Factorizing an Algebraic Function

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SUMMARY

The discussion focuses on factorizing the cubic polynomial \(x^3 - 4x^2 + x - 1\) using the division algorithm and the factor theorem. The quotient is given as \(x - 6\) with a remainder of \(10x + 17\). The solution involves rewriting the polynomial as \((x - 6)(ax^2 + bx + c)\) and determining the coefficients \(a\), \(b\), and \(c\) through equating coefficients, ultimately finding \(a = 1\) and \(c = 3\). The factor theorem confirms that \(x - 6\) is indeed a factor of the polynomial.

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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.
 
Last edited:

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