This answer may contain some information which you already know but I have added it for the sake of completeness. This has also caused the solution to APPEAR very long although the solution to the problem itself is quite easy and concise . So do give it a read.
Suggestions are welcome.
Let us factorize all the four polynomials to see what we get.
To factorize a polynomial of type #ax^2+bx+c#, where #a, b, c# are constants, try to find two numbers, say #d# & #e# (by trial and error) such that:-
#i.# #b = d+e# &
#ii.# #d xx e = a xx c#
then #ax^2+bx+c# can be written as;
#ax^2+(d+e)x+c = ax^2+dx+ex+c#
# = ax(x+d/a)+e(x+c/e)#
[but from condition #ii.#, #d xx e = a xx c implies d/a=c/e#]
#=ax(x+d/a)+e(x+d/a)# OR #ax(x+c/e)+e(x+c/e)#
#=color(red)((x+d/a)(ax+e))# OR #color(red)((x+c/e)(ax+e))#
1. #x^2+5x-14#
[Here, #a=1, b=5, c=-14#; also observe that #7+(-2)=7-2=5# & #7 xx (-2) = -14=1 xx (-14)#]
So, #x^2+5x-14 = x^2+(7-2)x-14#
#=x^2 +7x -2x -14 #
#= x(x+7)-2(x+7)#
#=color(red)((x-2)(x+7))#
2. #x^2-9x+20#
[Here, #a=1, b=-9, c=20#; and #-5+(-4)=-5-4=-9# & #(-5) xx (-4) = 20=1 xx 20#]
So, #x^2-9x+20 = x^2+(-5-4)x+20#
#=x^2-5x-4x+20#
#=x(x-5)-4(x-20)#
#=color(red)((x-4)(x-5))#
Following the similar procedure we can factorize rest two.
3. #x^2-3x-10#
#=x^2-5x+2x-10#
#=x(x-5)+2(x-5)#
#=color(red)((x+2)(x-5))#
4. #x^2+9x+14#
#=x^2+7x+2x+14#
#=x(x+7)+2(x+7)#
#=color(red)((x+2)(x+7))#
Substituting all the values in the given equation;
#(x^2+5x-14)/(x^2-9x+20)*(x^2-3x-10)/(x^2+9x+14)#
#=((x-2)cancel((x+7)))/((x-4)cancel((x-5)))*(cancel((x+2))cancel((x-5)))/(cancel((x+2))cancel((x+7)))#
#=color(red)((x-2)/(x+4))#
