Grade 11 Math · Problem type

Sum and difference of cubes

Use this for “Factor 27x³ + 8” or “Factor x³ − 125.”

Chart

f(x) = 27x³ + 8−3−2−10123−2.5−2−1.5−1−0.500.511.522.5xy(−2/3, 0)
Zero: x = −2/3 ≈ −0.6667
This is an example. Type your own values to start a new problem.
A
x³
+
B
=
(
a
x
+
b
)(
m
x²
+
(
n
)
x
+
k
)
z
Real zero, −b ÷ a
worked out (not typed)

Formulas

A
=
a
3
27
=
3
3
B
=
b
3
8
=
2
3
m
=
a
2
9
=
3
2
n
=
−1
×
a
×
b
−6 = −1 × 3 × 2
k
=
b
2
4
=
2
2
z
=
−1
×
b
a
−
2
3
=
−1
×
2
3

Assumptions

•
A sum of cubes factors as a³x³ + b³ = (ax + b)(a²x² − abx + b²).
•
A difference of cubes is the same with a negative b: x³ − 125 = (x − 5)(x² + 5x + 25).
•
The quadratic factor has no real zeros, so the only real zero is x = −b ÷ a.

Step-by-step

WE KNOW
A = 27, B = 8
FIND
cube root of A (a), cube root of B (b), x² coefficient of the second factor, a² (m), x coefficient of the second factor, −ab (n), constant of the second factor, b² (k), real zero, −b ÷ a (z)
Step 1 · Find cube root of A (a)
Use A = a³
The
number
whose
cube
is
A
.
a
=
∛(
A
)
a
=
∛(27)
a
=
3
Step 2 · Find cube root of B (b)
Use B = b³
The
number
whose
cube
is
B
;
a
negative
B
has
a
negative
b
.
b
=
∛(
B
)
b
=
∛(8)
b
=
2
Step 3 · Find x² coefficient of the second factor, a² (m)
Use m = a²
The
first
term
of
the
second
factor:
(ax)
2
.
m
=
3
2
m
=
9
Step 4 · Find x coefficient of the second factor, −ab (n)
Use n = −1 × a × b
The
middle
term
has
the
opposite
sign
of
b
:
−abx.
n
=
−1
×
3
×
2
n
=
−3
×
2
n
=
−6
Step 5 · Find constant of the second factor, b² (k)
Use k = b²
The
last
term
of
the
second
factor
is
b
2
,
always
positive.
k
=
2
2
27x³ + 8 = (3x + 2)(9x² − 6x + 4)
k
=
4
Step 6 · Find real zero, −b ÷ a (z)
Use z = −1 × b ÷ a
ax
+
b
=
0
here:
the
graph
crosses
the
x-axis
once.
z
=
−1
×
2
3
z
=
−
2
3
z
=
−
2
3
Check
Put the numbers back in. Both sides should match.
27
=
3
3
✓
8
=
2
3
✓
9
=
3
2
✓
−6 = −1 × 3 × 2 ✓
4
=
2
2
✓
−
2
3
=
−1
×
2
3
✓
Topic check
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