Question

Factor by grouping the equation
r^2-2r+1-4s^2

Answer 1

r2-2r2+1-4s2 i think...

Answer 2

wait i did this wrong

Answer 3

hmmm is that supposed to be an \(\large \bf s^2 \ or \ r^2?\)

Answer 4

for the -4 term?

Answer 5

i would say s2?

Answer 6

looks like an "s"... just wondering =)

Answer 7

@jdoe0001 the equation above is the exact equation it is s^2 on the -4 variable

Answer 8

ya i would say "s" to

Answer 9

hmm ok... do you know what a "perfect square trinomial" is?

Answer 10

yes

Answer 11

why?

Answer 12

hehh I meant samstopp =)

Answer 13

no i dont know

Answer 14

I know how to factor a trinomial square if that is what you are asking

Answer 15

"s"

Answer 16

I know that to factor an equation of x2+4x+4 is (x+2)^2 for an example

Answer 17

ok.... well.... I thinik that's all you can really squeeze out of that
because the "s" is not quite groupable per se

thus \(\bf r^2-2r+1-4s^2\implies (r^2-2r+1)-4s^2\implies ({\color{brown}{ r}}^2-2r+{\color{blue}{ 1}}^2)-4s^2
\\ \quad \\
\begin{array}{cccccllllll}
{\color{brown}{ a}}^2& - &2{\color{brown}{ a}}{\color{blue}{ b}}&+&{\color{blue}{ b}}^2\\
\downarrow && &&\downarrow \\
{\color{brown}{ a}}&& &&{\color{blue}{ b}}\\
&\to &({\color{brown}{ a}} - {\color{blue}{ b}})^2&\leftarrow
\end{array}\qquad thus
\\ \quad \\

({\color{brown}{ r}}^2-2r+{\color{blue}{ 1}}^2)-4s^2\implies (r-1)^2-4s^2\implies (r-1)(r-1)-4s^2\)

Answer 18

keep in mind that \(\bf 1^2 = 1\quad 1^{6,000} = 1 \quad ...\)