1. ## Rearranging Algebra

Can someone show me how $\displaystyle \frac{1}{4}k^2(k+1)^2+(k+1)^3$ can be factorise or rearrange into $\displaystyle \frac{1}{4}(k+1)^2(k^2+4(k+1))$

thanks

2. Sure.

$\displaystyle \frac{k^2}{4}(k+1)^2 + (k+1)^3 = \frac{k^2}{4}(k+1)^2 + (k+1)(k+1)^2$ $\displaystyle = (k+1)^2(\frac{k^2}{4} + (k+1)) = \frac{(k+1)^2}{4}(k^2 + 4(k+1))$

3. Originally Posted by Anonymous1
Sure.

$\displaystyle \frac{k^2}{4}(k+1)^2 + (k+1)^3 = \frac{k^2}{4}(k+1)^2 + (k+1)(k+1)^2$ $\displaystyle = (k+1)^2(\frac{k^2}{4} + (k+1)) = \frac{(k+1)^2}{4}(k^2 + 4(k+1))$
how did you go from

Code:
$\displaystyle \frac{k^2}{4}(k+1)^2 + (k+1)(k+1)^2$
to

Code:
$\displaystyle = (k+1)^2(\frac{k^2}{4} + (k+1))$
and what did you do this bit

$\displaystyle = (k+1)^2(\frac{k^2}{4} + (k+1)) = \frac{(k+1)^2}{4}(k^2 + 4(k+1))$

im quite un sure what's going on.

thanks

4. Hi BabyMilo,

$\displaystyle (k+1)^2$ is common to both terms, hence

$\displaystyle \frac{k^2}{4}\color{red}(k+1)^2\color{black}+(k+1) \color{red}(k+1)^2\color{black}=\left(\frac{k^2}{4 }+[k+1]\right)\color{red}(k+1)^2\color{black}$

and for the last piece

$\displaystyle k+1=\frac{4}{4}(k+1)=4\left(\frac{k+1}{4}\right)$

5. Originally Posted by Archie Meade
Hi BabyMilo,

$\displaystyle (k+1)^2$ is common to both terms, hence

$\displaystyle \frac{k^2}{4}\color{red}(k+1)^2\color{black}+(k+1) \color{red}(k+1)^2\color{black}=\left(\frac{k^2}{4 }+[k+1]\right)\color{red}(k+1)^2\color{black}$

and for the last piece

$\displaystyle k+1=\frac{4}{4}(k+1)=4\left(\frac{k+1}{4}\right)$

i get this bit now.

$\displaystyle \frac{k^2}{4}\color{red}(k+1)^2\color{black}+(k+1) \color{red}(k+1)^2\color{black}=\left(\frac{k^2}{4 }+[k+1]\right)\color{red}(k+1)^2\color{black}$

but still quite unsure how you get the last part.

thanks as always.

6. If you examine with numbers first, then it will be clearer with "k" introduced.

$\displaystyle 3=\frac{4}{4}(3)=\frac{4(3)}{4}=\frac{12}{4}$

If we divide any value by 4 and then multiply by 4, that will bring us back to the original value.

Hence

$\displaystyle \frac{4x}{4}=x$

In terms of the example you have

$\displaystyle \frac{1}{4}k^2(k+1)^2+(k+1)^3=\frac{1}{4}k^2(k+1)^ 2+(k+1)(k+1)^2$

$\displaystyle (k+1)^2$ is a common factor,

and to make $\displaystyle \frac{1}{4}$ a common factor also, we write $\displaystyle (k+1)$ as $\displaystyle \frac{1}{4}\left(4[k+1]\right)$

Hence we get $\displaystyle \frac{1}{4}(k+1)^2$ as a common factor

$\displaystyle \color{red}\frac{1}{4}(k+1)^2\color{black}k^2+\col or{red}\frac{1}{4}(k+1)^2\color{black}(4[k+1])=\color{red}\frac{1}{4}(k+1)^2\color{black}\left( k^2+4[k+1]\right)$

7. Originally Posted by Archie Meade
If you examine with numbers first, then it will be clearer with "k" introduced.

$\displaystyle 3=\frac{4}{4}(3)=\frac{4(3)}{4}=\frac{12}{4}$

If we divide any value by 4 and then multiply by 4, that will bring us back to the original value.

Hence

$\displaystyle \frac{4x}{4}=x$

In terms of the example you have

$\displaystyle \frac{1}{4}k^2(k+1)^2+(k+1)^3=\frac{1}{4}k^2(k+1)^ 2+(k+1)(k+1)^2$

$\displaystyle (k+1)^2$ is a common factor,

and to make $\displaystyle \frac{1}{4}$ a common factor also, we write $\displaystyle (k+1)$ as $\displaystyle \frac{1}{4}\left(4[k+1]\right)$

Hence we get $\displaystyle \frac{1}{4}(k+1)^2$ as a common factor

$\displaystyle \color{red}\frac{1}{4}(k+1)^2\color{black}k^2+\col or{red}\frac{1}{4}(k+1)^2\color{black}(4[k+1])=\color{red}\frac{1}{4}(k+1)^2\color{black}\left( k^2+4[k+1]\right)$
thanks made it much easier to understand.