SME
Grade 12 Math · Problem type
A quadratic factor
Use this for “Write (3x² − 2x + 3) ÷ ((x − 1)(x² + 1)) as partial fractions.”
Chart
Vertical asymptote x = 1
Horizontal asymptote y = 0
f(2) = 2.2
This is an example. Type your own values to start a new problem.
x²
+
(
)
x
+
(x
−
)(x²
+
x
+
)
=
A
x
−
+
x
+
(
)
x²
+
x
+
A coefficient of 1 is not written: 1x is written x.
x
An x to check
example
y
The fraction’s value at x
worked out (not typed)
Formulas
A
=
a
×
p
2
+
b
×
p
+
c
p
2
+
j
×
p
+
k
2
=
3 ×
1
2
+ (−2) × 1 + 3
1
2
+ 0 × 1 + 1
B
=
a
−
A
1 = 3 − 2
C
=
b
+
B
×
p
−
A
×
j
−1 = −2 + 1 × 1 − 2 × 0
y
=
a
×
x
2
+
b
×
x
+
c
(
x
−
p
) × (
x
2
+
j
×
x
+
k
)
2.2
=
3 ×
2
2
+ (−2) × 2 + 3
(2 − 1) × (
2
2
+ 0 × 2 + 1)
Assumptions
•
A factor x² + jx + k that doesn’t split (j² < 4k) gets a linear top: A ÷ (x − p) + (Bx + C) ÷ (x² + jx + k).
•
Cover up (x − p) and put x = p for A; then match the x² and x terms for B and C.
•
The number terms must match too: Ak − Cp = c is the check.
Step-by-step
WE KNOW
a = 3, b = −2, c = 3, p = 1, j = 0, k = 1, x = 2
FIND
top over (x − p) (A), number before x over the quadratic (B), number alone over the quadratic (C), the fraction’s value at x (y)
Step 1 · Find top over (x − p): A
Use A = (a × p² + b × p + c) ÷ (p² + j × p + k)
Cover
up
(
x
−
p
)
and
put
x
=
p
:
the
top
over
what
is
left
of
the
bottom.
A
=
3 ×
1
2
− 2 × 1 + 3
1
2
+ 0 × 1 + 1
A
=
3 − 2 + 3
1 + 1
A
=
1 + 3
2
A
=
4
2
A
=
2
Step 2 · Find number before x over the quadratic (B)
Use B = a − A
The
x
2
terms:
A
(
x
2
+
jx
+
k
)
+
(Bx
+
C
)(
x
−
p
)
has
(
A
+
B
)
x
2
,
so
A
+
B
=
a
.
B
=
3
−
2
B
=
1
Step 3 · Find number alone over the quadratic (C)
Use C = b + B × p − A × j
The
x
terms:
Ajx
from
the
first
fraction
and
(
C
−
Bp)
x
from
the
second,
so
Aj
+
C
−
Bp
=
b
.
C
=
−2
+
1
×
1
−
2
×
0
C
=
−2
+
1
C
=
−1
→
check:
2
×
1
−
(−1)
×
1
=
3
Step 4 · Find the fraction’s value at x (y)
Use y = (a × x² + b × x + c) ÷ ((x − p) × (x² + j × x + k))
The
fraction
at
one
x
;
the
two
partial
fractions
add
to
the
same
number
there.
y
=
3 ×
2
2
− 2 × 2 + 3
(2 − 1) × (
2
2
+ 0 × 2 + 1)
Top: 3 − 2 × 2 + 3 × 4 = 11
Bottom: 1 × 5 = 5
y
=
11
5
y
=
2.2
→
A
x
−
p
+
Bx
+
C
x
2
+
jx
+
k
=
2
1
+
1 × 2 + (−1)
5
=
2.2
Check
Put the numbers back in. Both sides should match.
2
=
3 ×
1
2
− 2 × 1 + 3
1
2
+ 0 × 1 + 1
✓
1 = 3 − 2 ✓
−1 = −2 + 1 × 1 − 2 × 0 ✓
2.2
=
3 ×
2
2
− 2 × 2 + 3
(2 − 1) × (
2
2
+ 0 × 2 + 1)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.