How do you simplify #\frac { 4} { 3x - 6} - \frac { 5} { 4x + 4} + \frac { 2} { 5x + 10}#?

1 Answer
ErmWhat
Sep 20, 2017

#(29x^2+216x+412)/((4x+4)(5x+10)(3x-6))#

Explanation:

You first have to make every denominator the same, so multiply all the fractions by another fraction that equals one.

#=4/(3x-6)*((4x+4)(5x+10))/((4x+4)(5x+10)) - 5/(4x+4)((3x-6)(5x+10))/((3x-6)(5x+10))+2/(5x+10) ((3x-6)(4x+4))/((3x-6)(4x+4))#

#= (4(4x+4)(5x+10)-5(5x+10)(3x-6)+2(3x-6)(4x+4))/((4x+4)(5x+10)(3x-6))#

You get a really large fraction, but from here you just expand simplify. You don't have to expand the denominator because nothing factors easily yet.

#=(4(20x^2+60x+40)-5(15x^2-60)+2(12x^2-12x-24))/((4x+4)(5x+10)(3x-6))#
#=(80x^2+240x+160-75x^2+300+24x^2-24x-48)/((4x+4)(5x+10)(3x-6)#
#=(29x^2+216x+412)/((4x+4)(5x+10)(3x-6))#

Using the quadratic formula, we can tell what the roots are for the quadratic in the numerator.

#x=(-b+-sqrt(b^2-4ac))/(2a) = (-216+-sqrt(216^2-4(29)(412)))/(2(29))=(-216+-sqrt(46656-47792))/58=(-216+-sqrt(-1136))/58#

Since taking the square root of a negative number gives us imaginary answers, we can conclude that the quadratic cannot factor further, and this is the final, simplified fraction:
#(29x^2+216x+412)/((4x+4)(5x+10)(3x-6))#

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