Hi,
Given x³ + 2x² - 5x - 6 = 0, and x = -1 is a root, find the others.
. ____________
-1)1 . 2. .-5 . .-6
. ___-1_-1___6_
. . 1 . 1 -6 . . 0
1 . 1 -6 are the quadratic x² + x - 6 which factors into (x + 3)(x - 2) = 0.
These solve to x = -3 and x = 2.
The roots are x = -3, x = -1, and x = 2. <==ANSWER
I hope that helps!! :-)
Given x³ + 2x² - 5x - 6 = 0, and x = -1 is a root, find the others.
. ____________
-1)1 . 2. .-5 . .-6
. ___-1_-1___6_
. . 1 . 1 -6 . . 0
1 . 1 -6 are the quadratic x² + x - 6 which factors into (x + 3)(x - 2) = 0.
These solve to x = -3 and x = 2.
The roots are x = -3, x = -1, and x = 2. <==ANSWER
I hope that helps!! :-)
-
note that highest power x-term is 3 so there are three solutions in form
(x-x1)(x-x2)(x-x3) where x is variable x, and x1,x2, x3 are solutions
in your case you know that x1=-1 so above factored form is then
(x-(-1))(x-x2)(x-x3)
(x+1)(x-x2)(x-x3)
to find remaining two terms we just do polynomial division
(x^3+2x^2-5x-6) / (x+1) = x^2+x-6
and that can be factored as (x-2)(x+3)
which brings last two solutions:
x2=2
x3=-3
(x-x1)(x-x2)(x-x3) where x is variable x, and x1,x2, x3 are solutions
in your case you know that x1=-1 so above factored form is then
(x-(-1))(x-x2)(x-x3)
(x+1)(x-x2)(x-x3)
to find remaining two terms we just do polynomial division
(x^3+2x^2-5x-6) / (x+1) = x^2+x-6
and that can be factored as (x-2)(x+3)
which brings last two solutions:
x2=2
x3=-3
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Two ways of doing this.
The best is to see that as x=-1 is a solution, then (x+1) is a factor. Divide x^3+2x^2-5x-6 by x+1 (see http://en.wikipedia.org/wiki/Polynomial_…
To give x^2 + x - 6 which factorises to (x+3)(x-2), so the other roots are -3 and 2.
The other method is to try other integers, and when you try x=-3 and x=2, the expression will equal 0. So you know (x+3) and (x-2) are factors. This can waste time or save time, depending on whether the other factors are integers, normally they won't be.
The best is to see that as x=-1 is a solution, then (x+1) is a factor. Divide x^3+2x^2-5x-6 by x+1 (see http://en.wikipedia.org/wiki/Polynomial_…
To give x^2 + x - 6 which factorises to (x+3)(x-2), so the other roots are -3 and 2.
The other method is to try other integers, and when you try x=-3 and x=2, the expression will equal 0. So you know (x+3) and (x-2) are factors. This can waste time or save time, depending on whether the other factors are integers, normally they won't be.
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x ^ 2 + x - 6 [After synthetic division]
(x + 3 ) ( x - 2 )
x = -3 , x = 2
(x + 3 ) ( x - 2 )
x = -3 , x = 2