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Illustrative
Mathematics
A-APR Zeroes and factorization
of a quadratic polynomial I
Alignments to Content Standards: A-APR.B.2
Task
Suppose f is a quadratic function given by the equation f (x) = ax 2
a, b, c are real numbers and we will assume that a is non-zero.
a. If 0 is a root of f explain why c
by x.
+ bx + c where
= 0 or, in other words, ax 2 + bx + c is evenly divisible
b. If 1 is a root of f explain why ax 2
+ bx + c is evenly divisible by x − 1.
c. Suppose r is a real number. If r is a root of f explain why ax 2
divisible by x − r.
+ bx + c is evenly
IM Commentary
For a polynomial function p, a real number r is a root of p if and only if p(x) is evenly
divisible by x − r. This fact leads to one of the important properties of polynomial
functions: a polynomial of degree d can have at most d roots. This is the first of a
sequence of problems aiming at showing this fact. The teacher should pay close
attention to the logic used in the solution to part (c) where the divisibility of
ax 2 + bx + c by x − r is obtained not by performing long division but by using the
result of long division of these polynomials; namely, that said division will result in an
expression of the following form:
a
2
+ bx + c = (x − r)ℓ(x) + k
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Illustrative
Mathematics
ax 2 + bx + c = (x − r)ℓ(x) + k
where ℓ is a linear polynomial and
k is a number.
This task could be used either for assessment or for instructional purposes. If it is used
for assessment, parts (a) and (b) are more suitable than part (c). Each of the questions
in this task could be formulated as an if and only if statement but the other implication,
namely that f (x) is divisible by x − r if and only if r is a root of f . The direction not
presented in this task is more straightforward and so has been left out.
Edit this solution
Solution
a. If 0 is a root of the function f this means that f (0)
formula f (x) = ax 2 + bx + c this means that
= 0. Since f is given by the
a ⋅ 02 + b ⋅ 0 + c = 0.
From here we see that c
= 0. Since c = 0, we have
f (x) = ax 2 + bx = x(ax + b)
so f (x) is evenly divisible by x.
b. If 1 is a root of the function f this means that f (1)
formula f (x) = ax 2 + bx + c this means that
= 0. Since f is given by the
a ⋅ 12 + b ⋅ 1 + c = 0.
So we have a + b + c = 0. If we perform long division of ax 2 + bx + c by (x − 1) we
find that x − 1 goes in ax + a + b times with a remainder of a + b + c. But we just saw
that, when 1 is a root of f , a + b + c is zero so this means that (x − 1) divides
ax 2 + bx + c evenly.
c. If r is a root of f this means that f (r) = 0. We could perform long division as in part
(b) but it is complicated because we have an extra variable r. The result of performing
long division will be a linear polynomial function l and a constant k so that
ax 2 + bx + c = (x − r)l(x) + k.
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Illustrative
Mathematics
To evaluate k we could perform the long division as in part (b) or we can plug r into the
above equation to find
ar 2 + br + c = (r − r)l(r) + k
= k.
= ar 2 + br + c = f (r). But we know that f (r) = 0 because r is a root of f . Hence
ax 2 + bx + c = (x − r)l(x) and so x − r divides ax 2 + bx + c evenly as desired.
So k
Notice in part (c) that it was not necessary to calculate l(x) explicitly, just to know that
such a linear polynomial exists and can be found by performing long division of
polynomials. This is important because the division process is tricky to perform when
there are so many variables and it becomes even more difficult when the polynomials
involved are of larger degree.
A-APR Zeroes and factorization of a quadratic polynomial I
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