How can I factorise $c^4+4c^3+5c^2+8c+6$?

Question

How can I factorise $c^4+4c^3+5c^2+8c+6$?

Answer 1

When all polynomial coefficients are integers, any rational root (if it exists at all) $\frac p q$ of $a_nx^n + a_{n-1}x^{n-1} +\ldots + a_1x + a_0$ must satisfy: $$ p | a_0,\ q | a_n$$

In this particular case, we have $a_n = 1$, so $q = 1$, i.e., any rational root of $x^4 + 4x^3 +5x^2 +8x + 6$ must be integer that divides $6$, i.e. $\frac p q\in\{-6,-3,-2,-1,1,2,3,6\}$. Since all coefficients of your polynomial are positive, $\frac p q$ can't be positive.

One can then quickly check that $-1$ and $-3$ are indeed roots. That means that $$x^4 + 4x^3 +5x^2 +8x + 6 = (x+1)(x+3)r(x)$$ and if you use long division, you can find that $r(x) = x^2 +2$, which cannot be factored further over $\Bbb R$ (but can over $\Bbb C$, $x^2 + 2 = (x - i\sqrt 2)(x + i\sqrt 2)$).

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Alternatively, you can note that $$x^4 + 4x^3 + 5x^2 +8x + 6 = (x^4 + 4x^3 + 3x^2) + (2x^2 +8x + 6) =\\ =x^2(x^2 + 4x + 3) + 2(x^2 + 4x + 3) = (x^2 + 2)(x^2 + 4x + 3)$$

and find the roots by qudratic formula. Though, this is often harder to do in practice then the above method.

Answer 2

If $c \geq 0$, then the polynomial is positive.

Hence the root should be complex or negative.

Try the factors of $6$, you can find that $-1$ and $-3$ are indeed the roots.

Hence some of the factors are $(x+1)$ and $(x+3)$, divide these terms and obtain the remaining quadratic terms.

Answer 3

$$c^4+4c^3+5c^2+8c+6$$ $$c^4+4c^3+6c^2+4c+1-6c^2-4c-1+5c^2+8c+6$$ $$(c+1)^4-c^2+4c+5$$ $$(c+1)^4-(c+1)(c-5)$$ $$(c+1)((c+1)^3-c+5))$$ $$(c+1)(c^3+3c^2+2c+6)$$ $$(c+1)(c^2(c+3)+2(c+3))$$ $$(c+1)(c^2+2)(c+3)$$