Solve the following equation:x^3-2x^2-5x+6=0

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- Feb 25th 2007, 09:33 AMgracyquadratic equation
Solve the following equation:x^3-2x^2-5x+6=0

- Feb 25th 2007, 09:57 AMRanger SVO
It factors to ( x - 1 )* ( x + 2 )* ( x - 3 ) = 0

I bet you can get it from there - Feb 25th 2007, 11:07 AMticbol
What is that property or rule or "definition"? Anyway, always test the sum of all the numerical coefficients first. If they add up to zero, then x = 1 is a root.

x^3 -2x^2 -5x +6 = 0

1 -2 -5 +6 = 0 -----------------so 1 is a root, and so (x-1) is a factor.

Then divide the x^3 -2x^2 -5x +6 by (x-1) by long division or synthetic division. The quotient will be quadratic. Easy by then. Factor or use the Quadratic formula to find the other two factors or two roots.

After division, you'd get x^2 -x -6 as the quotient.

So,

x^2 -x -6 = 0

(x -3)(x+2) = 0

Hence the other two facxtors are (x-3) and (x+2)

Therefore, the original equation factors to

(x-1)(x-3)(x+2) = 0

Meaning,

x = 1 or 3 or -2 ------------answer. - Feb 25th 2007, 12:59 PMtopsquark
The "Rational Roots Theorem" states that if we have a polynomial equation:

ax^n + bx^{n-1} + ... + cx + d = 0

then any rational roots of the equation (if they exist, they many not!) will be of the form:

x = (+/-)[factor of d]/[factor of a]

In your case, the possible rational roots are going to be (+/-)1, 2, 3, 6. When we test each of these we find that x = -2, 1, 3 are solutions. Since this is a cubic equation, we know that 3 complex roots exist. Thus we have found all of them.

-Dan