How do you simplify #(x+2) /(4x^2 - 14x + 6)-(x+4)/(x^2 + x -12)#?

Question

1778 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-16T15:34:09

#= [-3x^2- 16x + 16]/[(x-3)(4x-2)(x+4)]#

Factor first.

#(x+2) / (4x^2-14x+6) -(x+4)/(x^2+x-12)#

= #(x+2)/((x-3)(4x-2)) - (x+4)/((x+4)(x-3))#

Change to similar fractions by getting #LCD = (x -3)(4x-2)(x-3)#

= #[(x+2)(x+4)]/((x-3)(4x-2)(x+4))- [(x +4)(4x - 2)]/((x-3)(4x-2)(x+4))#

= #((x+2)(x+4)-(x+4)(4x-2))/((x-3)(4x-2)(x+4))#

Use distributive property of multiplication.

#=[(x^2 +4x + 2x + 8)-(4x^2 -2x +16x -8)]/[(x-3)(4x-2)(x+4)]#

Combine like terms.

#= [-3x^2- 8x + 16]/[(x-3)(4x-2)(x+4)]#