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I'm looking at Calculus by Adams, 7th Ed. page 43 Example 3b. It shows how to find the roots of $9x^2-6x+1$ by using the quadratic formula to arrive at the double root $\frac{1}{3}$. That I understand.

However, I don't understand how the factoring (?) is carried out in the next line:

$9x^2-6x+1 = 9\left(x - \frac{1}{3}\right)^2 = (3x - 1)^2$

I thought the factor theorem says something along the lines of that if we have roots such as $\frac{1}{3}$, we can write the factors $(x - \frac{1}{3})(x - \frac{1}{3}) = \left(x - \frac{1}{3}\right)^2$. I don't get where the $9$ comes from after the first equal sign. Are they using a different method than completing the square?

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    $\begingroup$ the factor theorem says ...that if $\frac{1}{3}$ is a root, then $x-\frac{1}{3}$ is a factor, therefore $9x^2-6x+1$ must be a scalar multiple of $\left(x-\frac{1}{3}\right)^2\,$. The value of the scalar can de determined by comparing the leading coefficients (or the free terms, for that matter) and, in this case, it turns out to be $9\,$. $\endgroup$
    – dxiv
    Mar 16 '17 at 4:33
  • $\begingroup$ Interesting! So to determine the value of the scalar, it seems I can look at $x^2$ and compare it to $9x^2$, which is $9$ times larger. Or I can look at $\left(\frac{1}{3}\right)^2 = \frac{1}{9}$ and comare it to 1, which is $9$ times larger! Amazing. $\endgroup$ Mar 16 '17 at 4:47
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    $\begingroup$ Would it be the same as using completing the square like this: $\frac{9x^2}{9}-\frac{6x}{9}+\frac{1}{9} = 0 => x^2-\frac{2}{3}x + \left(\frac{1}{3}\right)^2 = -\frac{1}{9} + \frac{1}{9} => \left(x - \frac{1}{3}\right)^2 = 0$ then multiply by 9: $9\left(x - \frac{1}{3}\right)^2 = 0$ $\endgroup$ Mar 16 '17 at 5:01
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    $\begingroup$ Not much left to complete: $\,9x^2-6x+1=(3x)^2 - 2 \cdot (3x) \cdot 1 + 1^2 = (3x-1)^2\,$. $\endgroup$
    – dxiv
    Mar 16 '17 at 5:02
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The factor theorem says only that $x-c$ is a factor of a polynomial $p(x)$ if and only if $p(c)=0$. What this exactly means is that there is a polynomial $q(x)$ such that $p(x) = (x-c)q(x)$. The theorem doesn't say, but it's required that $q(x)$ has degree one less than $p(x)$.

Applied to a quadratic polynomial such as your polynomial. We set $p(x) = 9x^2-6x+1$ has the only roots at $x=1/3$. This means that $x-1/3$ is a factor of it which means that is there's a polynomial $q(x)$ such that $p(x) = (x-1/3)q(x)$. Now $q(x)$ must be of degree $1$ that is on the form $q(x) = ax+b = a(x+b/a)$, we see from this that $x+b/a$ must be a factor of $p(x)$. So we know that by the factor theorem that $p(-b/a) = 0$ which we now requires $b/a$ to be $1/3$. We now know that:

$$9x^2-6x+1 = a(x+b/a)(x-1/3) = a(x-1/3)(x-1/3) = a(x-1/3)^2 = a\left(x^2-{2\over3}x + {1\over 9}\right)$$

With $a$ needing to be $9$ in order to the LHS and RHS to be equal.

Another approach would be to complete the square (this is how the quadratic equation is solved without the formula), we know that $9x^2-6x = (3x-1)^2 -1$ (since $(3x-1)^2 = 9x^2-6x+1$, the $3x-1$ is chosen so that the $x^2$ and $x$ terms match). This means that

$$9x^2-6x+1 = (3x-1)^2-1 + 1 = (3x-1)^2$$

If you want to bring that into the other form you just rewrite the factors using $(3x-1) = 3(x-1/3)$

Note that one often shortcuts this reasoning. However you come to the conclusion you must of course end up with a factorization that is correct. You can just rely on that you by expanding $9(x-1/3)^2$ via the square rule come to the correct polynomial.

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  • $\begingroup$ Thank you so much, this gives me new insight into factoring and the relationship between a polynomial $p(x)$ and its factor $x-c$. I didn't quite understood the part with "the $3x − 1$ is chosen so that the $x^2$ and $x$ terms match". Are you using a technique Khan Academy calls "factoring by grouping" possibly? Like this: $9x^2-6x+1$ We want to find two numbers that multiply to $1 * 9$ and whose sum is $-6$. That is $-3$ and $-3$. We rewrite the equation as $=9x^2-3x-3x+1$. Now we factor $ = 3x(3x-1)-(3x-1) = (3x-1)(3x-1)=(3x-1)^2$ $\endgroup$ Mar 17 '17 at 3:21
  • $\begingroup$ I get it! (The general method for factoring quadratics with leading coefficient not being 1 uses factoring by grouping. Khan) $\endgroup$ Mar 17 '17 at 4:31

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