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
how did you go from
toCode:$\displaystyle \frac{k^2}{4}(k+1)^2 + (k+1)(k+1)^2$
and what did you do this bitCode:$\displaystyle = (k+1)^2(\frac{k^2}{4} + (k+1))$
$\displaystyle = (k+1)^2(\frac{k^2}{4} + (k+1)) = \frac{(k+1)^2}{4}(k^2 + 4(k+1))$
im quite un sure what's going on.
thanks
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
$\displaystyle k+1=\frac{4}{4}(k+1)=4\left(\frac{k+1}{4}\right)$
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
$\displaystyle \color{red}\frac{1}{4}(k+1)^2\color{black}k^2+\col or{red}\frac{1}{4}(k+1)^2\color{black}(4[k+1])=\color{red}\frac{1}{4}(k+1)^2\color{black}\left( k^2+4[k+1]\right)$