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I have some issues understanding factorization.

If I have the expression $x^{2}-x-7$ then (I was told like this) I can put this expression equal to zero and then find the solutions with the quadratic formula, so it gives me $x_{0,1}= 1 \pm 2\sqrt{2}$ then $$x^{2}-x-7 = (x-1-2\sqrt{2})(x-1+2\sqrt{2}).$$

That is correct I have checked it.

Now for the expression $3x^{2}-x-2$ if I do the same I have $x_{0} = 1$ and $x_1=\frac{-2}{3}$ so I would have $$3x^{2}-x-2 = (x-1)(x+\frac{2}{3})$$ but this is not correct since $(x-1)(x+\frac{2}{3}) = \frac{1}{3}(3x^{2}-x-2)$,

the correct factorization is $3x^{2}-x-2 = (3x+2)(x-1)$.

So I guess finding the roots of a quadratic expression is not sufficient for factorizing.

Novato
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    If $a$ is the lead coefficient and the roots are $r$ and $s$, the factorization is $a(x-r)(x-s)$. If you feel like, you can absorb the $a$ into one of $x-r$ and $x-s$, whichever gives you the "nicer" expression. Sometimes it is useful to absorb part of $a$ into one and part into the other. So the quadratic formula is sufficient. – André Nicolas Feb 04 '16 at 04:00

5 Answers5

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Look at what happens when we complete the square:

$$\begin{array}{lll} ax^2+bx+c &=& \frac{1}{4a}(4a^2x^2 + 4abx + 4ac)\\ &=& \frac{1}{4a}((2ax)^2 + 2b(2ax) + 4ac)\\ &=& \frac{1}{4a}((2ax)^2 + 2b(2ax) +b^2-b^2+ 4ac)\\ &=& \frac{1}{4a}((2ax+b)^2-(b^2 - 4ac))\\ &=& \frac{1}{4a}((2ax+b)^2-\bigg(\sqrt{b^2 - 4ac}\bigg)^2)\\ &=& \frac{1}{4a}(2ax+b+\sqrt{b^2-4ac})(2ax+b-\sqrt{b^2-4ac})\\ &=& \frac{4a^2}{4a}\bigg(\frac{2ax+b+\sqrt{b^2-4ac}}{2a}\bigg)\bigg(\frac{2ax+b-\sqrt{b^2-4ac}}{2a}\bigg)\\ &=& a\bigg(x-\frac{-b-\sqrt{b^2-4ac}}{2a}\bigg)\bigg(x-\frac{-b+\sqrt{b^2-4ac}}{2a}\bigg)\\ \end{array}$$

John Joy
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Factoring tells you the roots, but the roots don't describe any function completely, which is where the discrepancy comes from.

What's going on can be understood perhaps best from plotting both of the functions you used to represent the roots, $3x^2 - x - 2$ and $(x-1)(x+\frac23)$:

enter image description here

Both pass through $(-\frac23, 0)$ and $(1, 0)$, but besides that they are out of sync -- in fact, one is simply a "stretched" version of the other--the blue curve is obtained by multiplying the $y$-coordinate of the red curve at every point by 3. The stay aligned at 0 because the $y$-coordinate there is 0--and of course, multiplying 0 by 3 makes no difference.

In fact, there is an infinitude of parabolas that share those roots, the family of which is indexed by the coefficient $a$ alluded to by AndreNicolas above: $a(x-1)(x+\frac23)$.

Here are those curves for $a = -5, \ldots, 5$:

enter image description here

Note: Plots produced using R, a variation on the following:

xx <- seq(1/6 - 1.5, 1/6 + 1.5, length.out = 1000)
yy1 <- (xx - 1) * (xx + 2/3)
matplot(xx, yy <- sapply(aa <- c(-5:-1, 1:5), function(a) a * yy1),
        lty = 1, type = "l", lwd = 3, xlab = "", ylab = "",
        col = c("brown", "cyan", "green", "orange", "darkgrey",
                "red", "magenta", "blue", "orchid", "darkgreen"),
        ylim = range(yy[ , 3:8]),
        main = expression(paste(plain("Several values of "), a)))
abline(h = 0, v = 0, lwd = 1, col = "black")
text(1/6, yy[500,], labels = paste0("a = ", aa), pos=3)
MichaelChirico
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The quadratic formula only gives the roots of the equation, which are unaltered by the constant factor.

Let $\ P(x)=3x^{2}-x-2 \ \text{ and } \ Q(x) = x^2 -\frac{x}{3}-\frac{2}{3}$

The roots of both the polynomials are the same, as

$$\begin{align}3x^{2}-x-2&=0 \\ \implies x^2 -\frac{x}{3}-\frac{2}{3}&=0\end{align}$$

but $P(x)$ and $Q(x)$ differ by a constant factor $3$, the leading coefficient of $P(x)$.

$$P(x) = 3\times Q(x)$$

Max Payne
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  • One-liners rarely make good answers. Please fill out the details. I don't think the quadratic formula should be a cause for cryptic breadcrumbs. – hardmath Feb 04 '16 at 04:04
  • As I learned the quadratic formula , it allowed a coefficient on the $x^2$ term. – Ross Millikan Feb 04 '16 at 15:06
  • @RossMillikan That is the case for finding the roots. But when factoring, it doesn't give the constant factor (coefficient of $x^2$), since the roots are unaltered by it. – Max Payne Feb 04 '16 at 15:26
  • The "a" in the quadration is the coefficient. The presumed factorization however is: $a(x - r_1)(x - r_2)$ But setting $a(x-r_1)(x - r_2) = 0$ the "a" (if $a\ne 0$) becomes irrelevant. – fleablood Feb 04 '16 at 21:19
  • @fleablood yes thats the point. $a$ becomes irrelevant to the roots. So it gets lost. in reality,$a$ is a factor of the polynomial. – Max Payne Feb 05 '16 at 02:58
  • I've edited my answer to remove the line which was the root of much agony. – Max Payne Feb 05 '16 at 03:12
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For equations of the form $ax^2 + bx + c$ you can use the quadratic equation $x = \frac{-b \pm \sqrt{b^2-4ac }}{2a}$ to help you factor, like you were doing. The adjustment you need to make is to multiply your factorization by a.

In the example you had, $3x^2 -x - 2$, you found the roots were x = 1 and x $= \frac{-2}{3}$. If you do $3*(x+\frac{2}{3})(x−1)$, then you have (3x+2)(x−1) when you multiply. This works in general. I hope this helps.

user3016098
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$3(x - 1)(x + {2 \over 3}) = (x -1)(3x + 2) = (3x - 3)(x + {2 \over 3}) =...$ etc. are all valid factoring. The leading coefficient is just a constant.

And if $(x - 1)(x + {2 \over 3}) = 0$ then $3(x - 1)(x + {2 \over 3}) = 0 = (x - 1)(x + {2 \over 3}) $.

If you are concerned about going from roots to factoring think of it this way:

If the roots of $P(x) = ax^n + ...... + c$ are $r_1, .... ,r_n$ then the polynomial factors as $a(x - r_1)(x - r_2).....(x - r_3) = P(x)$ or $(x - r_1)(x - r_2).....(x - r_3) = P(x)/a$. When setting to 0 the $a$ doesn't matter as if $y = 0$ then $a*y = 0$ also no matter what the $a$ is.

Solving for 0, the $a$ gets lost. Going from the roots back, we only take $(x - r_i)$ so the resulting coefficient will always be $1$. The resulting polynomial will be $1/a$ of the original.

fleablood
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