368 Quadratic Equations with Real Roots
The fact that the above general equation is a quadratic might not appear obvious, but
when we multiply out the left side of the equation using the product of sums rule, the truth
is revealed. Start with the expression
(px+q)(rx+s)
Multiplying this out according to the product of sums rule gives us
pxrx+pxs+qrx+qs
Using the commutative law for multiplication in the first two terms and then simplifying the
first term, we get
prx^2 +psx+qrx+qs
The distributive law “in reverse” allows us to rewrite this as
prx^2 + (ps+qr)x+qs
When we set this equal to 0, we get an equation in polynomial standard form:
prx^2 + (ps+qr)x+qs= 0
Are you confused?
If you have trouble seeing why the above equation is in polynomial standard form, you can rename the
coefficients and constants like this:
pr=a
ps+qr=b
qs=c
By substitution, you get
ax^2 +bx+c= 0
If you wonder about the requirement that p≠ 0 and r≠ 0 in the binomial factor form, the reason for this
restriction should be clear now. If p= 0 or r= 0, then pr= 0, the term containing x^2 vanishes, and you
have the one-variable first-degree equation
(ps+qr)x+qs= 0
What are the roots?
If we find a quadratic equation in binomial factor form and we want to find the roots, we’re
in luck! Here is the binomial factor form again, for a quadratic in the variable x:
(px+q)(rx+s)= 0