How do you multiply #(x^2-x-6)/(x^2+4x+3) *( x^2-x-12)/(x^2-2x-8)#?

1 Answer
Apr 5, 2016

Answer:

#(x^2-x-6)/(x^2+4x+3)*(x^2-x-12)/(x^2-x-8)=(x-3)/(x+1)#

Explanation:

To solve the multiplication, we should first factorize each quadratic polynomial. Hence,

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

#x^2+4x+3=x^2+3x+x+3=x(x+3)+1(x+3)=(x+1)(x+3)#

#x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x+3)(x-4)#

#x^2-x-8=x^2-4x+2x-8=x(x-4)+2(x-4)=(x+2)(x-4)#

Hence #(x^2-x-6)/(x^2+4x+3)*(x^2-x-12)/(x^2-x-8)#

= #((x+2)(x-3))/((x+1)(x+3))*((x+3)(x-4))/((x+2)(x-4))#

= #((cancel(x+2))(x-3))/((x+1)(cancel(x+3)))*((cancel(x+3))(cancel(x-4)))/(cancel((x+2))(cancel(x-4)))#

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