How do you multiply $(-13u+v^2)^2$ ?

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Answer 1 · 2018-03-08T14:13:33

$v^4-26uv^2+169u^2$

$(v^2-13u)^2$
$(v^2)^2-2(v^2)(13u)+(13u)^2$
$v^4-26uv^2+169u^2$

Answer 2 · 2018-03-08T14:15:57

The formula for $(a+b)^2$ is $a^2+2ab+b^2$ . Let's apply this to our problem.

$(-13u+v^2)^2$ = $(-13u)^2+2(-13u)(v^2)+(v^2)^2$

$-13u*-13u = 169u^2$
$2(-13u)(v^2) = -26uv^2$
$v^2*v^2 = v^4$

Therefore, our answer is $169u^2-26uv^2+v^4$ .

Answer 3 · 2018-03-08T14:16:54

$169u^2-26uv^2+v^4$

In general
$color(white)("XXX")(color(red)a+color(blue)b)^2=color(red)a^2+2color(red)acolor(blue)b+color(blue)b^2$

So
$color(white)("XXX")(color(red)(-13u)+color(blue)v^2))^2$

$color(white)("XXXXXXXX")=(color(red)(-13u))^2+2 * (color(red)(-13u)) * color(blue)(v^2) + (color(blue)(v^2))^2$

$color(white)("XXXXXXXX")=169u^2-26uv^2+v^4$

How do you multiply (-13u+v^2)^2(-13u+v^2)^2?