Question

can someone explain to me the process how to Factor?
example factor 27y^3+x^9.
and Factor by grouping: xw+3x−6w−18

Answer 1

sum of cubes ..... seems to be the first one.

and ive never got the hang of grouping :/

Answer 2

\[a^3+b^3=(a+b)(a^2-ab+b^2)\]
this is just one of those rules that you have to memorize

Answer 3

xw+3x−6w−18; this is usually accomplished by forming an addition of 2 groups:

(xw+3x) + (−6w−18), take out common factors from each group
^^ ^^
x -6


x (w+3) -6 (w+3) ; with any luck, this produces another
common factor of the 2 groups

(w+3) (x-6)

Answer 4

I got it up to here x(w+3) -6(w+3) , but how do you get (w+3) (x-6)?

Answer 5

what is the common factor of: x(w+3) -6(w+3)

we can clean this up by letting a = w+3

xa -6a, do you agree that combining like terms of a, we get: (x-6) a ??

Answer 6

if so, then we can just replace a with what we defined it to be to begin with

(x-6) a = (x-6) (w+3)

Answer 7

OK i get it, the common factor for xa-6a is a so a(x-6)

Answer 8

correct

Answer 9

amistre64 could you help i got lost in factoring 27y^3+x^9.
i used a3+b3=(a+b)(a2−ab+b2)

but i'm not getting anywhere

Answer 10

a = 3y, a^3 = 27y^3

b = x^3, b^3 = x^9

thats all that amounts to

Answer 11

recall that when we raise a product to an exponent, we raise all the parts as well.

\[(kx)^3=kn*kn*kn=k^3n^3\]

Answer 12

lol, umm either change x to n, or n to x i seem to have been thinking of 2 different ideas at ones

Answer 13

\[(kx)^n=kx*kx*kx*...*kx:n~times=k^nx^n\]
\[(kx)^n=k^nx^n\]

Answer 14

ok thanks