Question

Divide (6x^2-y^2-z^2+xy+xz-2yz) by (2x+y+z)​

Answer 1

Given: The equation is$(6x^2-y^2-z^2+xy+xz-2yz) \div (2x+y+z)$.

We have to solve the above equation.

• By using the Bodmas rule, we are solving the above equation.
• As we know that the Bodmas rule is used to remember the order of operations to be followed while solving expressions in mathematics.

Where,

$\begin{array}{l}\mathrm{B}=\text{brackets}\\\mathrm{O}=\text { order of powers or rules } \\\mathrm{D}=\text { division } \\\mathrm{M}=\text { multiplication } \\\mathrm{A}=\text { addition } \\\mathrm{S}=\text { subtraction }\end{array}$

We are solving in the following way:

We have,

$(6x^2-y^2-z^2+xy+xz-2yz) \div (2x+y+z)$

$= \frac{6 x^{2}-y^{2}-z^{2}+x y+x z-2 y z}{(2 x+y+z)}\\\\=\frac{6 x^{2}-y^{2}-z^{2}+x y+x z-2 y z}{2 x+y+z}\\\\=\frac{6 x^{2}-y^{2}-z^{2}+x y+x z-2 y z}{2 x+y+z}$

Hence, the solution of the equation is$\frac{6 x^{2}-y^{2}-z^{2}+x y+x z-2 y z}{2 x+y+z}$.