Algebra/Factoring Polynomials

=Formulas= Computing factors of polynomials requires knowledge of different formulas and some experience to find out which formula to be applied. Below, we give some important formulas:

$${x^2-y^2=(x+y)(x-y)}$$

$${x^2+2xy+y^2=(x+y)^2}$$

$${x^2-2xy+y^2=(x-y)^2}$$

$${x^3-y^3=(x-y)(x^2+xy+y^2)}$$

$${x^3+y^3=(x+y)(x^2-xy+y^2)}$$

$${a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)}$$

Examples
:$${a(a + b)}$$

$${x^2+3x+2}$$ $${ x^2 + y^2 = 2xy }$$ $${x(x + 1) + 2(x + 1)}$$ $${(x + 1)(x + 2)}$$

:$${(a^2 - 2ax) + (2ax - 4x^2)}$$ :$${a(a - 2x) + 2x(a - 2x)}$$ :$${(a - 2x)(a + 2x)}$$

$${x^3+8y^3}$$ $${(x^3 - 2x^2y + 4xy^2) + (2x^2y - 4xy^2 + 8y^3)}$$ $${x(x^2 - 2xy + 4y^2) + 2y(x^2 - 2xy + 4y^2)}$$ $${(x^2 - 2xy + 4y^2)(x + 2y)}$$ $${(x + 2y)(x^2 - 2xy + 4y^2)}$$

:$${(x^3 + 4x^2 - 5x) - (2x^2 + 8x - 10)}$$ :$${x(x^2 + 4x - 5) - 2(x^2 + 4x - 5)}$$ :$${(x^2 + 4x - 5)(x - 2)}$$ :$${(x - 2)(x^2 + 4x - 5)}$$ :$${(x - 2)[(x^2 - x) + (5x - 5)]}$$ :$${(x - 2)[x(x - 1) + 5(x - 1)]}$$ :$${(x - 2)(x - 1)(x + 5)}$$

$${3x^4 - 3x^3 - 2x^2 - x - 1}$$ $${(3x^4 - 3x^3 - 3x^2) + (x^2 - x - 1)}$$ $${3x^2(x^2 - x - 1) + 1(x^2 - x - 1)}$$ $${(x^2 - x - 1)(3x^2 + 1)}$$

:$${(x^4 + 4x^2 + 4) - 4x^2}$$ :$${(x^2 + 2)^2 - (2x)^2}$$ :$${(x^2 + 2x + 2)(x^2 - 2x + 2)}$$

(name thorem)

write out the coefficients and if the end is equal to zero, than it is a root

example: $${9x^2 - 6x + 9}$$

$${4r^2 - 12 + 9^2}$$

= Possible Factors =

To factor we must first look for possible factors. Possible factors are any number that might be a factor. Once we have a possible factor then we divide that number into the number we are factoring. If they divide evenly then we have a factor! The factor is the possible factor we found and the result of the division problem. Here is an example. Let's say the number we are factoring is 20. 2 is the possible factor. 20 / 2 = 10. They divide evenly which means we have a factor. The factors are 2 (the possible factor), and 10 (the result of the division problem). Now that we have a factor we start over with a new possible factor and find all of the factors.

Examples
Factor 12

First find all the possible factors

The possible factors are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12

Next we will try them one by one

12/1 = 12 (1 and 12 are factors)

12/2 = 6 (2 and 6 are factors)

12/3 = 4 (3 and 4 are factors)

12/4 = 3 (we already have the factors 3 and 4)

Once we get a factor we already have then we know we have all the factors.

So the factors for 12 are 1, 2, 3, 4, 6, and 12.

Factor 54

First find all the possible factors

The possible factors are {1, 2, 3 ... 52, 53, 54}

Do not worry this is not as much work as it seems!

54/1 = 54   (1 and 54 are factors)

54/2 = 27   (2 and 27 are factors)

54/3 = 18   (3 and 18 are factors)

54/4 = 13r2 (4 is not a factor)

54/5 = 10r4 (5 is not a factor)

54/6 = 9    (6 and 9 are factors)

54/7 = 7r5  (7 is not a factor)

54/8 = 6r6  (8 is not a factor)

54/9 = 6    (we already have the factors 9 and 6)

So the factors for 54 are 1, 2, 3, 6, 9, 18, 27, and 54

Factor 180

First find all the possible factors

The possible factors are {1, 2, 3 ... 178, 179, 180}

Do not worry this is not as much work as it seems!

180/1 = 180   (1 and 180 are factors)

180/2 = 90    (2 and 90 are factors)

180/3 = 60    (3 and 60 are factors)

180/4 = 45    (4 and 45 are factors)

180/5 = 36    (5 and 36 are factors)

180/6 = 30    (6 and 30 are factors)

180/7 = 25r5  (7 is not a factor)

180/8 = 22r4  (8 is not a factor)

180/9 = 20    (9 and 20 are factors)

180/10 = 18   (10 and 18 are factors)

180/11 = 16r4 (11 is not a factor)

180/12 = 15   (12 and 15 are factors)

180/13 = 13r11 (13 is not a factor)

180/14 = 12r12 (14 is not a factor)

180/15 = 12   (we already have the factors 15 and 12)

So the factors for 180 are 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, and 180.

= Dividing polynomials =

The process of factoring will require dividing polynomials. This form of division is not too different from long division method, and is known as synthetic division.

Consider the polynomial x3 - 21x2 + 143x - 315. In this case, determining factors may require trial and error (until you learn of alternate techniques), and when you do, you will need to divide the polynomial with the discovered factor.

In this example, we will divide by (x-5). The full division starts like this:

1x -5 | 1x^3 -21x^2 +143x - 315

As with long division, you need to find the number used for subtraction and place it on top - in this case, you need to make sure the left-most term becomes zero. Next, multiply the newly added top-most term with the left hand side to get the amount to subtract, and perform the subtraction.

1x^2 1x -5 | 1x^3 -21x^2 +143x - 315 1x^3  -5x^2 -16x^2 +143x - 315

Repeat until the division is complete:

1x^2 -16x + 63 1x -5 | 1x^3 -21x^2 +143x - 315 1x^3  -5x^2 -16x^2 +143x - 315 -16x^2 + 80x -                       63x - 315 63x - 315 -                               0

(If there is a remainder at this point, place it as the numerator over the term being factored out.)

Some people may find writing the x3 and other variables to be bulky - if writing on pen and paper, they can be omitted as part of shorthand.

1 -16 + 63 1 -5 | 1 -21 +143 - 315        1  -5        -          -16 +143 - 315          -16 + 80          -                63 - 315                63 - 315                       0

In this case, factoring is straight forward since you can easily determine the number to use for the next step in division.

=See also=
 * Factoring polynomials - example of use