Ask your own question, for FREE!
Mathematics 15 Online
OpenStudy (anonymous):

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

OpenStudy (anonymous):

I got 5 for one of the answers, and I got it down to the quadratic \[(2x ^{3}+5x-3)\] but now Im stuck.

OpenStudy (anonymous):

Try reading this! http://www.wikihow.com/Factor-a-Cubic-Polynomial

OpenStudy (anonymous):

use the quadratic formula

OpenStudy (anonymous):

ah, its multiple choice too. the other two factors are +/-3 and +/- (1/2) but I dont know how to find that.

OpenStudy (anonymous):

sorry. I meant the other 2 zeros.

OpenStudy (anonymous):

Did you read my link? If you follow the same process, you'll get your answers

OpenStudy (anonymous):

yeah, but how would I use that system with a fraction?

OpenStudy (anonymous):

What are you talking about?

OpenStudy (anonymous):

hmm... I need to know how to factor the quadratic I got it down to. Its relatively basic, but its confusing me.

OpenStudy (anonymous):

\[\large \text{Quadratic Formula}: \huge x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] If you unsure how to factor a second degree polynomial with 3 terms, always resort to the quadratic formula. The more you do it, the faster you'll get with it, promise.

OpenStudy (anonymous):

what does b stand for again?

OpenStudy (anonymous):

Correction: It doesn't have to be with 3 terms. Can't be with any amount of terms but a cannot be 0 (or else it wouldn't be a quadratic anymore!) \[\huge y=ax^2+bx+c\]

OpenStudy (anonymous):

thanks.

OpenStudy (anonymous):

Also, if you're certain that 5 is a solution, there is a theorem that lets you solve the rest of the roots via synthetic division! If you're taking Algebra 2, you will probably learn this (unless you already have). http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-cubicequations-2009-1.pdf Section 3

OpenStudy (anonymous):

yeah. thats what this problem is about.

OpenStudy (anonymous):

im on the last step.

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!