document.write( "Question 137613: Can 4u^2-9u+4 be factored? If so could you please show me how? \n" ); document.write( "

Algebra.Com's Answer #100661 by solver91311(24713) $\"\"$ $\"About$

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Well, the answer to your first question is yes and no. In fact, all quadratic polynomials can be factored, but the question becomes what is the nature of those factors. In fact, given $\"alpha%5B0%5Dx%5En%2Balpha%5B1%5Dx%5E%28n-1%29\"$ +...+ $\"alpha%5Bn-1%5Dx%2Balpha%5Bn%5D\"$ , $\"x-a\"$ is a factor if and only if $\"a\"$ is a root of $\"alpha%5B0%5Dx%5En%2Balpha%5B1%5Dx%5E%28n-1%29\"$ +...+ $\"alpha%5Bn-1%5Dx%2Balpha%5Bn%5D=0\"$ .\r
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\n" ); document.write( "\n" ); document.write( "So all you need to do is find the two roots of $\"4u%5E2-9u%2B4=0\"$ , let's say they are $\"a%5B1%5D\"$ and $\"a%5B2%5D\"$ , then your factors of $\"4u%5E2-9u%2B4\"$ are $\"u-a%5B1%5D\"$ and $\"u-a%5B2%5D\"$ .\r
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\n" ); document.write( "\n" ); document.write( "On the other hand, I think you meant to ask if $\"4u%5E2-9u%2B4\"$ is factorable over the integers or perhaps the rationals. If that is what you meant, then the answer is most certainly no. That is because the roots of $\"4u%5E2-9u%2B4=0\"$ are irrational, specifically $\"a%5B1%5D=%289%2Bsqrt%2817%29%29%2F2\"$ and $\"a%5B2%5D=%289-sqrt%2817%29%29%2F2\"$ a conjugate pair of irrational numbers. \n" ); document.write( "

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