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)}
(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.
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)}