Rearranging Algebra

**BabyMilo** · March 18th 2010, 10:36 AM

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

thanks

**Anonymous1** · March 18th 2010, 10:40 AM

Sure.

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

**BabyMilo** · March 18th 2010, 12:59 PM

Originally Posted by Anonymous1

Sure.

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

how did you go from

Code:

to

Code:

and what did you do this bit

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

im quite un sure what's going on.

thanks

**Archie Meade** · March 18th 2010, 02:24 PM

Hi BabyMilo,

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

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

and for the last piece

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

**BabyMilo** · March 18th 2010, 02:57 PM

Originally Posted by Archie Meade

Hi BabyMilo,

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

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

and for the last piece

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

i get this bit now.

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

but still quite unsure how you get the last part.

thanks as always.

**Archie Meade** · March 18th 2010, 03:12 PM

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

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

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

Hence

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

In terms of the example you have

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

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

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

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

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

**BabyMilo** · March 19th 2010, 12:10 AM

Originally Posted by Archie Meade

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

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

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

Hence

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

In terms of the example you have

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

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

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

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

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

thanks made it much easier to understand.

Thread: Rearranging Algebra

LinkBack

Thread Tools

Display

Rearranging Algebra

Similar Math Help Forum Discussions

Rearranging

help rearranging this

Rearranging Algebra

Rearranging Algebra to find Expression for Slope

Help With Rearranging!

Search Tags