# Factoring a Quintic Polynomial [closed]

I need help in factoring this polynomial:

$$2x^5-10x^4+7x^3+13x^2+3x+9$$

## closed as off-topic by Simply Beautiful Art, user21820, Namaste, Dave, José Carlos SantosAug 12 '17 at 17:16

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• is it $2x^5-10x^4+7x^3+13x^2+3x+9$? – Dr. Sonnhard Graubner Aug 12 '17 at 16:56
• Yes it is.;)))) – Clarence Feliciano Aug 12 '17 at 16:57
• Use geogebra!!! – Aqua Aug 12 '17 at 16:58
• what is geogebra? – Dr. Sonnhard Graubner Aug 12 '17 at 16:59
• Google it...... – Aqua Aug 12 '17 at 16:59

the simple way is first to try hit and trial method and find one root after that just try multiplication between one power less polynomial and $(x-root)$

so $-1$ was my root.now multiply $(x+1)$ with $x^{5-1}$
now what you will you get: $x^5+x^4$

but you want $-10x^4$ in your polynomial

so now add $-11x^4$ in your polynomial

after repeating this for several times you will get

$(x + 1) (x - 3)^2 (2 x^2 + 1)$

Using the rational root theorem, you can see that it has two rational roots: $-1$ and $3$. Dividing your polynomial by $(x+1)(x-3)$, you'll get $2x^3-6x^2+x-3$, of which $3$ is a root (again). Dividing it by $x-3$, you'll get $2x^2+1$. Therefore, in $\mathbb{Q}[x]$, you can factor your polynomial as$$2(x+1)(x-3)^2\left(x^2+\frac12\right).$$