Hello!
The answer is:
[tex](3xy-1)(4xy+2)=12(xy)^{2}+2xy-2[/tex]
or
[tex](3xy-1)(4xy+2)=12x^{2}y^{2}+2xy-2[/tex]
Why?
To solve the problem and identify the resultant expression, we need to apply the distributive property, and the, add or subtract like terms.
Describing the distributive property:
[tex](a+b)(c+d)=a*c+a*d+b*c+b*d[/tex]
Remember, like terms are the terms that share the same variable and same exponent, for example:
[tex]a^{2} +3a^{2}+a^{3}=4a^{2}+a^{3}[/tex]
We were able to add only the terms that have the same exponent (2).
We are given the expression:
[tex](3xy-1)(4xy+2)[/tex]
Now, solving we have:
[tex](3xy-1)(4xy+2)=(3xy*4xy)+(3xy*2)-(1*4xy)-2\\\\(3xy*4xy)+(3xy*2)-(1*4xy)-2=12(xy)^{2}+6xy-4xy-2\\\\12(xy)^{2}+6xy-4xy-2=12(xy)^{2}+2xy-2[/tex]
Hence, we have that:
[tex](3xy-1)(4xy+2)=12(xy)^{2}+2xy-2[/tex]
or
[tex](3xy-1)(4xy+2)=12x^{2}y^{2}+2xy-2[/tex]
Have a nice day!
