Question

Calcula el cuadrado de cada binomio: A. (9 + 4m)^2
B. (X^10 - 5y^2)^2
C. (2x - 3z)^2
D. (4m^5 + 5n^3)^2
E. (3/6w - 1/2y)^2
F. (X10 - 5y2)2​

Answer 1

Answer:

Part A) $(9+4m)^{2}=81+72m+16m^{2}$

Part B) $(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}$

Part C) $(2x-3z)^{2}=4x^{2}-12xz+9z^{2}$

Part D) $(4m^{5}+5n^{3})^{2}=16m^{10}+40m^{5}n^{3}+25n^{6}$

Part E) $(\frac{3}{6}w-\frac{1}{2}y)^{2}=\frac{9}{36}w^{2}-\frac{3}{6}wy+\frac{1}{4}y^{2}$

Part F) $(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}$

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

$(a+b)^{2}=a^{2}+2ab+b^{2}$

and

$(a-b)^{2}=a^{2}-2ab+b^{2}$

Part A) we have

$(9+4m)^{2}$

Applying the formula

$(9+4m)^{2}=(9)^{2}+2(9)(4m)+(4m)^{2}$

$(9+4m)^{2}=81+72m+16m^{2}$

Part B) we have

$(x^{10}-5y^{2})^{2}$

Applying the formula

$(x^{10}-5y^{2})^{2}=(x^{10})^{2}-2(x^{10})(5y^{2})+(5y^{2})^{2}$

$(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}$

Part C) we have

$(2x-3z)^{2}$

Applying the formula

$(2x-3z)^{2}=(2x)^{2}-2(2x)(3z)+(3z)^{2}$

$(2x-3z)^{2}=4x^{2}-12xz+9z^{2}$

Part D) we have

$(4m^{5}+5n^{3})^{2}$

Applying the formula

$(4m^{5}+5n^{3})^{2}=(4m^{5})^{2}+2(4m^{5})(5n^{3})+(5n^{3})^{2}$

$(4m^{5}+5n^{3})^{2}=16m^{10}+40m^{5}n^{3}+25n^{6}$

Part E) we have

$(\frac{3}{6}w-\frac{1}{2}y)^{2}$

Applying the formula

$(\frac{3}{6}w-\frac{1}{2}y)^{2}=(\frac{3}{6}w)^{2}-2(\frac{3}{6}w)(\frac{1}{2}y)+(\frac{1}{2}y)^{2}$

$(\frac{3}{6}w-\frac{1}{2}y)^{2}=\frac{9}{36}w^{2}-\frac{3}{6}wy+\frac{1}{4}y^{2}$

Part F) we have

$(x^{10}-5y^{2})^{2}$

Applying the formula

$(x^{10}-5y^{2})^{2}=(x^{10})^{2}-2(x^{10})(5y^{2})+(5y^{2})^{2}$

$(x^{10}-5y^{2})^{2}=x^{20}-10x^{10}y^{2}+25y^{4}$

Note the problem F is the same problem B