Square of a Binomial

$(x+3{)}^2$ means $(x+3)(x+3)$, so every term in the first bracket must multiply every term in the second.

Using $(a+b{)}^2=a^2+2ab+b^2$ with $a=x$ and $b=3$: square x to get $x^2$; multiply x by 3 and double it, giving $6x$; square 3 to get 9.

So $(x+3{)}^2=x^2+6x+9$.

The middle term is the one most often lost: the two cross-products $3x$ and $3x$ add to $6x$, which is why squaring the terms separately, $x^2+9$, is wrong. The constant is $3\times 3=9$, not $3+3$.

Use $(a-b{)}^2=a^2-2ab+b^2$ with $a=x$ and $b=4$.

Square the first term to get $x^2$. The middle term is $-2ab=-2\times x\times 4=-8x$. Squaring the last term gives $(-4{)}^2=+16$, which is positive because a negative multiplied by a negative is positive.

So $(x-4{)}^2=x^2-8x+16$.

Use $(a+b{)}^2=a^2+2ab+b^2$ with $a=2x$ and $b=5$.

Square the first term: $(2x{)}^2=4x^2$. The 2 is squared as well as the x. The middle term is $2ab=2\times 2x\times 5=20x$; there are two identical cross products of $10x$ each, so they must be added together. Square the last term: $5^2=25$.

So $(2x+5{)}^2=4x^2+20x+25$.

Use $(a-b{)}^2=a^2-2ab+b^2$ with $a=2x$ and $b=7$.

Square the first term: $(2x{)}^2=4x^2$. The 2 is squared as well as the x. The middle term is $-2ab=-2\times 2x\times 7=-28x$; there are two identical cross products of $-14x$ each. Square the last term: $(-7{)}^2=+49$, positive because a negative times a negative is positive.

So $(2x-7{)}^2=4x^2-28x+49$.

This is a difference of two squares: $(a+b)(a-b)=a^2-b^2$, with $a=3x$ and $b=5$.

Expanding in full gives $9x^2-15x+15x-25$, and the two middle terms cancel out. Square the first term: $(3x{)}^2=9x^2$. The 3 is squared as well as the x. Square the second term: $5^2=25$, and it is subtracted because $(+5)\times (-5)=-25$.

So $(3x+5)(3x-5)=9x^2-25$.

Use $(a-b{)}^2=a^2-2ab+b^2$ with $a=3x$ and $b=2y$.

Square the first term: $(3x{)}^2=9x^2$. The 3 is squared as well as the x. There are two identical cross products of $3x\times (-2y)=-6xy$, so the middle term is $-12xy$. Square the last term: $(-2y{)}^2=4y^2$.

So $(3x-2y{)}^2=9x^2-12xy+4y^2$.