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I am studying for finals and in the review packet is shown this problem: $$P(x)=2x^4 + 5x^3 + 5x^2 + 20x - 12$$ I don't know what to do, I have already tried looking in the textbook and Khan academy.

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  • $\begingroup$ I am looking for the roots by the way. $\endgroup$ – Patrick Vance-colby Dec 30 '15 at 19:01
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    $\begingroup$ The Rational Root Theorem solves this. $-3$ and $1/2$ are roots of $P(x)$. Then divide $P(x)$ by the polynomial $(x+3)(x-1/2)$ and solve the quadratic equation. $\endgroup$ – user236182 Dec 30 '15 at 19:05
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The rational root test theorem says that, if rational factors of a polynomial exist, then they are always in the form of

$\pm$(factor of last coefficient) / (factor of first coefficient)

In this case, the factors you can try are: $\pm 12, \pm 6, \pm 4, \pm 3, \pm 2, \pm 1, \pm 1.5, \pm 0.5$

Plug these in to see which one gives you 0. Once you find one that does, then use synthetic division to divide your polynomial by ($x-$that factor). In this case, the factor that comes out is $x=-3$, so you would divide by (x+3).

Repeat until you've reduced to a quadratic, and then use the quadratic formula to find the irrational coefficients (if any).

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Simple factorisation solves your problem.

$$P(x)=2x^4+5x^3+5x^2+20x-12$$ $$=2x^4+5x^3-3x^2+8x^2+20x-12$$ $$=x^2(2x^2+5x-3)+4(2x^2+5x-3)$$ $$=(x^2+4)(2x^2+5x-3)$$ $$=(x^2+4)(2x-1)(x+3)$$

Hence the $4$ roots are $x=\frac{1}{2}$,$x=-3$,$x=2i$ and $x=-2i$.

Hope this helps.

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Have you heard of the rational root test?

It implies that rational roots to this are of the form $p/q$ where $p$ divides $12$ and $q$ divides $2$. There are only a few numbers like this so you can check them.

Once you have a few roots, you can divide and get a smaller polynomial that is easier to manage.

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