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Is there an easier/different way to do it? We have some monsterous funcitons like for example 7x^9+3x^4-3x^2+2x+8 which we'd need to factor... I don't think this method works very well on those. lol I know there are online calculators or better and newer calculators that can do it but I have only TI 84 plus.
Thank you very much, Susan! Your video helped me a lot. Would you please tell me how did you get your last formula of the last question when you divided each term by 1/k? Again, thank you so much.
fahad... maybe ur missing the point? this wasn't an algebra lesson. i mean, she was demonstrating Ti-84 capability not teach us how to factor... we have to bring something to the table too :)
Why not just enter Y2 = 0 and then just find the intersection of the quadratic and the x-axis to find any real roots? It's much easier than having to go left of the root and then right of the root. Just a suggestion.
"But this video doesn't explain why you can simply multiply an expression by 2 and then by 3 and it's mathematically valid. That's a serious problem." It didn't directly state it, but when she said that x = -3.5 and x = 1/3 she then showed that implies x+3.5 = 0 and x - 1/3 = 0 Multiplying those two together gives you (x+3.5)(x - 1/3) which is equal to 0*0 = 0. Then since 3.5 = 7/2 she wrote (x+7/2)(x - 1/3) But keep in mind these are both still = 0 as mentioned above. (x + 7/2)(x - 1/3) = 0 Multiply both sides of the equation by 2 and the equal sign stays true. 2(x + 7/2)(x - 1/3) = 2*0 (2x + 7)(x - 1/3) = 0 Now multiply both sides by 3 and the equation still stays true because, again, you're doing the same thing to both sides of an equal expression. 3(2x+7)(x - 1/3) = 3*0 Commutative property (2x+7)*3(x - 1/3) = 0 (2x+7)(3x-1) = 0 And it is still equal to zero just as (x + 7/2)(x - 1/3) form was. It is two versions of the same factored form, one just has integers which is the way we traditionally think of "factoring".
Thanks , this helped me a lot . The chapter I was reviewing in my book was a bit confusing.. I had to sketch a graph by using y=f(x) for y=2f(1/2x)+1 and it was confusing me Because i had two separate C's in one equation. (C•y) and (cx) . Since the first one was (2f(1/2x)+1), I decided to do get my y value first to multiple c and add 1. Then for the second one f(1/2x) , I did the reciprocal times x to get my x values . I obviously need more practice, but this video helped me understand more with what I am trying to learn . Thanks
Theres an easier way under the APPS section. Check it out. Someone else has a video on youtube. This way works but pain too much button action...Just sayin..
If you are using the TI 83/84 family of calculators, you should be able to enter your number in decimal form, hit MATH ENTER ENTER, and it will convert to a fraction.
