Intermediate Algebra/Factoring Polynomials

Template:Algebra Page 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:

:$a^2+ab$
:$a(a+b)$

$x^2+3x+2$
$(x^2+x)+(2x+2)$
$x(x+1)+2(x+1)$
$(x+1)(x+2)$

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

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

:$x^3+2x^2-13x+10$
:$(x^3+4x^2-5x)-(2x^2+8x-10)$
:$x(x^2+4x-5)-2(x^2+4x-5)$
:$(x^2+4x-5)(x-2)$
:$(x-2)(x^2+4x-5)$
:$(x-2)[(x^2-x)+(5x-5)]$
:$(x-2)[x(x-1)+5(x-1)]$
:$(x-2)(x-1)(x+5)$

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

:$x^{4}5+4$
:$(x^4+4x^2+4)-4x^2$
:$(x^2+2{)}^2-(2x{)}^2$
:$(x^2+2x+2)(x^2-2x+2)$

(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.

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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