How do you add #\frac { 7x } { 3x ^ { 2} - 19x + 6} + \frac { 5} { x ^ { 2} - 2x - 24}#?

1 Answer
George C.
Mar 4, 2017

#(7x)/(3x^2-19x+6)+5/(x^2-2x-24) = (7x^2+43x-5)/(3x^3-7x^2-70x+24)#

Explanation:

In order to add two fractions they need to have the same denominator. What is true for numbers is true for polynomials...

Note that:

#3x^2-19x+6 = (3x^2-18x)-(x-6) = (3x-1)(x-6)#

#x^2-2x-24 = (x-6)(x+4)#

So the LCM of #3x^2-19x+6# and #x^2-2x-24# is:

#(3x-1)(x-6)(x+4) = (3x^2-19x+6)(x+4)#

#color(white)((3x-1)(x-6)(x+4)) = 3x^3-7x^2-70x+24#

So we find:

#(7x)/(3x^2-19x+6)+5/(x^2-2x-24) = (7x(x+4)+5(3x-1))/(3x^3-7x^2-70x+24)#

#color(white)((7x)/(3x^2-19x+6)+5/(x^2-2x-24)) = (7x^2+28x+15x-5)/(3x^3-7x^2-70x+24)#

#color(white)((7x)/(3x^2-19x+6)+5/(x^2-2x-24)) = (7x^2+43x-5)/(3x^3-7x^2-70x+24)#

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