Strange quadratic manipulation

[shirozack]

TL;DR Summary

Why can y = ax^2 + bx +c be express as y = (a)(x - intercept1)(x - intercept2) ?

For example, let's say the quadratic curve ax^2 + bx + c intersects the x-axis at x=-5, x = 3
Why is it we can say the equation of the curve is then (a)(x+5)(x-3) ?
how does this manipulation come about, in particular, the (a) coefficient. Why can we just slot (a) into the factors (x+5)(x-3) ?

THanks!

 

[anuttarasammyak]

$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$
where
$$\alpha,\beta=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
You can confim it as
$$a(x-\alpha)(x-\beta)=a[x^2-(\alpha+\beta)x+\alpha\beta]=ax^2+bx+c$$
by calculation.
quadratic equation
$$ax^2+bx+c=a(x-\alpha)(x-\beta)=0$$
has solution
$$x=\alpha,\beta$$
so crossing points of curve y=ax^2+bx+c and line y=0 is
$$(\alpha,0)(\beta,0)$$

 

[fresh_42]

TL;DR Summary: Why can y = ax^2 + bx +c be express as y = (a)(x - intercept1)(x - intercept2) ?

For example, let's say the quadratic curve ax^2 + bx + c intersects the x-axis at x=-5, x = 3
Why is it we can say the equation of the curve is then (a)(x+5)(x-3) ?
how does this manipulation come about, in particular, the (a) coefficient. Why can we just slot (a) into the factors (x+5)(x-3) ?

THanks!

If you have any polynomial $y(x)=a_nx^n+\ldots+a_1x+a_0$ then we can perform a long division dividing it by $x-c.$ This results in an expression $y(x)=q_1(x)\cdot (x-c) + r_1(x)$ where the remainder polynomial has a degree of at most $n-1$ and $q_1(x)=a_n.$

Now, if $x=c$ is an intersection point, i.e. $0=y(c)$ then $$0=y(c)=q_1(c)\cdot(c-c)+r_1(c)=q_1(c)\cdot 0+r_1(c)=r_1(c) .$$
This means, that the remainder is either zero or has an intersection point at $x=c,$ too. Now, we can do the same long division again for $r_1(x)$ and get an expression $r_1(x)=q_2(x)\cdot(x-c)+r_2(x)$ where the remainder polynomial has this time at most a degree of $n-2$ and
\begin{align*}
y(x)&=q_1(x)\cdot(x-c)+r_1(x)\\&=q_1(x)\cdot(x-c)+q_2(x)\cdot(x-c)+r_2(x)\\&=(q_1(x)+q_2(x))\cdot (x-c) +r_2(x).
\end{align*}
We proceed this algorithm of long divisions until $r_m(x)=0.$ Since the degree of the remainders shrinks with every step, this has to be the case at some point. In the end, we have an expression
$$
y(x)=(q_1(x)+\ldots +q_m(x))\cdot (x-c) +0.
$$
This proves, that if $x=c$ is an intersection point of the polynomial $y(x)$ then $(x-c)\,|\,y(x).$

The other direction is easy: if $(x-c)\,|\,y(x)$ then $y(x)=q(x)\cdot (x-c)$ and $y(c)=0.$ Together, we have shown that $x=c$ is an intersection point of $y(x)$ if and only if $(x-c)$ divides $y(x).$

There are, however, polynomials like for example $y(x)=x^2+1$ that do not have real intersection points at all. Those polynomials only split over the complex numbers: $x^2+1=(x+\boldsymbol i)(x- \boldsymbol i).$

Finally, here is an example how long division for polynomials is done:
https://www.physicsforums.com/threa...r-a-polynomial-over-z-z3.889140/#post-5595083

 

[shirozack]

$$ax^2+bx+c=a(x-\alpha)(x-\beta)$$
where
$$\alpha,\beta=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
You can confim it as
$$a(x-\alpha)(x-\beta)=a[x^2-(\alpha+\beta)x+\alpha\beta]=ax^2+bx+c$$
by calculation.
quadratic equation
$$ax^2+bx+c=a(x-\alpha)(x-\beta)=0$$
has solution
$$x=\alpha,\beta$$
so crossing points of curve y=ax^2+bx+c and line y=0 is
$$(\alpha,0)(\beta,0)$$

so b is -a(α+β)
and c is aαβ ?

 

[fresh_42]

so b is -a(α+β)
and c is aαβ ?

Yes. It's called Vieta's formulas.

 

[PeroK]

TL;DR Summary: Why can y = ax^2 + bx +c be express as y = (a)(x - intercept1)(x - intercept2) ?

For example, let's say the quadratic curve ax^2 + bx + c intersects the x-axis at x=-5, x = 3
Why is it we can say the equation of the curve is then (a)(x+5)(x-3) ?
how does this manipulation come about, in particular, the (a) coefficient. Why can we just slot (a) into the factors (x+5)(x-3) ?

THanks!

Here's another way to look at it.

Suppose $p(x) = x^2 + bx + c$ has a y-intercept at $x = \alpha$. Note that $p(\alpha) = 0$.

Let $z = x - \alpha$, then:
$$p(x) = p(z+ \alpha) = (z + \alpha)^2 + b(z + \alpha) + c = z^2 + (2\alpha + b)z + \alpha^2 + b\alpha + c$$Now, we know that $\alpha^2 + b\alpha + c = p(\alpha) = 0$. So:
$$p(x) = z^2 + (2\alpha + b)z = z(z + 2\alpha + b) = (x - \alpha)(x + \alpha + b)$$And, we see that $p(x) = 0$ when $x = -\alpha - b$. That gives us a second y-intercept. Let's call this $\beta = -\alpha - b$. This completes the factorisation:
$$p(x) = (x - \alpha)(x - \beta)$$Where $\alpha, \beta$ are the y-intercepts. Note that it's possible, of course, that $\alpha = \beta$. That happens when $\alpha = -\frac b 2$.

Finally, we consider the general quadratic
$$p(x) = ax^2 + bx + c = a(x^2 + \frac b a x + \frac c a) = a(x- \alpha)(x - \beta)$$Where $\alpha, \beta$ are the y-interecpts of the quadratic $x^2 + \frac b a x + \frac c a$. But, this has precisely the same y-intercepts as $p(x)$. One quadratic is just a scalar multiple of the other. Hence $\alpha, \beta$ are also the y-intercepts of $p(x)$. QED

 

[julian]

Given the values of a quadratic polynomial, $ax^2+bx+c$, at three points $x_0,x_1$, and $x_2$, which are $y_0,y_1$, and $y_2$ respectively, you can construct a quadratic polynomial $p(x)$ that matches these values as follows:

\begin{align*}
p(x) = y_0 \frac{x-x_1}{x_0-x_1} \frac{x-x_2}{x_0-x_2} + y_1 \frac{x-x_0}{x_1-x_0} \frac{x-x_2}{x_1-x_2} + y_2 \frac{x-x_0}{x_2-x_0} \frac{x-x_1}{x_2-x_1} \qquad (1)
\end{align*}

This quadratic polynomial $p(x)$ uniquely fits the given values: $p(x_0)=y_0 , p(x_1) = y_1$, and $p(x_2)=y_2$. To show its uniqueness, assume there is another quadratic polynomial $\tilde{p}(x)$ that also fits these values. Consider the difference $q(x)=p(x) - \tilde{p}(x)$. This difference $q(x)$ is a quadratic polynomial:

\begin{align*}
q(x) = p (x) - \tilde{p} (x) = ex^2+fx+g
\end{align*}

which has the three distinct real roots $x_0,x_1,x_2$. This implies that $q(x)$ is actually the zero polynomial - i.e. $e=f=g=0$. To understand why, note that as $q(x)$ has three distinct real roots, it would have a shape with both a minimum and a maximum, as illustrated in the graph.

At the minimum and a maximum the derivative $q' (x)$ vanishes. However, the derivative of $q(x)$, which is $q'(x) = 2ex + f$, can only have one root. Therefore, it cannot provide the necessary two critical points for $q(x)$ to have both a minimum and a maximum. This contradiction implies that $q(x)$ must be identically zero, and that $\tilde{p}(x)=p(x)$.

Now let us assume that $p(x)$ has two distinct real roots: $x_1$ and $x_2$. Then $y_1=0$ and $y_2=0$ and using this in Eq. (1):

\begin{align*}
p(x) = y_0 \frac{x-x_1}{x_0-x_1} \frac{x-x_2}{x_0-x_2} = \frac{y_0}{(x_0-x_1)(x_0-x_2)} (x-x_1)(x-x_2)
\end{align*}

As we have shown that the quadratic polynomial is unique, and we know it is given by $ax^2+bx+c$, we have $a=\frac{y_0}{(x_0-x_1)(x_0-x_2)}$, and so:

\begin{align*}
p(x) = a (x-x_1)(x-x_2) .
\end{align*}