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Algebra 20 Online
OpenStudy (anonymous):

Factor 8y 2 + 17y + 2.

OpenStudy (anonymous):

32y+2 I think

OpenStudy (anonymous):

(4n - 3)(n + 2) (4n + 3)(n - 2) (4n + 2)(n - 3) here is anwsers

OpenStudy (anonymous):

i get that but i can finish it

OpenStudy (anonymous):

^^^ please do not give out direct answers :(

OpenStudy (anonymous):

What do you think the answer is?

OpenStudy (anonymous):

(4n - 3)(n + 2)

OpenStudy (anonymous):

Trying to factor by splitting the middle term 2.1 Factoring 8y2+17y+2 The first term is, 8y2 its coefficient is 8 . The middle term is, +17y its coefficient is 17 . The last term, "the constant", is +2 Step-1 : Multiply the coefficient of the first term by the constant 8 • 2 = 16 Step-2 : Find two factors of 16 whose sum equals the coefficient of the middle term, which is 17 . -16 + -1 = -17 -8 + -2 = -10 -4 + -4 = -8 -2 + -8 = -10 -1 + -16 = -17 1 + 16 = 17 That's it Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 1 and 16 8y2 + 1y + 16y + 2 Step-4 : Add up the first 2 terms, pulling out like factors : y • (8y+1) Add up the last 2 terms, pulling out common factors : 2 • (8y+1) Step-5 : Add up the four terms of step 4 :

OpenStudy (anonymous):

Read through this then let me know what you the the answer is :)

OpenStudy (anonymous):

By the way… none of the answer choices shown include the correct answer.

OpenStudy (anonymous):

that is what my list gave

OpenStudy (anonymous):

Factor 4n 2 - 5n - 6. (4n - 3)(n + 2) (4n + 3)(n - 2) (4n + 2)(n - 3)

OpenStudy (anonymous):

I just showed you the step-by-step process to solve.

OpenStudy (anonymous):

their is whole problm

OpenStudy (anonymous):

Ah, that was not the original Q above.

OpenStudy (anonymous):

I was solving the first question. Sorry.

OpenStudy (anonymous):

oh shoot i screwed up sorry

OpenStudy (anonymous):

You should open up a new thread for new questions to give the opportunity to others to be able to receive a medal.

OpenStudy (anonymous):

Welcome to OpenStudy! :)

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