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Factorize by remarkable identities:49(x+1)squared minus 25

 Feb 3, 2017
 #1
avatar+33653 
0

"Factorize by remarkable identities:49(x+1)squared minus 25"

 

\(49(x+1)^2-25\rightarrow(7x+7)^2-5^2\rightarrow(7x+7+5)(7x+7-5)\rightarrow(7x+12)(7x+2)\)

 Feb 3, 2017
 #2
avatar+37084 
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49 (x+1)^2 = 25

(x+1)^2 = 25/49

x+1 = + - 5/7

x = 5/7 -1    or  -5/7 -1

x = -2/7 , -12/7

 Feb 3, 2017
 #3
avatar+14985 
0

49(x+1)squared minus 25

 

\(\large 49(x+1)^2-25=7^2(x+1+5)(x+1-5)\) 

 

\(=7^2(x^2+x-5x+x+1-5+5x+5-25)\)

 

\(=7^2(x^2+2x-24) \)  

             \(x=1\pm\sqrt{1+24} \ \Rightarrow x_1=6 \ \wedge \ x_2=-4\)

 

\(\large =7\times 7 \times (x-6)\times (x+4) \)

 

laugh !

 Feb 3, 2017
edited by asinus  Feb 3, 2017
 #4
avatar+14985 
0

Not correct! Here comes the correction.

 

49(x+1)squared minus 25  Factorize!

 

\(\large 49(x+1)^2-25\) 

 

\((7(x+1))^2-25=(7(x+1)+5)\times (7(x+1)-5)\)

 

\(=(7x+12)\times (7x+2)\)

 

smiley !

asinus  Feb 3, 2017

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