1. [ 1 / sqroot(x)] - 1 / (x-1)

2. [x*sqroot(x)] -8 / (x-4)

Factor:

a) x^6+8

b)x^4-16

c)r^8-1

d) x^3-x^2-16x+16

f)x^4-3x^3-7x^2+27x-18

are there any easy ways to factor theses without graphing, algebraically. I only know of synthetic division.

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- Feb 6th 2013, 07:23 PMskg94Rationalizing the numerator and complex factoring
1. [ 1 / sqroot(x)] - 1 / (x-1)

2. [x*sqroot(x)] -8 / (x-4)

Factor:

a) x^6+8

b)x^4-16

c)r^8-1

d) x^3-x^2-16x+16

f)x^4-3x^3-7x^2+27x-18

are there any easy ways to factor theses without graphing, algebraically. I only know of synthetic division. - Feb 6th 2013, 08:34 PMibduttRe: Rationalizing the numerator and complex factoring
There are certain formulas which one must learn. (a^3+b^3) = (a+b)(a^2-ab+b^2); (a^3-b^3) = (a+b)(a^2+ab+b^2); a^2-b^2 = (a+b)(a-b); in case of cubic and bi-quadratic polynomials we can find linear factor by checking the sum of the coefficients, if the sum of the coefficients is zero then x = 1 is a solution and (x-1) a factor. Lastly if the sum of the coefficients of the even powers of the variable is equal to the sum of the coefficients of the odd powers of the variables then x = -1 is a solution and (x+1) is a factor.

- Feb 6th 2013, 08:47 PMSorobanRe: Rationalizing the numerator and complex factoring
Hello, skg94!

You are expected to know basic factoring and these three formulas:

. . Difference of squares: .

. . Sum/difference of cubes: .

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Here's where it gets sneaky . . . is a.*difference of squares*

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. sum of cubes

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. diff. of squares

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.diff. of squares

. . . . . . . . . .

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. . . . . .

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We have: .

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. .

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