Factoring by grouping does work here, unless I messed up somewhere. x^4 - 1 + 2x^3 - 2x (x^2 + 1)(x^2 - 1) + 2x(x^2-1) You can factor out (x^2-1) and proceed
In the first method, once you get to a quartic, the first step should be to look at the sum of coefficients. If it's zero (which it is), we know immediately that 1 is a solution. We can then factor by grouping, with that in mind. Similarly, if we compare the sum of coefficients for the even exponents to that for the odd, we find that they are equal, meaning that -1 is also a solution. That immediately eliminates the need to reduce the quartic. (By the way, I didn't pay close attention, but I think you dropped a 2y term.)
factorization does work, we just need a different way of grouping: x^4 + 2x^3 - 2x - 1 = 0 x^4 - 1 + 2x^3 - 2x (x^2+1)(x^2-1) + 2x(x^2-1) = 0 by the distribution rule, the LHS is (x^2+2x+1)(x^2-1) (x^2+2x+1)(x^2-1) = 0 (x^2+2x+1) = (x+1)^2 (x^2-1) = (x+1)(x-1) so the equation can be factorized as (x-1)(x+1)^3 = 0 so x can be 1 or -1
MERRY CHRISTMAS 🎁🎄 🎅🏻 ❤❤ problem (2x + 1) / (x + 2) = x³ Domain: x ≠ -2 Multiply by x + 2 . 2x + 1 = x³ ( x + 2 ) Expand. Put in standard form. x⁴ + 2 x³- 2 x - 1 = 0 Since the sum of the coefficients is 0, x = 1 is a root and x -1 is a factor. Factor. (x - 1) x³ + 3 (x - 1) x² + 3 (x - 1) x + (x-1) = 0 x⁴ + 2 x³- 2 x - 1 = (x-1)(x³ +3x² +3x+1) Use the zero product property. x³ +3x² +3x+1 = 0 x = -1 is a root, so x+1 is a factor. Factor. (x+1)x² + 2(x+1)x+(x+1) = 0 x³ +3x² +3x+1 = (x+1)(x² + 2x +1) x⁴ + 2 x³- 2 x - 1 = (x-1)(x+1)(x² + 2x +1) x² + 2x +1 = (x+1)² Factored form of x⁴ + 2 x³- 2 x - 1 = 0 is (x-1)(x+1)³ = 0 x = 1 is a root and x = -1 is a triple root. Discussion: The graph is interesting because it is the intersection of the cubic y = x³ with an hyperbola having vertical asymptote at x = -2 and horizontal asymptote of y = 2. The cubic intersects only the branch of the 2 branched hyperbola with x > -2. answer x ∈ { -1, 1 }
Factoring by grouping does work here, unless I messed up somewhere.
x^4 - 1 + 2x^3 - 2x
(x^2 + 1)(x^2 - 1) + 2x(x^2-1)
You can factor out (x^2-1) and proceed
This works, you can simply use the expand function on Symbolab to confirm it
Very good!
(x-1)(x+1)^3=0
In the first method, once you get to a quartic, the first step should be to look at the sum of coefficients. If it's zero (which it is), we know immediately that 1 is a solution. We can then factor by grouping, with that in mind. Similarly, if we compare the sum of coefficients for the even exponents to that for the odd, we find that they are equal, meaning that -1 is also a solution.
That immediately eliminates the need to reduce the quartic. (By the way, I didn't pay close attention, but I think you dropped a 2y term.)
factorization does work, we just need a different way of grouping:
x^4 + 2x^3 - 2x - 1 = 0
x^4 - 1 + 2x^3 - 2x
(x^2+1)(x^2-1) + 2x(x^2-1) = 0
by the distribution rule, the LHS is (x^2+2x+1)(x^2-1)
(x^2+2x+1)(x^2-1) = 0
(x^2+2x+1) = (x+1)^2
(x^2-1) = (x+1)(x-1)
so the equation can be factorized as
(x-1)(x+1)^3 = 0
so x can be 1 or -1
Good thinking!
X=1, X=-1 (triple root)
Got 'em both!
It's pretty shoddy to upload unchecked (wrong) solutions.
x⁴ + x³ + x³ - x - x - 1 = 0
⇒ x³(x + 1) + x(x² - 1) - (x + 1) = 0
⇒ (x + 1)( x³ - 1 + x(x - 1) ) = 0
⇒ (x + 1)( (x - 1)(x² + 1 + x) + x(x - 1) ) = 0
⇒ (x + 1)(x - 1)(x² + 2x + 1) = 0
⇒ (x + 1)³(x - 1) = 0
Happy Christmas!
x^4+2x^3-2x-1=0
x^4-1+2x(x^2-1)=0
(x^2-1)(x^2+1)+2x(x^2-1)=0
(x^2-1)(X^2+1+2x)=0
(X-1)(x+1)(x+1)^2=0
(x-1)(x+1)^3=0
Yo, I need help proving there is a unique prime p such that p^3 - p + 9 is a perfect square.
Too long answer, there is shorter way that guys suggested.
Interesting! Maybe I can make a video about that!
You mixed up x^2 & x^3 in the middle of method 1
x = 1 or -1
x=1 0:33 guessing x=-1
👍
x^4+2x^3-2x-1 = 0
consider
(x+1)^4 = x^4+4 x^3+6 x^2+4 x+1
2(x+1)^3 = 2x^3+6 x^2+6 x+2
hence
(x+1)^4 - 2 (x+1)^3 = 0
(x+1)^3 (x-1) = 0
x = -1 ,-1 , -1 , 1
ua-cam.com/users/shortsQql-4XAAfyQ?si=ShpTEQ2fx2Z6Fu3F
x=±1 at a glance. The rest is not interesting
😁😄😁😄😁😄😜😂😂
MERRY CHRISTMAS 🎁🎄 🎅🏻 ❤❤
problem
(2x + 1) / (x + 2) = x³
Domain: x ≠ -2
Multiply by x + 2 .
2x + 1 = x³ ( x + 2 )
Expand. Put in standard form.
x⁴ + 2 x³- 2 x - 1 = 0
Since the sum of the coefficients is 0, x = 1 is a root and x -1 is a factor. Factor.
(x - 1) x³ + 3 (x - 1) x² + 3 (x - 1) x + (x-1) = 0
x⁴ + 2 x³- 2 x - 1 = (x-1)(x³ +3x² +3x+1)
Use the zero product property.
x³ +3x² +3x+1 = 0
x = -1 is a root, so x+1 is a factor. Factor.
(x+1)x² + 2(x+1)x+(x+1) = 0
x³ +3x² +3x+1 = (x+1)(x² + 2x +1)
x⁴ + 2 x³- 2 x - 1 = (x-1)(x+1)(x² + 2x +1)
x² + 2x +1 = (x+1)²
Factored form of
x⁴ + 2 x³- 2 x - 1 = 0
is
(x-1)(x+1)³ = 0
x = 1 is a root and x = -1 is a triple root.
Discussion:
The graph is interesting because it is the intersection of the cubic y = x³ with an hyperbola having vertical asymptote at x = -2 and horizontal asymptote of y = 2. The cubic intersects only the branch of the 2 branched hyperbola with x > -2.
answer
x ∈ { -1, 1 }
Merry Christmas! Happy Holidays!!! 🎁🎄 😍
@@SyberMath thank you 🙏