Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

# how to factor these problems- math help?

2c^2-242

r^3-2r^2-12r

thanks!

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• Dr. J.
Lv 6
9 years ago

2. r (r^2-2r-12)=

r (r-6)(r+2)

• 9 years ago

1.) 2c^2-242 = 2(c^2-121) = 2(c+11)(c-11)

2.) r^3-2r^2-12r = r(r^2-2r-12)

To factor further, we want to factor r^2-2r-12. Use the quadratic formula to find the zeros.

The zeros are 2+sqrt(13) and 2-sqrt(13). Therefore, the factors are (r-(2+sqrt(13))) and (r-(2-sqrt(13)))

so the complete factorization is r(r-(2+sqrt(13)))(r-(2-sqrt(13))) - there won't be so many parentheses when you simplify and write out the sort using the normal symbol.

Hope this helps! :)

• 9 years ago

2c^2-242

2 ( c^2 - 121)

USE FORMULA a^2 - b^2 = ( a+b) ( a-b)

2 ( c + 11) ( c- 11) ANSWER

r^3-2r^2-12r

r ( r^2 - 2r - 12) ANSWER

• Anonymous
9 years ago

1. 2(c^2 - 121) (just factor out two)

(c + sqrt(11)) (c - sqrt(11)) (Difference of squares).

2. r(r^2 - 2r - 12) (factor out r)

Find the roots using the quadratic formula (I don't think this is factorable)

and then write it in the form r(r + d)(r + e) where d and e are the roots. You should have a negative and a positive root, and the magnitude of the negative root should be larger.

Hope this helps