Part A) ![(9+4m)^{2}=81+72m+16m^{2}](/tpl/images/0143/7417/1b0ca.png)
Part B) ![(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}](/tpl/images/0143/7417/0eb82.png)
Part C) ![(2x-3z)^{2}=4x^{2}-12xz+9z^{2}](/tpl/images/0143/7417/30341.png)
Part D) ![(4m^{5}+5n^{3})^{2}=16m^{10}+40m^{5}n^{3}+25n^{6}](/tpl/images/0143/7417/35928.png)
Part E) ![(\frac{3}{6}w-\frac{1}{2}y)^{2}=\frac{9}{36}w^{2}-\frac{3}{6}wy+\frac{1}{4}y^{2}](/tpl/images/0143/7417/64b38.png)
Part F) ![(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}](/tpl/images/0143/7417/0eb82.png)
Step-by-step explanation:
The question in English is
Calculate the square of each binomial
we know that
The square of a binomial is always a trinomial
so
![(a+b)^{2}=a^{2}+2ab+b^{2}](/tpl/images/0143/7417/cc817.png)
and
![(a-b)^{2}=a^{2}-2ab+b^{2}](/tpl/images/0143/7417/9b1a2.png)
Part A) we have
![(9+4m)^{2}](/tpl/images/0143/7417/85660.png)
Applying the formula
![(9+4m)^{2}=(9)^{2}+2(9)(4m)+(4m)^{2}](/tpl/images/0143/7417/bea73.png)
![(9+4m)^{2}=81+72m+16m^{2}](/tpl/images/0143/7417/1b0ca.png)
Part B) we have
![(x^{10}-5y^{2})^{2}](/tpl/images/0143/7417/6cb9f.png)
Applying the formula
![(x^{10}-5y^{2})^{2}=(x^{10})^{2}-2(x^{10})(5y^{2})+(5y^{2})^{2}](/tpl/images/0143/7417/8a2ab.png)
![(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}](/tpl/images/0143/7417/0eb82.png)
Part C) we have
![(2x-3z)^{2}](/tpl/images/0143/7417/97dca.png)
Applying the formula
![(2x-3z)^{2}=(2x)^{2}-2(2x)(3z)+(3z)^{2}](/tpl/images/0143/7417/c021f.png)
![(2x-3z)^{2}=4x^{2}-12xz+9z^{2}](/tpl/images/0143/7417/30341.png)
Part D) we have
![(4m^{5}+5n^{3})^{2}](/tpl/images/0143/7417/23871.png)
Applying the formula
![(4m^{5}+5n^{3})^{2}=(4m^{5})^{2}+2(4m^{5})(5n^{3})+(5n^{3})^{2}](/tpl/images/0143/7417/83fae.png)
![(4m^{5}+5n^{3})^{2}=16m^{10}+40m^{5}n^{3}+25n^{6}](/tpl/images/0143/7417/35928.png)
Part E) we have
![(\frac{3}{6}w-\frac{1}{2}y)^{2}](/tpl/images/0143/7417/156d9.png)
Applying the formula
![(\frac{3}{6}w-\frac{1}{2}y)^{2}=(\frac{3}{6}w)^{2}-2(\frac{3}{6}w)(\frac{1}{2}y)+(\frac{1}{2}y)^{2}](/tpl/images/0143/7417/b9b67.png)
![(\frac{3}{6}w-\frac{1}{2}y)^{2}=\frac{9}{36}w^{2}-\frac{3}{6}wy+\frac{1}{4}y^{2}](/tpl/images/0143/7417/64b38.png)
Part F) we have
![(x^{10}-5y^{2})^{2}](/tpl/images/0143/7417/6cb9f.png)
Applying the formula
![(x^{10}-5y^{2})^{2}=(x^{10})^{2}-2(x^{10})(5y^{2})+(5y^{2})^{2}](/tpl/images/0143/7417/8a2ab.png)
![(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}](/tpl/images/0143/7417/0eb82.png)
Note the problem F is the same problem B