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+cax2+bx+c, where a, b, ca,b,c are constants, try to find two numbers, say dd & ee (by trial and error) such that:-
i.i. b = d+eb=d+e &
ii.ii. d xx e = a xx cd×e=a×c
then ax^2+bx+cax2+bx+c can be written as;
ax^2+(d+e)x+c = ax^2+dx+ex+cax2+(d+e)x+c=ax2+dx+ex+c
= ax(x+d/a)+e(x+c/e)=ax(x+da)+e(x+ce)
[but from condition ii.ii., d xx e = a xx c implies d/a=c/ed×e=a×c⇒da=ce]
=ax(x+d/a)+e(x+d/a)=ax(x+da)+e(x+da) OR ax(x+c/e)+e(x+c/e)ax(x+ce)+e(x+ce)
=color(red)((x+d/a)(ax+e))=(x+da)(ax+e) OR color(red)((x+c/e)(ax+e))(x+ce)(ax+e)
1. x^2+5x-14x2+5x−14
[Here, a=1, b=5, c=-14a=1,b=5,c=−14; also observe that 7+(-2)=7-2=57+(−2)=7−2=5 & 7 xx (-2) = -14=1 xx (-14)7×(−2)=−14=1×(−14)]
So, x^2+5x-14 = x^2+(7-2)x-14x2+5x−14=x2+(7−2)x−14
=x^2 +7x -2x -14 =x2+7x−2x−14
= x(x+7)-2(x+7)=x(x+7)−2(x+7)
=color(red)((x-2)(x+7))=(x−2)(x+7)
2. x^2-9x+20x2−9x+20
[Here, a=1, b=-9, c=20a=1,b=−9,c=20; and -5+(-4)=-5-4=-9−5+(−4)=−5−4=−9 & (-5) xx (-4) = 20=1 xx 20(−5)×(−4)=20=1×20]
So, x^2-9x+20 = x^2+(-5-4)x+20x2−9x+20=x2+(−5−4)x+20
=x^2-5x-4x+20=x2−5x−4x+20
=x(x-5)-4(x-20)=x(x−5)−4(x−20)
=color(red)((x-4)(x-5))=(x−4)(x−5)
Following the similar procedure we can factorize rest two.
3. x^2-3x-10x2−3x−10
=x^2-5x+2x-10=x2−5x+2x−10
=x(x-5)+2(x-5)=x(x−5)+2(x−5)
=color(red)((x+2)(x-5))=(x+2)(x−5)
4. x^2+9x+14x2+9x+14
=x^2+7x+2x+14=x2+7x+2x+14
=x(x+7)+2(x+7)=x(x+7)+2(x+7)
=color(red)((x+2)(x+7))=(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)x2+5x−14x2−9x+20⋅x2−3x−10x2+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))
