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I came across the equation in my maths book. I wonder why it is as such. Can it be equal to $b(x-\alpha)(x-\beta)$ or $c(x-\alpha)(x-\beta)$?

Kindly help me out.

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    $\begingroup$ No because if you expand the RHS expression you'll see that the leading coefficient needs to be $a$. $\endgroup$ – Peter Foreman Jun 27 '19 at 6:32
  • $\begingroup$ I think the following can help: math.stackexchange.com/questions/2464409 $\endgroup$ – Michael Rozenberg Jun 27 '19 at 6:43
  • $\begingroup$ You might get the idea by reading about rational root theorem/integral root theorem, and why the leading coefficient is there while finding the expression for zeros of a polynomial. $\endgroup$ – Roopesh Singh Jun 27 '19 at 7:04
  • $\begingroup$ Yes, it can, if for example $a=b$. $\endgroup$ – Michael Hoppe Jun 27 '19 at 11:04
  • $\begingroup$ Shortly: the product of two polynomials of the first degree is a polynomial of the second degree, and this operation is reversible. $\endgroup$ – Yves Daoust Jul 3 '19 at 7:17
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$$p(x-\alpha)(x-\beta)=px^2-p(\alpha+\beta)x+p\alpha\beta.$$

If you want this quadratic polynomial to equal $$ax^2+bx+c$$ for all $x$, the respective coefficients must be equal:

$$\begin{cases}p&=a,\\-p(\alpha+\beta)&=b,\\p\alpha\beta&=c.\end{cases}$$

As you see, there is no other option than $p=a$.


Addendum:

From the above we draw

$$\begin{cases}a(\alpha+\beta)+b&=0,\\a\alpha\beta&=c.\end{cases}$$

Then, multiplying the first by $\alpha$,

$$a\alpha^2+a\alpha\beta+b\alpha=0,$$

and substituting the second,

$$a\alpha^2+b\alpha+c=0.$$

This shows that $\alpha$ must be a root of the given polynomial (indeed, $a\alpha^2+b\alpha+c=a(\alpha-\alpha)(\beta-\alpha)=0$). And so is $\beta$, by symmetry.

Note that if the polynomial has no real root, you can't use this factorization (in the reals).


E.g.

$$2x^2+3x-2$$ has the roots $$-2,\dfrac12$$ (check by plugging).

Then

$$2x^2+3x-2=2(x+2)\left(x-\frac12\right).$$


Note that this generalizes to higher degrees,

$$ax^n+bx^{n-1}+\cdots c=a(x-\alpha)(x-\beta)\cdots(x-\gamma).$$

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The titular equation is true if and only if $\alpha$ and $\beta$ are roots of the expression on the left. We may then express them in terms of $a\ne 0,b,c$ by the quadratic formula $$\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

I originally interpreted your question as

Why can we always factor every expression of the form $ax^2+bx+c,$ with $a\ne0$?

If this is indeed what you meant to ask, then there are many ways to see how (and many levels of understanding why so). One reason that that is always possible is simply to note that such expressions can always be expressed as a difference of two squares, as follows:

$$ax^2+bx+c=a\left(x^2+\frac ba x+\frac ca\right)=a\left(x^2+2x\frac{b}{2a}+\left(\frac{b}{2a}\right)^2-\left(\frac{b}{2a}\right)^2+\frac ca\right)=a\left(x^2+2x\frac{b}{2a}+\left(\frac{b}{2a}\right)^2\right)-a\left(\left(\frac{b}{2a}\right)^2-\frac ca\right).$$

Now the first summand is a complete square, and the second is just a real number, and as such can always be written as a square of some complex number. Thus by setting $$a\left(x^2+2x\frac{b}{2a}+\left(\frac{b}{2a}\right)^2\right)=a\left(x+\frac{b}{2a}\right)^2$$ and $$a\left(\left(\frac{b}{2a}\right)^2-\frac ca\right)=A,$$ the last expression from the above sequence becomes $$a\left(x+\frac{b}{2a}\right)^2-A=(\sqrt a)^2\left(x+\frac{b}{2a}\right)^2-(\sqrt A)^2,$$ which you can always factor since it is a difference of two squares.

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  • $\begingroup$ I guess that the question is "can the leading coefficient be $b$ or $c$ ?" $\endgroup$ – Yves Daoust Jul 3 '19 at 6:48
  • $\begingroup$ @YvesDaoust The best I can say is that OP is not quite clear. I don't see, for example, a clear relationship between the question in the title (which I take to be the main question bothering them) and what they write in the body of the post (which I can't claim to understand). $\endgroup$ – Allawonder Jul 3 '19 at 6:58
  • $\begingroup$ You are right. The main question seems to be "why is it so", and an ancillary one is about the leading coefficient. $\endgroup$ – Yves Daoust Jul 3 '19 at 7:00
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Two polynomials can only be equal to each other if 1) their degrees are the same and 2) if the coefficients for $x$ to each power, are the same for both polynomials.

$a(x-\alpha)(x-\beta) = ax^2 - a(\alpha + \beta)x + a\alpha\beta$.

This can only equal $ax^2 + bx + c$ if:

1) $2= 2$. It does.

2) $a = a$. It does.

3) $-a(\alpha+\beta) = b$. It might. There are an infinite number of $\alpha, \beta$ and it's very easy to find some where this is true.

4) $a\alpha\beta = c$. It might.There are an infinite number of $\alpha, \beta$ and it's very easy to find some where this is true. BUT we also need $-a(\alpha+\beta) = b$. Now it is often the case we can find one pair that satisfy both. But not always.

This idea they are getting at is that if sometimes there are $\alpha, \beta$ so that $(\alpha + \beta) = -\frac ba$ and $\alpha\beta = c$. When there are such $\alpha$ and $\beta$ you will be able to factor the polynomial as $a(x-\alpha)(x - \beta)$.

You will find out if you keep reading that if $ax^2 + bx + c=0$ has solutions then those $\alpha, \beta$ will be the solutions and that if $ax^2 + bx + c=0$ doesn't have solutions then you will not be able to factor it thus.

And if you read even further they will tell you how two find those $\alpha$ and $\beta$ and when they exist.

Now to your question:

If $ax^2 + bx + c = b(x-\alpha)(x - \beta) = bx^2 - b(\alpha + \beta) + b\alpha \beta$ then that would mean that $a = b$.

... which could happen.... but doesn't happen in general and isn't important.

The idea is that $a$ is the leading coeficient and will have to be the leading coefficient in in the factored version as well.

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Oh what the heck:

$ax^2 + bx + c =$

$a (x^2 + \frac ba x +\frac ca) = $

$a(x^2 + 2\frac b{2a} +\frac {b^2}{4a^2} - \frac {b^2}{4a^2} + \frac ca) = $

$a[(x + \frac b{2a})^2 -(\frac {b^2}{4a^2} - \frac ca)]=$

$a[(x+\frac b{2a})^2 - (\frac {b^2 - 4ac}{4a^2})]=$

$a[(x + \frac b{2a})^2 - (\frac {\sqrt{b^2 - 4ac}}{2a})^2] = $

$a[(x+\frac b{2a} + \frac {\sqrt{b^2 -4ac}}{2a})(x+\frac b{2a} - \frac {\sqrt{b^2 -4ac}}{2a})] = $

$a(x -\frac {-b-\sqrt{b^2 - 4ac}}{2a})(x- \frac {-b + \sqrt{b^2 -4ac}}{2a}) =$

$a(x -\alpha) (x - \beta)$

if $\alpha = \frac {-b-\sqrt{b^2 - 4ac}}{2a}$ and $\beta = \frac {-b + \sqrt{b^2 -4ac}}{2a}$.

This assumes that

1) $a \ne 0$.

2) $b^2 - 4ac \ge 0$.

We couldn't write this unless those were both true.

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