factoring polynomial 2nd and 3rd degree

Question

how to factor a polynomial 2nd and 3rd degree :

$$a = 6x^3+36x^2+66x+36$$

$$b = 4x^2+12x+8$$

I have try with $b$ and I got $x_1=-1$ & $x_2=-2$

but $(x+1)\cdot(x+2)$ different to $4x^2+12x+8$

Answer 1

We can write $$b=4(x^2+3x+2)$$

Now, we need to factorise $x^2+3x+2$, which gives us $(x+2)(x+1)$

So, we have $$b=4(x+1)(x+2)$$

Similarly for $a$:

$$a=6(x^3+6x^2+11x+6)$$

Now we need to factorise $x^3+6x^2+11x+6$, which gives us $(x+1)(x+2)(x+3)$

So, we have $$a=6(x+1)(x+2)(x+3)$$

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Steps to factorise degree $3$ polynomial

\begin{align}x^3+6x^2+11x+6&=(x+\alpha)(x+\beta)(x+\gamma)\\ &=(x^2+\alpha x+\beta x+\alpha\beta)(x+\gamma)\\ &=x^3+\alpha x^2+\beta x^2+\alpha\beta x+\gamma x^2+\alpha\gamma x+\beta\gamma x+\alpha\beta\gamma\\ &=x^3+(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\alpha\gamma)x+\alpha\beta\gamma\end{align}

Now we have \begin{align}\alpha+\beta+\gamma&=6\\ \alpha\beta+\alpha\gamma+\beta\gamma&=11\\ \alpha\beta\gamma&=6\end{align}

The only solution to this is $(\alpha,\beta,\gamma)=(1,2,3)$

We could also guess at factors (by putting in small integers and finding the ones which give a result of $0$), and find that $f(-1)=0$, which would then give us

\begin{align}x^3+6x^2+11x+6&=(x+1)(ax^2+bx+c)\\ &=ax^3+bx^2+cx+ax^2+bx+c\\ &=ax^3+(a+b)x^2+(b+c)x+c\end{align}

Equating coefficients then gives us:

\begin{align}a&=1\\ a+b&=6\implies b=5\\ b+c&=11\\ c&=6\end{align}

So now we have $$x^3+6x^2+11x+6=(x+1)(x^2+5x+6)$$ which you should now be able to factorise

Answer 2

$$6x^3+36x^2+66x+36=6(x^3+6x^2+11x+6)=$$ $$=6(x^3+3x^2+3x^2+9x+2x+6)=6(x+3)(x^2+3x+2)=$$ $$=6(x+3)(x^2+2x+x+2)=6(x+3)(x+2)(x+1)=6(x+1)(x+2)(x+3).$$

If $a\neq0$ and $\Delta\geq0$ then $$ax^2+bx+c=a(x-x_1)(x-x_2)$$