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\({4(x^{2} - 9)^{2}-(x+3)^{2} \over x^{2} + 6x +9}\)

 

the textbook answer is \({(2x - 7)(2x - 5), x≠ - 3}\). how should i get this?

Guest Jul 23, 2018
 #1
avatar+20025 
+1

Simplifying this rational expression
{4(x^{2} - 9)^{2}-(x+3)^{2} \over x^{2} + 6x +9}

\(\dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {x^{2} + 6x +9}\)

 

\(\begin{array}{|rcll|} \hline && \dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {x^{2} + 6x +9} \quad | \quad x^{2} + 6x +9 = (x+3)(x+3)=(x+3)^2 \\\\ &=& \dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {(x+3)^2} \quad | \quad x+3 \ne 0 \Rightarrow x\ne -3 \\\\ &=& \dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {(x+3)^2} \\\\ && 4(x^{2} - 9)^{2}-(x+3)^{2} = \Big(2(x^2-9)-(x+3)\Big)\Big(2(x^2-9)+(x+3)\Big) \\\\ &=& \dfrac{\Big(2(x^2-9)-(x+3)\Big)\Big(2(x^2-9)+(x+3)\Big)} {(x+3)^2} \\\\ &=& \dfrac{ (2x^2-x-21)(2x^2+x-15)} {(x+3)^2} \\\\ &=& \dfrac{ 2(x-\frac72)(x+3)2(x-\frac52)(x+3)} {(x+3)^2} \\\\ &=& 2(x-\frac72)2(x-\frac52) \\\\ &=& (2x-7)(2x-5) \\ \hline \end{array}\)

 

laugh

heureka  Jul 23, 2018
 #2
avatar+89953 
+1

Here's one more way :

 

4(x^2 - 9)^2  - ( x + 3)^2        factors as

4 [ (x + 3)(x - 3)] ^2  - (x + 3)^2 =

[(x + 3}^2  [ 4 ( x + 3)^2 ] - 1 ]

 

And x^2 + 6x + 9   factors as  ( x + 3)(x + 3)  = (x + 3)^2

 

So we have

 

[(x + 3}^2  [ 4 ( x - 3)^2 ] - 1 ]

_______________________         we can "cancel the ( x + 3)^2  terms  and we are left with

   (x + 3)^2

 

4( x - 3)^2 - 1  =

4(x^2 - 6x + 9)  - 1 =

4x^2 - 24x + 36 - 1 =

4x^2 - 24x + 35       

We need two factors that multiply to 4  and 35  with the two factors multiplying to 35 both being negative

We might try  4,1  and  -5 , -7

(4x - 5) ( x - 7) =  4x^2 -5x - 28x + 35  = 4x^2 - 33x + 35....no good

Next...we might guess 2,2  and  -5, -7

(2x - 5) (2x - 7)   = 4x^2 - 10x - 14x + 35  =  4x^2 - 24x + 35   which is what we  need

Also....x  cannot equal -3  because this would make  the origianl denominator   = 0 

 

cool cool cool

CPhill  Jul 23, 2018

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