I'm stuck on this one....
6r^2-r-2=0
Any assistance appreciated.
Thanks
Thomas
Hi tbradnc,
Here's a technique you can use.
Multiply the leading coefficient (6) by the constant (-2).
This gives you -12.
The coefficient of the middle term is -1.
You need a pair of numbers whose product is -12 and whose sum is -1.
After thinking a bit on this, you should come up with the only combination that works, namely -4 and 3.
Now restate your quadratic replacing the middle coefficient with these two.
$\displaystyle 6r^2{\color{red}-4r +3r}-2$
Now group the first two terms, and then the last two terms.
$\displaystyle (6r^2-4r)+(3r-2)$
Factor each group.
$\displaystyle 2r(3r-2)+1(3r-2)$
$\displaystyle (3r-2)$ is a common factor, so
$\displaystyle (3r-2)(2r+1)$
All done.
Thank you so much....I'm so glad I found this site. ;-)
I'm an older adult going back to school and it's been 30 years since I've played with algebra. I leave every class barely understanding what we're doing and I have to drill, drill, drill to get the hang of it.