Quandaries and Queries


I am having a really hard time with the factor theorem.
Can you please help me with the following?
(a'3 means a to the exponent 3)

Thank you very much



Hi Marcin,

In many instances factoring is recognizing patterns. For example in your first problem I see that 8, a3, 64, b3 and c6 are all cubes.Hence I can write

8a3 + 64b3c6 = (2a)3 + (4bc2)3

Can you factor it now as a sum of cubes?

In b), all the coefficients are divisible by 4 and there is an x common factor in each term, thus

8x4 - 12x3 - 44x2 + 24x = 4x(x3 -3x2 -11x + 6)

Hence you are left with the cubic x3 -3x2 -11x + 6 which is a candidate for the factor theorem.

To use the factor theorem you need to find a root, that is a number x for which

x3 -3x2 -11x + 6 = 0.

Usually we look for a root that is an integer. Such an integer must divide the constant term (6 in this case). Thus the only candidates for x are 1,2,3,6,-1,-2,-3 and -6. If you try these numbers (start with the small numbers to make the calculation easy) you will find that x=-2 is a root. The factor theorem then implies that (x+2) is a factor of the polynomial.

Now divide (x+2) into x3 -3x2 -11x + 6. to get

x3 -3x2 -11x + 6 = (x + 2)(a quadratic)

In this example you can also factor the quadratic.



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