SME
Grade 11 Math · Problem type
A polynomial from its zeros
Use this for “Write the polynomial of least degree with zeros 1 and 2 + 3i.”
Diagram
The conjugate z̄ = 2 − 3i: z reflected across the real axis.
Real zero: 1
p² + q²: 13
This is an example. Type your own values to start a new problem.
(x
−
r
)(x
−
(
p
+
q
i))
×
(x
−
(
p
−
q
i))
=
x³
+
(
b
)
x²
+
c
x
+
(
d
)
s
p² + q²
worked out (not typed)
Formulas
s
=
p
2
+
q
2
13
=
2
2
+
3
2
b
=
−2
×
p
−
r
−5 = −2 × 2 − 1
c
=
s
+
2
×
p
×
r
17 = 13 + 2 × 2 × 1
d
=
−
r
×
s
−13 = −1 × 13
Assumptions
•
A degree-n polynomial has exactly n roots, counting complex roots and repeats (the fundamental theorem of algebra).
•
Real coefficients bring complex roots in conjugate pairs: 2 + 3i comes with 2 − 3i.
•
The pair multiplies to x² − 2px + (p² + q²), then times x − r.
Step-by-step
WE KNOW
r = 1, p = 2, q = 3
FIND
p² + q² (s), x² coefficient (b), x coefficient (c), constant (d)
Step 1 · Find p² + q² (s)
Use s = p² + q²
(
p
+
qi)(
p
−
qi)
=
p
2
−
q
2
i
2
=
p
2
+
q
2
.
s
=
2
2
+
3
2
s
=
4
+
9
s
=
13
Step 2 · Find x² coefficient (b)
Use b = −2 × p − r
x
2
−
2px
+
s
times
x
−
r
:
the
x
2
terms
are
−2px²
and
−rx².
b
=
−2
×
2
−
1
b
=
−4
−
1
b
=
−5
Step 3 · Find x coefficient (c)
Use c = s + 2 × p × r
The
x
terms:
s
×
x
and
−2px
×
(−
r
).
c
=
13
+
2
×
2
×
1
c
=
13
+
4
c
=
17
Step 4 · Find constant (d)
Use d = −r × s
The
constant:
s
×
(−
r
).
d
=
−1
×
13
(x
−
1)(
x
2
−
4x
+
13)
=
x
3
−
5x²
+
17x
−
13
d
=
−13
Check
Put the numbers back in. Both sides should match.
13
=
2
2
+
3
2
✓
−5 = −2 × 2 − 1 ✓
17 = 13 + 2 × 2 × 1 ✓
−13 = −1 × 13 ✓
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Factoring polynomials: GCF, trinomials and special products
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