document.write( "Question 133080: Ok I need some help square 2x^2+5x+6=x If I subtract x I get 2x^2+4x+6=0 If I do that then it can't be factored right? \n" ); document.write( "

Algebra.Com's Answer #97300 by solver91311(24713) $\"\"$ $\"About$

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Well, that is not precisely correct. ALL quadratic trinomials can be factored, just not necessarily to integers, rationals, or even real numbers. In other words, for any second degree polynomial $\"ax%5E2%2Bbx%2Bc\"$ there exist two numbers $\"p\"$ and $\"q\"$ such that $\"%28x-p%29%28x-q%29=ax%5E2%2Bbx%2Bc\"$ . It is just that in the case of your equation, $\"2x%5E2%2B4x%2B6=0\"$ , $\"p\"$ and $\"q\"$ are a conjugate pair of complex numbers, namely $\"-1%2B-i%2Asqrt%282%29\"$ , where $\"i%5E2=-1\"$ .\r
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\n" ); document.write( "\n" ); document.write( "That means your factors would be $\"%28x-%28-1%2Bi%2Asqrt%282%29%29%29%28x-%28-1-i%2Asqrt%282%29%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "Having said all of that, I suspect you meant that the polynomial cannot be factored over the integers. That is an absolutely correct statement. \n" ); document.write( "

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