Answer:
[tex](7m^2+10)^2[/tex]
Step-by-step explanation:
This is a quadratic in disguise. To visualize this better, let [tex]x=m^2[/tex]. We have:
[tex]49x^2+140x+100[/tex]
To factor, we can write out the format [tex](ax+y)(bx+z)[/tex]. We're looking for numbers [tex]a[/tex], [tex]b[/tex], [tex]y[/tex], and [tex]z[/tex] such that the following is true:
- [tex]a\cdot b=49[/tex]
- [tex]a\cdot z+b\cdot y=140[/tex]
- [tex]y\cdot z=100[/tex]
We find the following numbers:
[tex]a=7,\\y=10, \\b=7, \\z=10[/tex]
Thus, we have:
[tex](7x+10)(7x+10)[/tex]
Substitute back [tex]x=m^2[/tex] to get your final answer:
[tex](7m^2+10)(7m^2+10)=\boxed{(7m^2+10)^2}[/tex]
