Don't think I remember ever learning this synthetic division. Maybe I've forgotten. I can see *how* the synthetic division method is calculating 4×2³ - 2×2² + 1×2 + 5 but is that all it is? Is it just a different way to calculate f(2) given f(x)? Or is there more to it than that?
10/18/2024 10:11 PM Theorem: A theorem is a statement that has been proven to be true. The proof of a theorem involves using mathematical axioms and other previously proven theorems. Remainder Theorem f[x] = 4x³ - 2x² + x + 5 f[2] = 4[2]³ - 2[2]² + 2 + 5 f[2] = 4[8] - 2[4] + 2 + 5 = 32 - 8 + 2 + 5 = 31 [Using Synthetic Division] 2| 4 -2 1 5 ---> The coefficients of the polynomial 4 8 12 26 ---------------------- 4 6 13 31 → remainder is 31 which proves the calculation
He is nesting the x es. So he doesn't use exponents, because he applies that reapeted mulipication is exponentiation. f(x) = 2x**2 + 3x + 5 = ((2x)+3)x +5 So f(2) = 2(2)**2 + 3(2) + 5 = ((2*2)+3)*2 +5
That synthetic division is great - something I don't recall ever learning in all of my 86 years
Who discovered this theorem? Very smart short-cut!
i remember using this to factorize , finding the roots
Don't think I remember ever learning this synthetic division. Maybe I've forgotten.
I can see *how* the synthetic division method is calculating
4×2³ - 2×2² + 1×2 + 5
but is that all it is? Is it just a different way to calculate f(2) given f(x)? Or is there more to it than that?
f(2) = 32 − 8 + 2 + 5 = 31
10/18/2024 10:11 PM
Theorem: A theorem is a statement that has been proven to be true. The proof of a theorem involves using mathematical axioms and other previously proven theorems.
Remainder Theorem
f[x] = 4x³ - 2x² + x + 5
f[2] = 4[2]³ - 2[2]² + 2 + 5
f[2] = 4[8] - 2[4] + 2 + 5 = 32 - 8 + 2 + 5 = 31
[Using Synthetic Division]
2| 4 -2 1 5 ---> The coefficients of the polynomial
4 8 12 26
----------------------
4 6 13 31 → remainder is 31 which proves the calculation
Plugging 2 in for you get 31 4×8 32-2×4 8=24+2+=26+5=31
y = f(2) = 4 . (2)³ - 2 . (2)² + (2) + 5 = 32 - 8 + 2 + 5 = 31
Got 31 easy substitutions
Thanks for the fun
f(x)=9, under the condition, x=2...
Never heard of it. Almost see the logic.
He is nesting the x es.
So he doesn't use exponents, because he applies that reapeted mulipication is exponentiation.
f(x) = 2x**2 + 3x + 5 = ((2x)+3)x +5
So f(2) = 2(2)**2 + 3(2) + 5 = ((2*2)+3)*2 +5
f(2)= 31
31
Thank god for the skip forward button with these videos. Holy f.
I think you. Issued it with this one. Only someone who is confident in math would follow. Sorry 15:47
This is as clear as mud . Dreadful. Too many words going round in circles
F(x)=4x^3-2x^2+x+5
F(2)= 4*2^3-2*2^2+2+5
F(2)=4*8-2*4+2+5
F(2)=32 - 8 + 2 + 5 =31
F(2) = 31
f(2) = 32 - 8 + 2 + 5 = 31