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Mathematics 22 Online
OpenStudy (anonymous):

how would i factor 8x^2 + 2x -21

OpenStudy (sasogeek):

multiply the coefficient of the x^2 term by the constant and find 2 factors of the result that when you add or subtract will give you the x coefficient, replace the x coefficient with the two factors in a way that when you subtract or add them, you'll get the x coefficient. you'll get 4 terms. take the first two terms and factor them, take the last two terms and factor them. you're done. if you didn't understand what i've just written, read it carefully again, take your time, and solve it.

OpenStudy (king):

You can use the quadratic equation formula (-b +or-sqrt(b^2-4ac))/2a here a,b,c are the coefficients ax^2+bx+c

OpenStudy (king):

understood? @alexeis_nicole

OpenStudy (anonymous):

yep.. i got 24 and 26... :S ?

OpenStudy (anonymous):

I MEAN ... 1.5 AND -1.75?

OpenStudy (anonymous):

Quadratic Formula in an easier-to-read form: \[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\] Binomials will be written as (where a, b, and c are numerical constants) \[ax^2 + bx + c\] You can always check your factorization by multiplying it back together to get the original expanded form (what you have initially here).

OpenStudy (anonymous):

1.5 and -1.75 are correct, but put them in fraction form and you'll have a huge hint ;-)

OpenStudy (anonymous):

3/2 and -7/4 ?

OpenStudy (anonymous):

Huge hint: \[\frac {numerator}{demominator} = \frac {c}{a}\]

OpenStudy (anonymous):

(ax+c)(ax+c) (but different a's and c's, call them a1 & c1 and a2 & c2 if you want)

OpenStudy (anonymous):

Here's one of the two, so you're not getting confused: -7/4 --> (4x+7)

OpenStudy (anonymous):

Here's how to check, solve this: \[4x+7 = 0 \]

OpenStudy (anonymous):

OH !!!!! (2x-3) (4x+7)

OpenStudy (anonymous):

You got it!

OpenStudy (anonymous):

Do you see why the -/+ sign switches?

OpenStudy (anonymous):

So here's the next step they'll probably introduce you to later, they'll ask you to solve this: \[8x^2 + 2x -21 = (2x-3)(4x+7) = 0\]

OpenStudy (anonymous):

uhm,.... lol

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