Expanding Brackets: Math Help for 2nd Term Maths

[raechelc93]

Hi there,

I am currently doing an Ext Math 1 subject and haven't really come across any issues when needing to expand brackets, however, have come across the below equation I can't quite figure out... Any help would be greatly appreciated!

\( (p-q+r^2)(3-(p^2+q^2)) \)

The double brackets as well as the 3rd term in the first set of brackets has got me very confused and I can't seem to figure out how to even start with this one.

 

[skeeter]

you’ve posted an expression, not an equation (there is no equal sign)

$(p-q+r^2)(3-p^2-q^2)$

to expand, distribute the $p$, then the $-q$, and finally the $r^2$ to the three terms in the second set of parentheses ...

$3p-p^3-pq^2 -3q + p^2q +q^3 +3r^2 -p^2r^2 -q^2r^2$

note the expanded expression is less “simplified” than the original factored expression ... expanding doesn’t always yield a better representation

 

[raechelc93]

Hi Skeeter,

Thanks for responding so quick! There are 2 sets of brackets within the second set though.. \[ (3-(p^2+q^2)) \] would it still be the same process if this is the case?

It looks like you altered the second set of brackets from the orignal to get \[ (3-p^2-q^2) \] - how/why did you change it?

 

[raechelc93]

Nevermind - I just figured it out :)

 

[HOI]

$-(p^2+ q^2)$ is the same as $(-1)(p^2+ q^2)$ so we are multiplying $p^2+ q^2$ by -1. Using the fact that "multiplication distributes over addition", that is $(-1)(p^2)+ (-1)(q^2)= -p^2+ (-q^2)= -p^2- q^2$.