How do you factor #9x^3-x#?

3 Answers
Jan 25, 2017

Answer:

#x ( 3x + 1 )( 3x-1 )#

Explanation:

#x# is common to both terms, so you can take it out of both and put the other factor in a bracket.

#9x^3 -x =x ( 9x^2 - 1)#.

#(x^2 -1)# and similar expressions are known as the difference of two squares, and can be further factorised to give:

#x( 3x + 1 )( 3x - 1 )#.

Jan 25, 2017

Answer:

#x(3x-1)(3x+1)#

Explanation:

#9x^3-x#

#:.=x(9x^2-1)#

#:.=x(3x-1)(3x+1)#

Jan 25, 2017

Answer:

#x(3x-1)(3x+1)#

Explanation:

There is a #color(blue)"common factor"# of x in both terms.

#rArrx(9x^2-1)larrcolor(red)"remove common factor"#

#9x^2-1" is a"color(blue)" difference of squares"# and factorises, in general as shown.

#color(red)(bar(ul(|color(white)(2/2)color(black)(a^2-b^2=(a-b)(a+b))color(white)(2/2)|)))#

#"here " (3x)^2=9x^2" and " 1^2=1#

#rArra=3x" and " b=1#

#rArr9x^2-1=(3x-1)(3x+1)#

#rArr9x^3-1=x(3x-1)(3x+1)larrcolor(red)" fully factorised"#