9$ and quotient of $\frac{xyz(x+y)(y+z)(z+x)}{xy^2+xyz}$

Solution:

Quotient=$\frac{xyz(x+y)(y+z)(z+x)}{xy^2+xyz}$

             =$\frac{xyz(x+y)(y+z)(z+x)}{xy(y+z)}$

              =$z(x+y)(z+x)$

Quotient=$z(xz+x^2+yz+xy)$

Quotient=$xz^2+x^2z+yz^2+xyz$

Sum of $3x^2-2x+9$ and $xz^2+x^2z+yz^2+xyz$

=$3x^2-2x+9+xz^2+x^2z+yz^2+xyz$

=$x^2(3+z)+x(-2+z^2+yz)+yz^2+9$