f(x) = x3 + 4x2 + x - 6, Which is the factored form of f(x)?

1.) See "Vieta"

2.) You find $$x_1=1$$ do:

$$\begin{array}{ rlllr } ( x^3 & +4x^2 & +x & -6) & : (x-1) =\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6} \\ \textcolor[rgb]{1,0,0}{{\underline{-( x^3}} & \textcolor[rgb]{1,0,0}{\underline{-x^2)}}} & & & \\ 0 & +5x^2 & +x \\ & \textcolor[rgb]{0,0,1}{\underline{-(5x^2}} & \textcolor[rgb]{0,0,1}{\underline{-5x)}} \\ & 0 & +6x & -6 \\ && \textcolor[rgb]{0,1,0}{\underline{-(6x}} & \textcolor[rgb]{0,1,0}{\underline{-6)}} \\ && 0 & +0 \end{array}$$

\begin{array}{
rlllr
} (
x^3
&
+4x^2
&
+x
&
-6)
&
:
(x-1)
=\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6}
\\
\textcolor[rgb]{1,0,0}{{\underline{-(
x^3}}
&
\textcolor[rgb]{1,0,0}{\underline{-x^2)}}}
&
&
&
\\
0
&
+5x^2
&
+x
\\
&
\textcolor[rgb]{0,0,1}{\underline{-(5x^2}}
&
\textcolor[rgb]{0,0,1}{\underline{-5x)}}
\\
&
0
&
+6x
&
-6
\\
&&
\textcolor[rgb]{0,1,0}{\underline{-(6x}}
&
\textcolor[rgb]{0,1,0}{\underline{-6)}}
\\
&&
0
&
+0
\end{array}

so we have: $$x^3+4x^2+x-6=(x-1)(x^2+5x+6)$$

x^3+4x^2+x-6=(x-1)(x^2+5x+6)

We solve:

$$\\x^2+5x+6=0\\ x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\ \left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\ \left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\ x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\ x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\ x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\ x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\$$

\\x^2+5x+6=0\\
x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\
\left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\
\left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\
x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\
x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\
x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\
x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\

$$\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\ \boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}$$

\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\
\boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}

This is all we can do......no further factoring available....!!!

(x + 3)*(x - 1)*(x + 2) = x³ +4x² + x - 6

To find this, start by assuming (x-a)*(x-b)*(x-c) =  x³ +4x² + x - 6

The left-hand side can be expanded, to get: x³ - (a+b+c)*x² +(a*b+a*c+b*c)*x -a*b*c =  x³ +4x² + x - 6

By equating the coefficients of powers of x on both sides of this equation you get three simultaneous equations in a, b and c, which when solved, produce the result shown above.

By equating the coefficients I mean:

coeff of x²: -(a+b+c) = 4

coeff of x¹: a*b + a*c +b*c = 1

coeff of x⁰: -a*b*c = -6

Thanks, Alan...I haven't seen that one before....another tool for my chest !!

For some reason...this technique  - to my knowledge -isn't covered in Algebra in the States.....it should be!!

Thums Up from me !!

That's what I like about this site......something new to learn for evetyone!!!

Well, I should really add a warning here.  The method is not always very useful.  The danger is that, in general, by the time you do the various substitutions to isolate a, b or c, all you get is the original equation with x replaced by a, say!!

In this case it was fairly easy to guess that 1, 2 and 3 were involved, because their product had to be 6 (the other equations help determine the signs), and I started by assuming they would be integers (I reasoned the question probably wouldn't have been posed if they weren't).  Luckily, this reasoning panned out; however, in general it won't and one could guess forever without getting anywhere!

Ok My turn

the roots have to be factors of 6 so I'd look at 1 first

f(1)=1+4+1-6=0 therefore (x-1) is a factor    [I am using factor theorem but it is just logical really]

I could now do algebraic division or synthetic division to find its cofactor but I'll just try easy possibilities first.

f(-2)=-8+16+-2-6=0 therefore (x+2) is another factor   

(x-1)(x+2)=x²+x-2

$$(x^3+4x^2+x-6) \div (x^2+x-2) =x+3$$

I did the division but I don't know how to set it out in latex.  SO

$$(x^3+4x^2+x-6)=(x-1)(x+2)(x+3)$$

And that will work every time so long as the roots are integers. 

1.) See "Vieta"

2.) You find $$x_1=1$$ do:

$$\begin{array}{ rlllr } ( x^3 & +4x^2 & +x & -6) & : (x-1) =\textcolor[rgb]{1,0,0}{x^2}\textcolor[rgb]{0,0,1}{+5x}\textcolor[rgb]{0,1,0}{+6} \\ \textcolor[rgb]{1,0,0}{{\underline{-( x^3}} & \textcolor[rgb]{1,0,0}{\underline{-x^2)}}} & & & \\ 0 & +5x^2 & +x \\ & \textcolor[rgb]{0,0,1}{\underline{-(5x^2}} & \textcolor[rgb]{0,0,1}{\underline{-5x)}} \\ & 0 & +6x & -6 \\ && \textcolor[rgb]{0,1,0}{\underline{-(6x}} & \textcolor[rgb]{0,1,0}{\underline{-6)}} \\ && 0 & +0 \end{array}$$

so we have: $$x^3+4x^2+x-6=(x-1)(x^2+5x+6)$$

We solve:

$$\\x^2+5x+6=0\\ x^2+5x+(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2-(\textcolor[rgb]{1,0,0}{\frac{5}{2}})^2+6=0\\ \left( x+\frac{5}{2}\right)^2=\frac{25}{4}-6\\ \left( x+\frac{5}{2}\right)^2=\frac{1}{4}\qquad| \quad \pm\sqrt{}\\ x+\frac{5}{2}=\pm\sqrt{\frac{1}{4}}\\ x_{2,3}=-\frac{5}{2}\pm\frac{1}{2}\\ x_2=-\frac{5}{2}+\frac{1}{2}=-\frac{4}{2}=\underline{-2}\\ x_3=-\frac{5}{2}-\frac{1}{2}=-\frac{6}{2}=\underline{-3}\\$$

$$\\x_1=1 \qquad x_2=-2 \qquad x_3=-3\\ \boxed{x^3+4x^2+x-6=(x-1)(x+2)(x+3)}$$

Hi Heureka,

Could you please post that latex here without it being in latex.  I just want to look at all the code, especially for the division bit.

I would have done it with the align function but I don't think that works in here.

I see you have used an array.  I have only used equarrays and again I don't think that they work in here

Thank you.

Thank you Heureka