CA Algebra I: Factoring Quadratics
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16 videosCalifornia Standards Test: Algebra I- 13:48
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AC 8.1.2 pt 2 Factoring Quadratics w Diamonds and Rectanglesby djfreeman12,665 views
slow mind ! guessing answer amp !
ksf911 7 months ago
slow mind !
ksf911 7 months ago
more like, can't skip school
paulceltics 9 months ago
my teacher told me to check out your videos, these are really helpful thx for posting!
millerliteisgud 1 year ago
@kingofdice66 That is not a quadratic polynomial since the first term is degree 3. To factor this polynomial, you need to think of all the possible factors of 9 which are: (+/-1, +/-3, +/-9)...then plug each number into the polynomial until you find one that makes the polynomial 0. (I believe -3 does that) So (x - (-3)) is a factor. Then you do long division. x^3-6x+9 divided by x+3. That gives you a quadratic and you continue factoring from there.
SpoiledLogic 1 year ago
work it backwards:
4(4-x)(4+x) = FOIL 4(16+4x-4x-x^2)....
4x-4x=0 so now we have 4(16-x^2) multiply 4 in to paranthases = 64-4x^2.
gabrielduran 2 years ago
64-4x^2 is 4(4-x)(4+x) i know, but i donit entirely understand why. Please help me
nipplelevikingofjuic 2 years ago
Weird, on your previous video you didn't put a 2 at the front when you divided both sides by 2 cause 0 divide 2 equals 0. Why did you put a 3 there at the front -.- It doesn't fit with what you did in the quadratic factoring in your first video and here
furyofnasa 2 years ago
what about x^3-6x+9
kingofdice66 2 years ago
good video i understand it completely, and i love that a^2 - b^2 trick its cool. im gonna show this to my teacher lol (im in 9th grade)
Pers0n495 2 years ago