It's only a cubic, and x=3 is obviously a root. Shouldn't be too hard. Why is 3 a root? Looking at the LHS, it's dominated by the x^3. (x-1)^2 is relatively small. So, what perfect cube is closest to 23? Obviously 27. So, try x=3. It works!
I used the same method.Then x³-x²+2x-24=0 becomes (x-3)(x²+2x+8)=0 which is so easy (and quick) to resolve. I don't know the theory used in the 2 method used in this video (a link to an explanation can be useful, perhaps SyberMath can provide one?). It's usually a good thing to try to guess obvious solutions so we can reduce the degree of the polynomial.
I'm glad I was able to solve this with the smart (second) method myself. By the way, I don't count mistakes in arithmetic as mistakes. If there is logic issue or some case missed, then it is mistake.
It doesn't have one root. There are three roots, you should have watched the full video(two imaginary solutions and one real number solution)... Yes, of course you can guess and check the integer solutions, but not all of them.
Using Descartes' Rule of Signs you can limited the number of integer roots the rrt has you check, You find 3 and see rhere are no larger positive roots. and no negative roots.
3rd method: check for solutions mod small numbers: mod 2 doesn't help but mod 3 shows any solutions need to be divisible by 3. So 3 turns out to be a solution when testing, etc. etc. etc.
Same answer but a different technique using the expected format (x+a)(x+b)(x-3)=0 knowing that ab=8 since -3ab = -24. Then solving for a & b by expanding the expected form above. Expansion yields the coefficient for the x^2 term: -3+a+b. Since we know by expanding the original equation x^3-x^2+2x-24=0, the coefficient is -1. Hence -3+a+b=-1 so a+b=2. Solving for a and substituting into ab=8 yields a quadratic in b. The quadratic formula gives us b=1+/-sqrt(7)i . Using the coefficient of the x term yields the same result for verification. Solving for a is simple and yields the complex conjugate. When these values for a and b are inserted in the expanded equation above we easily see our other two roots.
@@user-jj4dw6kv5u -- Thanks! I also was thrown by this. I'm usually thinking of integer factors that are neither 1 nor the number itself, so I was getting 6. But for this case -- polynomial roots -- you're absolutely right: We need negatives and we need 1 and 24.
Can't we just get 3 by putting it in the equation, and by synthetic division, we can get the equation x2+2x+8, which gonna have imaginary roots, so 3 is the only answer? (Noob alert)
x^3 - (x - 1)^2 = 23
x^3 - (x^2 - 2x + 1) = 23
x^3 - x^2 + 2x - 1 = 23
x^3 - x^2 + 2x - 24 = 0
So we try the divisors of 24, which are plus or minus 1,2,3,4,6,8,12,24.
It is easier to write
x^3 + 2x = x^2 + 24
in order to find the positive integer solutions. We find x1 = 3, because
3^3 + 2*3 = 3^2 + 24
27 + 6 = 9 + 24
33 = 33
Then we divide the polynomials:
(x^3 - 1x^2 +2x - 24):(x - 3) = x^2 + 2x + 8
- (x^3 - 3x^2)
-----------------------------------
2x^2 + 2x - 24
- (2x^2 - 6x)
--------------------------
8x - 24
- (8x - 24)
---------------
0
The remaining two solutions are complex:
x^2 + 2x + 8 = 0
Subtract 7:
x^2 + 2x + 1 = -7
(x + 1)^2 = -7
x + 1 = +- sqrt(-7)
x_{2,3} = -1 +- sqrt(7)*i
use horner's method if you want factorize efficiently
GENIUS 💪
It's only a cubic, and x=3 is obviously a root. Shouldn't be too hard.
Why is 3 a root? Looking at the LHS, it's dominated by the x^3. (x-1)^2 is relatively small. So, what perfect cube is closest to 23? Obviously 27. So, try x=3. It works!
I used the same method.Then x³-x²+2x-24=0 becomes (x-3)(x²+2x+8)=0 which is so easy (and quick) to resolve. I don't know the theory used in the 2 method used in this video (a link to an explanation can be useful, perhaps SyberMath can provide one?). It's usually a good thing to try to guess obvious solutions so we can reduce the degree of the polynomial.
that perfect cube must be bigger than 23 because (x-1)^2>=0
I'm glad I was able to solve this with the smart (second) method myself. By the way, I don't count mistakes in arithmetic as mistakes. If there is logic issue or some case missed, then it is mistake.
You could use Descartes Rule of Sign to see that there must be a positive root. This narrows what to check somewhat.
You might also want to mention Descartes' Rule of Signs occasionally.
i like the bloopers
Third power equetion, with 1 root, that you can just guess and check - simple enough
It doesn't have one root. There are three roots, you should have watched the full video(two imaginary solutions and one real number solution)... Yes, of course you can guess and check the integer solutions, but not all of them.
@@kobalt4083 of course, it have 3 roots, but because 1 root you can gues - you can just extract it as a common factor, and solve quadratci equetion
i love this videos
Glad to hear that!
Using Descartes' Rule of Signs you can limited the number of integer roots the rrt has you check, You find 3 and see rhere are no larger positive roots. and no negative roots.
3rd method: check for solutions mod small numbers: mod 2 doesn't help but mod 3 shows any solutions need to be divisible by 3. So 3 turns out to be a solution when testing, etc. etc. etc.
Same answer but a different technique using the expected format (x+a)(x+b)(x-3)=0 knowing that ab=8 since -3ab = -24. Then solving for a & b by expanding the expected form above. Expansion yields the coefficient for the x^2 term: -3+a+b. Since we know by expanding the original equation x^3-x^2+2x-24=0, the coefficient is -1. Hence -3+a+b=-1 so a+b=2. Solving for a and substituting into ab=8 yields a quadratic in b. The quadratic formula gives us b=1+/-sqrt(7)i . Using the coefficient of the x term yields the same result for verification. Solving for a is simple and yields the complex conjugate. When these values for a and b are inserted in the expanded equation above we easily see our other two roots.
So easy question
x^3-x^2+2x-24=0
(x-3)(x^2+2x+8)=0
x=3 or x^2+2x+8=0(rejected as it has no real roots)
Hence answer: x=3
Got 'em all!
3 is an obvious solution.
divide by x-3, solve the quadratic.
My Solution:
x^3 - (x-1)^2 = 23 --> x^3 - 3^3 - ((x-1)^2 - 4) = 0
--> (x-3)(x^2+3x+9) - (x-1-2)(x-1+2) = 0
--> (x-3)(x^2+3x+9-x-1) = (x-3)(x^2+2x+8) = 0
--> (Apply quadratic formula):
x = {3, -1-sqrt(7)i, -1+sqrt(7)i}
I could solve this problem in less than a minute 🤩😍
Xis √24.
[27] - [4] = 23
x = (3)
(3) - 1 = 2
x^3 = [27]
(x(=3) - 1)^2 = [4]
first solution found by guess))
Rest we have a quadratic and it easy to solve
I solved it by simply guessing and way faster
3
all that work just to get 3.
total number of factors of 24 is 8 not 16
@@user-jj4dw6kv5u -- Thanks! I also was thrown by this. I'm usually thinking of integer factors that are neither 1 nor the number itself, so I was getting 6. But for this case -- polynomial roots -- you're absolutely right: We need negatives and we need 1 and 24.
❤❤❤
Can't we just get 3 by putting it in the equation, and by synthetic division, we can get the equation x2+2x+8, which gonna have imaginary roots, so 3 is the only answer? (Noob alert)
x=3, -1±√7i
👍
yo
easy peasy
3
3