# How to find $(2x+5)(2x-5)$

Given that $(x + y)•(x - y) = x^2 - y^2$
How do I do this: $(2x+y)(2x-y) = ?$
Not really sure about the tag I should use since English is not my native tongue

• You could always do it the hard way, by carrying out the multiplication as it stands. Then see what you get and try to see the connection between your answer and your given formula. Personally, would recommend that you didn't start using that formula until you're able to see that connection. Commented Jun 25, 2016 at 10:57

By setting $2x = z$ we have:

$(2x+y)(2x-y) = (z+y)(z-y)= z^2 - y^2$ by your assumption. Then by replacing $2x$ instead of $z$, we have $(2x+y)(2x-y) = 4x^2 - y^2$.

And for your question, just put 5 instead of $y$; which will give us $4 x^2 - 25$!

And we are done!

A change of variables might make things clearer.
$$(a + b)(a - b) = a^2 - b^2$$ To find the product $(2x + 5)(2x - 5)$, set $a = 2x$ and $b = 5$ to obtain $$(2x + 5)(2x - 5) = (2x)^2 - 5^2 = 4x^2 - 25$$ To find the product $(2x + y)(2x - y)$, set $a = 2x$ and $b = y$ to obtain $$(2x + y)(2x - y) = (2x)^2 - y^2 = 4x^2 - y^2$$

Note that we verify our answers by applying the distributive law repeatedly.
\begin{align*} (2x + 5)(2x - 5) & = 2x(2x - 5) + 5(2x - 5) && \text{since $a(b + c) = ab + ac$}\\ & = 4x^2 - 10x + 10x - 25 && \text{since $u(v + w) = uv + uw$}\\ & = 4x^2 - 25 && \text{combine like terms} \end{align*}

• how does $(2x)^2$ becomes $4x^2$ ?
– Yoav
Commented Jun 25, 2016 at 11:54
• $(2x)^2 = (2x)(2x) = 2 \cdot x \cdot 2 \cdot x = 2 \cdot 2 \cdot x \cdot x = 4x^2$. Commented Jun 25, 2016 at 11:56

You have to multiply "everything with everything". $(a+b)(c+d)=ac+ad+bc+bd$

Edit: Or since $(a+b)(a-b)=a^2-b^2$, write $a=2x$ and $b=y$ and switch the terms in the given formula.

You get $(2x+y)(2x-y)=4x^2-y^2$.

It's just $(2x)^2 - y^2$. I don't see any reason to ask these questions here.

• The site's description clearly says that Mathematics Stack Exchange is a question and answer site for people studying math at any level...
– Yoav
Commented Jun 25, 2016 at 11:00