OpenStudy (anonymous):

5x^2 - 3x - 2

5 years ago
OpenStudy (anonymous):

what do you want to do? what's the question?

5 years ago
OpenStudy (jwheele1):

solve for x?

5 years ago
OpenStudy (anonymous):

factor

5 years ago
OpenStudy (jwheele1):

\[5x ^{2} - 3x -2\]solve for x Must facor or use the quadratic formula...

5 years ago
OpenStudy (jwheele1):

factor it is

5 years ago
OpenStudy (anonymous):

15x^3 - 9x^2 -6x factor using grouping

5 years ago
OpenStudy (anonymous):

ohh lol duh ok so you have to factor first. this factors into (5x+2)(x-1) then x is anything that makes those equal to 0 so x = -2/5 and 1

5 years ago
OpenStudy (jwheele1):

our goal is to fill in these: ( somethign here +/- somethign else here) ( somethign here +/- somethign else here) when both sets are multiples together you should get your original equation.

5 years ago
OpenStudy (anonymous):

Thankyou Kcheung1

5 years ago
OpenStudy (jwheele1):

(5x+2)(x-1) because 5x times 1x = 5x^2 because 2 * -1 = -2 at the end of your equation and because 2*x = 2x and 5x * -1 = -5x and 2x -5x = -3x (your middle number)

5 years ago
OpenStudy (jwheele1):

Just for knowledge (5x+2)(x-1) set each to zero, and solve each for x 5x+2 = 0 5x = -2 x= -2/5 x-1 = 0 x = 1 answers for x = -2/5 and 1 ....but always plug them back in to be sure

5 years ago
OpenStudy (anonymous):

x^3 - 64 factor

5 years ago
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