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QUESTION: Solve 18x^(3) - 33x^(2) + 5x + 6 = 0

ANSWER: Factor theorem can be used to solve this.

If we substitute x= 2/3 into 18x^(3) - 33x^(2) + 5x + 6,

we get 18*(8/27) - 33* (4/9) +10/3 +6 = 16/3 - 44/3 +10/3 +6 = 0

Hence, 3x-2 is a factor of the equation.

18x^(3) - 33x^(2) + 5x + 6 = 0 can therefore be rewritten as

(3x-2)(Ax^2 + Bx +C )=0 ----------(1)

By comparison, 3A=18 ===> A =6 and -2C =6 =====>C=-3

Thus, (1) becomes

(3x-2)(6x^2 + Bx -3)=0 --------(2)

The coefficient of x^2 in the LHS expansion of (2) is 3B-12; comparing this with the coefficient of x^2 in LHS of the original equation, we have 3B-12 = -33 ===>B =-7

So, (2) becomes

(3x-2)(6x^2 -7x -3)=0

Further factorization gives

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

Hence, the roots of the equation are x=2/3, x=3/2 and x =-1/3 (shown)

Hope this helps. Peace.

---------- FOLLOW-UP ----------

QUESTION: But how do we know what to substitute, in the question case we substitute x=2/3, how do we know what to substitute for value of x. Thanks for the solution.

Typically, if you are required to discover the roots of a cubic equation via factorization, the equation would be designed such that the student can at least discover a root via trial and error. Such a root would usually be a low value integer (1, 2 or 3) or a fraction (1/2, 1/3, 2/3 or 1/4). Of course if these don't work you can try their negative counterparts. Hope this helps. Peace.

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