Hello
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)
a)8a'3+64b'3c'6
b)8x'4-12x'3-44x'2+24x
c)x'4-21x'2-100

Thank you very much
Marcin

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.

Cheers,
Penny

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