How do you divide #\frac { x ^ { 2} - 3x - 40} { x ^ { 2} + 2x - 35} \div \frac { x ^ { 2} + 2x - 48} { x ^ { 2} + 3x - 18}#?

Question

638 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-25T06:24:31

#(x^2-3x-40)/(x^2+2x-35)xx(x^2+3x-18)/(x^2+2x-48)=((x+5)(x-8)(x-3)(x+6))/((x-5)(x+7)(x-6)(x+8))#

#(x^2-3x-40)/(x^2+2x-35)-:(x^2+2x-48)/(x^2+3x-18)#

= #(x^2-3x-40)/(x^2+2x-35)xx(x^2+3x-18)/(x^2+2x-48)#

Factorizing each of the quadratic polynomial

#x^2-3x-40=x^2-8x+5x-48=x(x-8)+5(x-8)=(x+5)(x-8)#

#x^2+2x-35=x^2+7x-5x-35=x(x+7)-5(x+7)=(x-5)(x+7)#

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

#x^2+2x-48=x^2+8x-6x-48=x(x+8)-6(x+8)=(x-6)(x+8)#

Hence #(x^2-3x-40)/(x^2+2x-35)xx(x^2+3x-18)/(x^2+2x-48)#

= #((x+5)(x-8))/((x-5)(x+7))xx((x-3)(x+6))/((x-6)(x+8))#

Observe that there is no binomial common between numerator and denominator and hence you cannot simplify it further and

#(x^2-3x-40)/(x^2+2x-35)xx(x^2+3x-18)/(x^2+2x-48)=((x+5)(x-8)(x-3)(x+6))/((x-5)(x+7)(x-6)(x+8))#