How do you multiply and simplify \frac { x ^ { 2} + 5x - 14} { x ^ { 2} - 9x + 20} \cdot \frac { x ^ { 2} - 3x - 10} { x ^ { 2} + 9x + 14}?

1 Answer
Apr 7, 2017

color(red)((x-2)/(x+4))

Explanation:

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))