1

I am trying to factor $9x^2-6x+1$ after finding the roots, I am using the following formula $a(x-x_1)(x-x_2)$ in this case there is just one root ($\frac{1}{3}$)

How do I know that the answer is $(3x-1)^2$ and not just $(3x-1)$?

gbox
  • 12,867
  • Question is not clear to me..Can you help me?? – Empty Aug 16 '15 at 15:37
  • It's a quadratic, so you need two linear factors. – Daniel Fischer Aug 16 '15 at 15:39
  • There is not one root : there are two identical roots. Do you see the difference ? – Claude Leibovici Aug 16 '15 at 15:41
  • @DanielFischer ok but what happen if we have an n-degree, and we have only real solutions? – gbox Aug 16 '15 at 16:21
  • @ClaudeLeibovici 2 above the complex? the other one has just an imaginary part? – gbox Aug 16 '15 at 16:23
  • For a polynomial of degree $n$, you have $n$ linear factors. If you have more than one zero, and fewer than $n$ distinct zeros, you need to find out the multiplicity of each zero. (If there is only one zero, or there are $n$ distinct zeros, the multiplicity is clear, so no need to find it.) – Daniel Fischer Aug 16 '15 at 18:05

4 Answers4

3

Multiplying out you get: $$(3x-1)^2=(3x-1)(3x-1)=9x^2-6x+1$$

so the $\frac{1}{3}$ is a double root. The fundamental theorem of algebra ensures that this double root is the only root.

Addendum:

You can verify that $\frac{1}{3}$ is a double root by checking that it is also a root of the derivative $f'(x) = 18x - 6$.

Marconius
  • 5,635
  • Why can I check with the derivative? where can I read more about it? – gbox Aug 16 '15 at 16:24
  • See http://math.stackexchange.com/questions/127794/how-to-prove-that-r-is-a-double-root-if-and-only-if-it-is-a-root-of-a-polynomial (I will look for a more complete source - I think I saw it originally in a book on abstract algebra, but not sure if this can be had online). As to the derivative, you will need to know how to get the derivative of a polynomial (the main idea here is that the derivative is linear and $\frac{d}{dx}(x^n) = nx^{n-1}$). The product rule from calculus is used in the proof. – Marconius Aug 16 '15 at 19:29
2

Using the quadratic formula for $ax^2+bx+c=0$ we get: $$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

In your case we have $x_1=\frac{1}{3}$ and $x_2=\frac{1}{3}$, hence we can write $$9x^2-6x+1=9(x-x_1)(x-x_2)=9\left(x-\frac{1}{3}\right)\cdot\left(x-\frac{1}{3}\right)=9\left(x-\frac{1}{3}\right)^2.$$

Hirshy
  • 5,040
1

You know it first because of the identity $(a-b)^2=\dots$

Second it is $(3x-1)^{\color{red}2}$ for reasons of degree: $9x^2-6x+1$ has degree $2$, while $3x-1$ has degree $1$.

Bernard
  • 175,478
  • What should I do in the case of n degree? (above $\mathbb{R}$) – gbox Aug 16 '15 at 16:26
  • 1
    If the polynomial has $n$ distinct roots, say $;\alpha_1,\dots,\alpha_n$ the polynomial factors as $;a(x-\alpha_1)(\dotsc(x-\alpha_n)$. If it has only real roots, but less than $n$, it's a little more complexe. The notion of multiplicity of a root steps in. The conclusion is there are $n$ linear factors, counting multiplicities of the roots. – Bernard Aug 16 '15 at 16:32
  • How do this process or notation is called so I can look for more info about it? – gbox Aug 16 '15 at 16:34
  • 1
    I'd say ‘factorisation of polynomials’. An essential result is that $\alpha$ is a root of $p(x)$ if and only if $p(x)$ is divisible by $x-\alpha$. – Bernard Aug 16 '15 at 16:39
1

After factoring out 9 you get $f(x)=9\left( x^2-2\frac{1}{3}+\frac{1}{9}\right)=9\left( x-\frac{1}{3} \right)\cdot\left( x-\frac{1}{3} \right)=0$

A product is equal to zero, if at least one factor is equal to zero. Therefore there is only one root. You are right. Because of the quadratic attribute of the function this root is a double root. The function only touches the x-axis at one single point. It can be well seen by plotting the graph:

enter image description here

callculus42
  • 30,550