B Strange quadratic manipulation

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The discussion centers on the manipulation of quadratic equations, specifically how a quadratic curve defined by ax^2 + bx + c can be expressed as a(x + 5)(x - 3) when it intersects the x-axis at x = -5 and x = 3. This transformation is rooted in the fact that if a polynomial has real roots at certain points, it can be factored accordingly, with the coefficient 'a' scaling the polynomial. The relationship between the coefficients and the roots is further clarified through Vieta's formulas, which relate the sum and product of the roots to the coefficients. The discussion also touches on polynomial long division, demonstrating that if a polynomial has a root at x = c, then (x - c) is a factor of that polynomial. Overall, the manipulation of quadratics is a fundamental aspect of polynomial algebra.
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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!
 
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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)
 
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shirozack said:
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
 
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anuttarasammyak said:
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 ?
 
shirozack said:
so b is -a(α+β)
and c is aαβ ?
Yes. It's called Vieta's formulas.
 
shirozack said:
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
 
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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.

q(x).jpg


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*}
 
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