Factorizing an Algebraic Function

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To find the divisor of the polynomial x^3 - 4x^2 + x - 1, given the quotient x - 6 and remainder 10x + 17, the division algorithm is applied. The expression simplifies to (x^3 - 4x^2 - 9x - 18) / (x - 6). It is noted that x - 6 is a factor since f(6) equals zero. The factorization involves solving the equation x^3 - 4x^2 - 9x - 18 = (x - 6)(ax^2 + bx + c), with a determined to be 1 and c found to be 3. The method of equating coefficients is highlighted as a useful technique in this factorization process.
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Homework Statement


Find the divisor in each of the following.
Dividend = x^3 - 4x^2 + x - 1, quotient = x - 6, remainder = 10x + 17

Homework Equations


Dividend = divisor x quotient + remainder


The Attempt at a Solution


By division algorithm, we have
f(x) = [(x^3 - 4x^2 + x - 1) - (10x + 17)] / (x - 6)
= (x^3 - 4x^2 - 9x - 18) / (x - 6)

At this point I can't factorize the numerator. Any help would be appreciated. Thanks!
 
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So it is obvious from the question that the cubic should have a factor of x-6 and you can even check it by showing f(6)=0.

Anyway, to factorize it all you have to do is solve

x^3 - 4x^2 - 9x - 18=(x-6)(ax^2+bx+c)

You just need to find the values of the constants a,b,c and you can quickly find two by noting that the cubic power on the right will be (x)(ax2) and this is equivalent to x3 so of course a=1, and the constant on the right is (-6)(c) and this is equal to -18 from the left, so c=3. Now just find the value of b by expanding the right side.

This is called equating coefficients by the way.
 
Thanks! I just found out that the solution is absolutely obvious if we use factor theorem.
 
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