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I'm self-studying from Stroud & Booth's amazing "Engineering Mathematics" and I am stuck on factorizing a particular polynomial to a product of three linear factors. Through the trial-and-error method, I've gotten that the factors are $(x+2)$ and $(x+4)$ (although, it is my understanding that only quadratic polynomials are candidate for looking for multiple factors in this way, in case the factor isn't already given).

I've been banging my head against this, and gotten a different result every time, and the best I could come up with is: $$(x+1)(x+4)(2x+5).$$

It just seems that every time I try it out, a different result comes out.

Can anybody help me out or give me some pointers? Generally I have this down pretty well, but this particular one is doing my head in.

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    $\begingroup$ $x+1$ cannot be a factor of the polynomial as $-1$ is not a root of it. $\endgroup$
    – little o
    Apr 6 '19 at 18:18
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    $\begingroup$ The rational root theorem gives you a fast and easy way to identify all the possible rational roots. $\endgroup$
    – lulu
    Apr 6 '19 at 18:19
  • $\begingroup$ The tag self-learning is intended to be used for questions about the pedagogy involved in learning on your own such as "If I am teaching myself a subject, how can I get feedback like one might in a traditional course through tests and homework? How do I know I'm headed in the right direction and not making mistakes?" The tag is not meant to be used for questions like yours which is more about the math involved rather than the learning-style. $\endgroup$
    – JMoravitz
    Apr 6 '19 at 19:14
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It's actually $$2x^3+7x^2-14x-40=(2x\color{red}-5)(x+\color{red}2)(x+4)$$

Once you've identified that $(x+2)$ and $(x+4)$ are factors, use the polynomial long division to obtain the last factor.

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  • $\begingroup$ Why the downvote? $\endgroup$
    – Dr. Mathva
    Apr 6 '19 at 19:53
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A general way to find such a factorization for a polynomial $p(x)$ of third degree is:

1) Try to find "by hand" a root of the polynomial: let's call it $a$.

2) Since $a$ is a root, (x-a) divides your polynomial. Hence you can write it as $p(x)=(x-a)(ax^2+bx+c)$. Now all you have to do is finding the roots of the second degree polynomial we just found.

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Observe that $$2x^3+7x^2-14x-40 = (x+4) (2x^2-x-10).$$

Can you proceed now?

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  • $\begingroup$ Why the downvote? $\endgroup$
    – little o
    Apr 6 '19 at 19:58
  • $\begingroup$ I'm not the downvoter but can you elaborate how did you find -4 as root? $\endgroup$
    – kelalaka
    Apr 7 '19 at 10:30
  • $\begingroup$ Because $2(-4)^3 + 7(-4)^2-14(-4)-40=0.$ $\endgroup$
    – little o
    Apr 7 '19 at 10:32
  • $\begingroup$ Yes, that is clear. Did you use the Rational Root theorem or try an error? $\endgroup$
    – kelalaka
    Apr 7 '19 at 10:34
  • $\begingroup$ I basically did by trial and error. $\endgroup$
    – little o
    Apr 7 '19 at 10:35

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