Assuming that you want the product of the two factors given in each problem:
⇒ we solve this problem by timing each unit of a factor to every unit
of the other factor
⇒then add each one of them who shares a common like-term
⇒to get the answer
Let's solve:
![(4m+n)(m-2n)=4m*m+(-2n)*4m+n*m+n(-2n)\\=4m^2-7nm-2n^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/u858bnr8fgodl7tvk63gnzmws0a5nzyxaj.png)
![(2a-4b)(7a-2b)=2a*7a+(-2b)(2a)+(-4b)7a+(-4b)(-2b)\\=14a^2-4ab-28ab+8b^2\\=14a^2-32ab+8b^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/9h5h10uo1fwshhrs5jg00dqd223dj4zare.png)
![(4m+1)^2=(4m+1)(4m+1)=4m*4m+4m*1+1*4m+1\\=16m^2+4m+4m+1\\=16m^2+8m+1](https://img.qammunity.org/2023/formulas/mathematics/high-school/4gnz0b4ukde53yilbuwvsfn2ow1y558lym.png)
![(2w+1)(2w-1)=2w*2w+2w(-1)+1*(2w)+1(-1)\\=4w^2-1](https://img.qammunity.org/2023/formulas/mathematics/high-school/gt0ajf6ca7d2qziz4y3scfexnovqho2w9v.png)
![(m+3)(m^2+4m+7)\\=m*m^2+m*4m+m*7+3m^2+3*4m+3*7\\=m^3+4m^2+7m+3m^2+12m+21\\=m^3+7m^2+19m+21](https://img.qammunity.org/2023/formulas/mathematics/high-school/yytgdif3mkp32rhv5fbye7cmeyz9i212zb.png)
![(3x+2)(5x^2-12x-2)\\=3x*5x^2+3x(-12x)+3x(-2)+2(5x^2)+2(-12x)+2(-2)\\=15x^3-36x^2-6x+10x^2-24x-4\\=15x^3-26x^2-30x-4](https://img.qammunity.org/2023/formulas/mathematics/high-school/apbztkr4p5r1yk6vnv6nmbxlfqdiwvg10b.png)
Hope that helps!