To solve #(x^2-3x-10)/(x^2+2x-35)*(x^2+4x-21)/(x^2+9x+14)=(x-3)/(x+7)#,
we just need to factorize each of the quadratic polynomial.
#x^2-3x-10=x^2-5x+2x-10=x(x-5)+2(x-5)=(x+2)(x-5)#
#x^2+2x-35=x^2+7x-5x-35=x(x+7)-5(x+7)=(x-5)(x+7)#
#x^2+4x-21=x^2+7x-3x-21=x(x+7)-3(x+7)=(x-3)(x+7)#
#x^2+9x+14=x^2+7x+2x+14=x(x+7)+2(x+7)=(x+2)(x+7)#
Hence #(x^2-3x-10)/(x^2+2x-35)*(x^2+4x-21)/(x^2+9x+14)#
= #((x+2)(x-5))/((x-5)(x+7))xx((x-3)(x+7))/((x+2)(x+7))#
= #(cancel((x+2))cancel((x-5)))/(cancel((x-5))cancel((x+7)))xx((x-3)cancel((x+7)))/(cancel((x+2))(x+7))#
= #(x-3)/(x+7)#
