MHB Kkittiee's question at Yahoo Answers involving factoring a cubic polynomial

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The cubic polynomial x^3 + 11x^2 + 51x + 41 can be factored using the rational roots theorem, which indicates potential rational roots of ±1 and ±41. Testing these, it is found that f(-1) equals zero, confirming that x + 1 is a factor. Synthetic division reveals that the polynomial can be expressed as (x + 1)(x^2 + 10x + 41). The quadratic factor x^2 + 10x + 41 is prime, concluding the factorization process.
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Here is the question:

Math help: factoring?

what is x^3 + 11x^2 + 51x + 41 factored? thank you so much!

Here is a link to the question:

Math help: factoring? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello kkittiee,

We are given to factor:

$\displaystyle f(x)=x^3+11x^2+51x+41$

The rational roots theorem tells us that if this polynomial has any rational roots, they will come from the list:

$\displaystyle \pm(1,41)$

and in fact, we find:

$\displaystyle f(-1)=(-1)^3+11(-1)^2+51(-1)+41=0$

So, we know $\displaystyle x+1$ is a factor. Performing synthetic division, we find:

View attachment 622

And so we know:

$\displaystyle f(x)=x^3+11x^2+51x+41=(x+1)(x^2+10x+41)$

The quadratic factor is prime, so we are done.
 

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