Simplifying a rational expression

[hollywalker]

Hello,
I am having difficulty solving my math problems.

Simplify the expression:
(12+r-r^2)/(r^3 +3r^2)

The answer is (4-r)/r^2

I know that i can expand 12+r-r^2 as (-r+4)(r+3)
But i cannot figure out the rest. Please help me. Thanks

 

[evinda]

Hello,
I am having difficulty solving my math problems.

Simplify the expression:
(12+r-r^2)/(r^3 +3r^2)

The answer is (4-r)/r^2

I know that i can expand 12+r-r^2 as (-r+4)(r+3)
But i cannot figure out the rest. Please help me. Thanks

(Wave)

$$-r^2+r+12=0$$

$$\Delta=b^2-4ac=1-4 (-1 \cdot 12)=1+48=49$$

$$r_{1,2}=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-1 \pm 7}{-2}=4,-3$$

So,the expression can be written like that:

$$\frac{12+r-r^2}{r^3+3r^2}=\frac{-(r-4) \cdot (r+3)}{r^2 \cdot (r+3)}=\frac{-(r-4)}{r^2}=\frac{4-r}{r^2}$$

 

[soroban]

Hello, hollywalker!

Simplify: .$$\frac{12+r-r^2}{r^3 +3r^2}$$

Factor the numerator: $$12+r-r^2 \:=\: (3+r)(4-r)$$
Factor the denominator: $$r^3 + 3r^2 \:=\:r^2(r+3)$$

The fraction becomes: .$$\frac{(3+r)(4-r)}{r^2(r+3)}$$

Reduce: .$$\frac{\cancel{(3+r)}(4-r)}{r^2\cancel{(r+3)}} \;=\;\frac{4-r}{r^2}$$

 

[Dethrone]

Also note that if you're crossing $(r+3)$ from the top and bottom, then $r \ne -3$.

 

[anemone]

Hi hollywalker, welcome to MHB!:)

I have moved your topic to a more appropriate sub-forum, and given a new title to it to indicate the nature of the question being asked.