How do you factor #2x^3-3x^2-2x+3#?

2 Answers
Jul 4, 2017

Answer:

#(2x-3)(x-1)(x+1)#

Explanation:

#2x^3-3x^2-2x+3#

First, start off by factoring the first two terms.

#x^2(2x-3)-2x+3#

Next, factor out the last two terms.

#x^2(2x-3)-(2x-3)#

By doing these steps, you now have #(2x-3)# to factor out.

#(2x-3)(x^2-1)#

The last thing you can do is factor #(x^2-1)#.

#(2x-3)(x-1)(x+1)#

Jul 4, 2017

Answer:

#(x-1)(2x-3)(x+1)#

Explanation:

#"note that the coefficients sum to zero"#

#2-3-2+3=0#

#rArr(x-1)" is a factor"#

#rArrcolor(red)(2x^2)(x-1)color(magenta)(+2x^2)-3x^2-2x+3#

#=color(red)(2x^2)(x-1)color(red)(-x)(x-1)color(magenta)(-x)-2x+3#

#=color(red)(2x^2)(x-1)color(red)(-x)(x-1)color(red)(-3)(x-1)color(magenta)(-3)+3#

#=color(red)(2x^2)(x-1)color(red)(-x)(x-1)color(red)(-3)(x-1)+0#

#rArr2x^3-3x^2-2x+3#

#=(x-1)(color(red)(2x^2-x-3))#

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