\({4(x^{2} - 9)^{2}-(x+3)^{2} \over x^{2} + 6x +9}\)
the textbook answer is \({(2x - 7)(2x - 5), x≠ - 3}\). how should i get this?
Simplifying this rational expression
{4(x^{2} - 9)^{2}-(x+3)^{2} \over x^{2} + 6x +9}
\(\dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {x^{2} + 6x +9}\)
\(\begin{array}{|rcll|} \hline && \dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {x^{2} + 6x +9} \quad | \quad x^{2} + 6x +9 = (x+3)(x+3)=(x+3)^2 \\\\ &=& \dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {(x+3)^2} \quad | \quad x+3 \ne 0 \Rightarrow x\ne -3 \\\\ &=& \dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {(x+3)^2} \\\\ && 4(x^{2} - 9)^{2}-(x+3)^{2} = \Big(2(x^2-9)-(x+3)\Big)\Big(2(x^2-9)+(x+3)\Big) \\\\ &=& \dfrac{\Big(2(x^2-9)-(x+3)\Big)\Big(2(x^2-9)+(x+3)\Big)} {(x+3)^2} \\\\ &=& \dfrac{ (2x^2-x-21)(2x^2+x-15)} {(x+3)^2} \\\\ &=& \dfrac{ 2(x-\frac72)(x+3)2(x-\frac52)(x+3)} {(x+3)^2} \\\\ &=& 2(x-\frac72)2(x-\frac52) \\\\ &=& (2x-7)(2x-5) \\ \hline \end{array}\)
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