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Sagot :
Hello!
We have the expression:
[tex]\frac{4x^2+14x+6}{x+3}[/tex]Note that all numbers in the numerator are even. So, we can put 2 in evidence, look:
[tex]\frac{2(2x^2+7x+3)}{x+3}[/tex]Now, let's rewrite 7x as 6x+x:
[tex]\frac{2(2x^2+6x+x+3)}{x+3}[/tex]The first and second terms are multiples of 2x, so let's rewrite it putting it in evidence too:
[tex]\frac{2(2x(x+3)+x+3)}{x+3}[/tex]Another term appears twice: (x+3). So, we'll have:
[tex]\frac{2(x+3)(2x+1)}{x+3}[/tex]Canceling the common factors:
[tex]\frac{2\cancel{x+3}(2x+1)}{\cancel{x+3}}=2(2x+1)=\boxed{4x+2}[/tex]Answer:
4x +2.
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