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))