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Precalculus 25 Online
OpenStudy (anonymous):

how doi solve this problem? Use the Remainder Theorem to find the remainder when 2x^3 – x^2 – 3x + 7 is divided by x + 2. State whether the binomial is a factor of the polynomial.

terenzreignz (terenzreignz):

Well, the remainder theorem states that to get the remainder when a polynomial is divided by a linear divisor (x - a) then you simply evaluate f(a), and that's your remainder

OpenStudy (anonymous):

i dont understand

terenzreignz (terenzreignz):

You're dividing by x+2, are you not? So, if x+2 = x - a then a = ?

OpenStudy (anonymous):

2?

terenzreignz (terenzreignz):

Nope... x + 2 = x - 2 ? I don't think so :)

OpenStudy (anonymous):

i dont know then

terenzreignz (terenzreignz):

a is -2.

terenzreignz (terenzreignz):

So now, I want you to evaluate the polynomial, at x = -2, and tell me what you get.

OpenStudy (anonymous):

how?

terenzreignz (terenzreignz):

Well, replace all the x's with -2, and just solve.

OpenStudy (anonymous):

so. -64-4+6+7?

OpenStudy (anonymous):

this doesnt look right

OpenStudy (softballgirl372015):

OpenStudy (softballgirl372015):

does this help?

OpenStudy (anonymous):

YES

OpenStudy (softballgirl372015):

did you learn it this way or by synthetic division?

OpenStudy (anonymous):

my teacher did synthetic division but my my old tutor did this way so.

OpenStudy (softballgirl372015):

are you in alg 2?

OpenStudy (anonymous):

precal

OpenStudy (softballgirl372015):

can you help me with a question please?

OpenStudy (anonymous):

possibly what is it

OpenStudy (softballgirl372015):

how do you write a quadratic equation with integral coefficients having the roots -5/2 and 1

OpenStudy (softballgirl372015):

do you know how to do this?

OpenStudy (anonymous):

one sec

OpenStudy (softballgirl372015):

okai

OpenStudy (anonymous):

no i dont remember this sorry

OpenStudy (softballgirl372015):

its okay. thanks though :)

OpenStudy (anonymous):

np

OpenStudy (softballgirl372015):

thanks for the medal too!!

OpenStudy (anonymous):

:)

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