factor the polynomial
-r^2+19rs+90s^2

Question

factor the polynomial 
-r^2+19rs+90s^2

anonymous · Answer

$$-r^2+19rs+90s^2$$

zepdrix · Answer

I don't think this will factor.
Are you sure you typed it correctly?
The negative sign is causing a problem.

anonymous · Answer

I'm very sorry i just checked over it and its actually:
$$r^2+19rs+90s^2$$

zepdrix · Answer

Let's try to find factors of 90 which will add to 19.

90 =
2 * 45
6 * 15
5 * 18
9 * 10

Will any of those work for us?
2+45 = 47
hmm that pair didn't give us 19

anonymous · Answer

9+10

zepdrix · Answer

Ooo ok good.
Let's try something sneaky then.
We'll rewrite the middle term 19rs as the sum of 9rs and 10rs.$$\large\rm r^2\color{orangered}{+19rs}+90s^2$$$$\large\rm r^2\color{orangered}{+9rs+10rs}+90s^2$$

zepdrix · Answer

From there, we can factor by grouping,$$\large\rm (r^2+9rs)+(10rs+s^2)$$

anonymous · Answer

ah, ok
is that the end?

zepdrix · Answer

Woops typo, my bad,$$\large\rm (r^2+9rs)+(10rs+90s^2)$$From the first set of brackets,
what can you take out of each term?
They don't share any type of number,
but they both have an r in them, right?

zepdrix · Answer

Taking an r out of each term in the first brackets gives us$$\large\rm r(r+9s)+(10rs+90s^2)$$How bout the other brackets?
What do they share in common?

anonymous · Answer

ok...

anonymous · Answer

i think i got the answer