How do you reduce to lowest terms #(2x^2 - 5x + 3)/(2x^2-x-3)#?

2 Answers
Jim G.
Mar 13, 2018

#(x-1)/(x+1)#

Explanation:

#"Factorise numerator and denominator"#

#• " numerator "2x^2-5x+3#

#"The factors of + 6 which sum to - 5 are - 3 and - 2"#

#2x^2-2x-3x+3larrcolor(blue)"split the middle term"#

#=2x(x-1)-3(x-1)larrcolor(blue)"factor by grouping"#

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

#• " denominator "2x^2-x-3#

#"The factors of - 6 which sum to - 1 are - 3 and + 2"#

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

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

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

#rArr(2x^2-5x+3)/(2x^2-x-3)#

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

#=(x-1)/(x+1)#

#"with restriction "x!=-1#

LM
Mar 13, 2018

#(x-1)/(x+1)#

Explanation:

#2 * 3 = 6#

#-3 +-2 = -5#
#-3 * -2 = -6#

#2x^2 - 5x + 3 = 2x^2-2x-3x+3#

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

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

#2 * -3 = -6#

#-3 +2 = -1#
#-3 * 2 = -6#

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

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

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

#(2x^2-5x+3)/(2x^2-x-3) = ((2x-3)(x-1))/((2x-3)(x+1))#

#=(x-1)/(x+1)#

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