Ask your own question, for FREE!
Mathematics 14 Online
OpenStudy (anonymous):

factor 2x^2+5x-3

OpenStudy (phebe):

(2 x-1) (x+3)

OpenStudy (phebe):

thas the answer XD

OpenStudy (anonymous):

How did you get that?

OpenStudy (phebe):

I useed this aronym called FOIL k

OpenStudy (phebe):

then i combined like terms

OpenStudy (anonymous):

but you can't use FOIL on something that isn't already factored or in parenthesis

OpenStudy (phebe):

tht is tru but i put them in parenthesis

OpenStudy (anonymous):

but how did you do that with only 3 numbers XD

OpenStudy (phebe):

(2x^2)+(5x-3) like this

OpenStudy (phebe):

then i did the factoring by using FOIL

OpenStudy (anonymous):

if you foiled that you would get 10x^3-6x^2 -.-

OpenStudy (phebe):

2x^2 - 5x - 3 First multiply the coefficient of the first and last term and try to find a sum of that number which will equal the middle term. 2 x -3 = -6 What are the products of -6? 1 x (-6) = -6 2 x (-3) = -6 -1 x 6 = -6 -2 x 3 = -6 Which of these two products when added will equal -5? 1 + (-6) = -5 Replace the middle term with these two numbers in the equation. 2x^2 + 1x - 6x - 3 Now factor using the grouping method. (2x^2 + 1x) + (-6x + 3) Factor out your common factors x(2x + 1) -3(2x+1) Which will leave you with: (x - 3) (2x + 1) You can test you answer by using the FOIL method: x * 2x * x * 1 + (-3) * 2x * (-3) * 1 You will arrive back at your original answer 2x^2 - 5x - 3, meaning (x - 3) (2x + 1) is correct.

OpenStudy (phebe):

here this is what i did ok XD

OpenStudy (phebe):

can i get a medal lol after i typed all that

OpenStudy (aum):

2x^2 + 5x - 3 a = 2, b = 5, c = -3 AC Method: a * c = 2 * (-3) = -6 What two factors of -6 add to +5 (the b-value)? +6 and -1 add to +5 and multiply to -6 2x^2 + 5x - 3 = 2x^2 + 6x - 1x - 3 2x(x + 3) - 1(x + 3) (2x-1)(x+3)

OpenStudy (anonymous):

Thanks :) @aum

OpenStudy (aum):

You are welcome.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!