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.
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.
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.
Hello, skg94!
You are expected to know basic factoring and these three formulas:
. . Difference of squares: .
. . Sum/difference of cubes: .
Here's where it gets sneaky . . . is a difference of squares.
. .
The fraction becomes: .
We have: .
. sum of cubes
. diff. of squares
.diff. of squares
. . . . . . . . . .
. . . . . .
We have: .
. .
. .
. .
. .