How do i factorize x^3 -5x^2+8x-4?

[theakdad]

$$x^3 -5x^2+8x-4$$

What is the simplest way to factorize given equation? Thank you.

 

[chisigma]

$$x^3 -5x^2+8x-4$$

What is the simplest way to factorize given equation? Thank you.

By sight You notice that x = 1 and a solution of the equation...

$\displaystyle P(x) = x^{3} - 5\ x^{2} + 8\ x - 4 = 0\ (1)$

... so that (x-1) divides P(x)... then divide P(x) by (x-1) and obtain a second order polynomial that possibly can be further factorized...

Kind regards

$\chi$ $\sigma$

 

[theakdad]

By sight You notice that x = 1 and a solution of the equation...

$\displaystyle P(x) = x^{3} - 5\ x^{2} + 8\ x - 4 = 0\ (1)$

... so that (x-1) divides P(x)... then divide P(x) by (x-1) and obtain a second order polynomial that possibly can be further factorized...

Kind regards

$\chi$ $\sigma$

I do not understand the part "By sight"...

 

[Nono713]

You could always try plotting it, it's not very rigorous mathematically but since polynomials are pretty well-behaved it's easy to spot roots most of the time.

 

[theakdad]

You could always try plotting it, it's not very rigorous mathematically but since polynomials are pretty well-behaved it's easy to spot roots most of the time.

+I have to find the roots of the polynomial...so have to factorize it...

 

[RLBrown]

I have to find the roots of the polynomial...so have to factorize it...

Click Here

to see why plotting the polynomial is useful in finding the roots.
See there is one root at x=1
See there is a double root at x=2

Then the three roots can be used to factor P(x).
P(x) = (x - r1)(x - r2)(x - r3)

can you see why P(x) is zero at exactly the same x-values as (x - r1)(x - r2)(x - r3)?

 

[MarkFL]

$$x^3 -5x^2+8x-4$$

What is the simplest way to factorize given equation? Thank you.

What chisigma was getting at, is that if you observe that given:

$$P(x)=x^3-5x^2+8x-4$$

and you see that:

$$P(1)=1-5+8-4=0$$

then you know that:

$$P(x)=(x-1)f(x)\implies f(x)=\frac{P(x)}{x-1}$$

where $f$ is of degree 2. Thus we can use division to determine $f$:

$$\begin{array}{c|rr}& 1 & -5 & +8 & -4 \\ 1 & & +1 & -4 & 4 \\ \hline & 1 & -4 & 4 & 0 \end{array}$$

Thus, we have found:

$$P(x)=(x-1)\left(x^2-4x+4\right)$$

And so now you only have a quadratic left to factor...can you continue?

 

[theakdad]

What chisigma was getting at, is that if you observe that given:

$$P(x)=x^3-5x^2+8x-4$$

and you see that:

$$P(1)=1-5+8-4=0$$

then you know that:

$$P(x)=(x-1)f(x)\implies f(x)=\frac{P(x)}{x-1}$$

where $f$ is of degree 2. Thus we can use division to determine $f$:

$$\begin{array}{c|rr}& 1 & -5 & +8 & -4 \\ 1 & & +1 & -4 & 4 \\ \hline & 1 & -4 & 4 & 0 \end{array}$$

Thus, we have found:

$$P(x)=(x-1)\left(x^2-4x+4\right)$$

And so now you only have a quadratic left to factor...can you continue?

Yes,its obvious...

So i get $$(x-1) (x-2) (x-2)$$

i have three roots: X₁ = 1 and X_2,3 = 2

 

[MarkFL]

Yes,its obvious...

So i get $$(x-1) (x-2) (x-2)$$

i have three roots: X₁ = 1 and X_2,3 = 2

Yes, that's correct...you have the root $x=1$ and the root $x=2$, which is a repeated root, or of multiplicity 2.

 

[theakdad]

Yes, that's correct...you have the root $x=1$ and the root $x=2$, which is a repeated root, or of multiplicity 2.

I was just wondering how do you guess the first root...
In my example there is a simple polynomial...but what to do when you have harder cases?

 

[I like Serena]

$$x^3 -5x^2+8x-4$$

What is the simplest way to factorize given equation? Thank you.

I was just wondering how do you guess the first root...
In my example there is a simple polynomial...but what to do when you have harder cases?

Hi wishmaster,We just try all of $0,1,-1,2,-2$.
If none of those are roots we give up. Alternatively, we can use the Rational root theorem. (Nerd)
It states that if all coefficients are integers and if there is a "guessible" root, it has to be of the form $\frac pq$, where $p$ is a divider of the last coefficient and $q$ is a divider of the first coefficient.

In your example $x^3 -5x^2+8x-4$, the symbol $p$ represents the dividers of $-4$, which are $\pm 1, \pm 2, \pm 4$.
And $q$ is a divider of $1$, which is one of $\pm 1$.
So if there is a guessible root, it is one of $\frac{\pm 1}{\pm 1}, \frac{\pm 2}{\pm 1}, \frac{\pm 4}{\pm 1}$.
That is, one of $-4,-2,-1,1,2,4$.

As you can see, all the actual roots (that is $1$ and $2$) are in there. (Mmm)

 

[BestMethod]

As a result of the remainder theorem,if x=b makes any polynomial vanish,then (x-b) is a factor of it.

Now,how could you find any number such b?

According to the rational root theorem,a possible integer root of such an equation should be a divisor of the constant term(here is 4) which are ;
1,2,4,and their negatives.

So,we should substitute them,one by one,starting from x=1 to see which one makes it vanish.

Here,x=1&2 make it vanish,then (x-1)(x-2) is a factor of it.

The other factor is f(x)/(x-1)(x-2),which is again (x-2).

So,f(x)=(x-1)(x-2)(x-2)=(x-1)(x-2)^2

Here are short tutorials on division;
https://www.youtube.com/watch?v=4ahKcCSL6Mg&list=PLsMjNmqC7C2T66EgsXhbS-zGiCQB3i2NU&index=17&t=0s
https://www.youtube.com/watch?v=Cti9roxbP4Q&list=PLsMjNmqC7C2T66EgsXhbS-zGiCQB3i2NU&index=29&t=0s
https://www.youtube.com/watch?v=bSh8G5FXay4&list=PLsMjNmqC7C2T66EgsXhbS-zGiCQB3i2NU&index=30&t=0s
https://www.youtube.com/watch?v=EagRlnlgmtg&list=PLsMjNmqC7C2T66EgsXhbS-zGiCQB3i2NU&index=31&t=0s

Substituting can be annoying.I mean,how would you evaluate the polynomial for x=4?!
Let’s calculate it by a fast substituting method;

x^3=x^2.x

Now plugging 4 into x,

x^3=x^2.x=x^2.4=4x^2

So,f(x)=x^3-5x^2+8x-4=4x^2-5x^2+8x-4=-x^2+8x-4

Again,writing -x^2=-x.x,and setting x=4,

x^2=-x.x=-x.4=-4x

So,f(x)=-x^2+8x-4=-4x+8x-4=4x-4

Setting x=4,f(x)=4(4)-4=12

You could well learn this method by watching the short tutorials below;

https://www.youtube.com/watch?index=19&list=PLsMjNmqC7C2T66EgsXhbS-zGiCQB3i2NU&t=0s&v=Kq5AKAXOOSc
https://www.youtube.com/watch?index=20&list=PLsMjNmqC7C2T66EgsXhbS-zGiCQB3i2NU&t=0s&v=r2EXS-mjPC0
https://www.youtube.com/watch?index=26&list=PLsMjNmqC7C2T66EgsXhbS-zGiCQB3i2NU&t=0s&v=2BoGYi0p5N4

 

[Prove It]

Just because the Rational Roots Theorem is not entirely obvious, some reasoning behind it...

We would be expecting that your degree "n" polynomial would factorise to a form like

$\displaystyle \begin{align*} P\left( x \right) = \left( a\,x + b \right) \left( c\,x + d \right) \left( e\,x + f \right) \dots \left( w\,x + z \right) \end{align*}$

so that the roots are $\displaystyle \begin{align*} x = \left\{ -\frac{b}{a} , -\frac{d}{c} , -\frac{f}{e} , \dots , -\frac{z}{w} \right\} \end{align*}$

The polynomial expands out to

$\displaystyle \begin{align*} P\left( x \right) = ace\dots w\,x^n + \left( \textrm{middle terms} \right) + bdf\dots z \end{align*}$

and now it can clearly be seen that the NUMERATOR of each of the possible roots must be a factor of the constant term, while the DENOMINATOR must be a factor of the leading coefficient.

Now, I know that in cases where those particular numbers have a lot of factors, it would mean that there would be a large number of possible roots to test. But the beauty is that you do end up with a FINITE number of possibilities, which is ALWAYS better than an infinite number.