SME
Grade 11 Math · Problem type
Synthetic division and the remainder theorem
Use this for “Divide 2x³ − 3x² + x − 5 by x − 2” or “Find P(2) by the remainder theorem.”
Chart
Zero: r ≈ 1.9186
P(2) = 1
Quotient’s x² coefficient: 2
Quotient’s x coefficient: 1
Quotient’s constant: 3
This is an example. Type your own values to start a new problem.
(
a
x³
+
(
b
)
x²
+
c
x
+
(
d
)
)
÷
(x
−
r
)
P(
r
)
=
R
q₂
Quotient’s x² coefficient
worked out (not typed)
q₁
Quotient’s x coefficient
worked out (not typed)
q₀
Quotient’s constant
worked out (not typed)
Formulas
q₂
=
a
2 = 2
q₁
=
b
+
r
×
q₂
1 = −3 + 2 × 2
q₀
=
c
+
r
×
q₁
3 = 1 + 2 × 1
R
=
d
+
r
×
q₀
1 = −5 + 2 × 3
Assumptions
•
Divide P(x) = ax³ + bx² + cx + d by x − r: bring down a, then multiply by r and add, column by column.
•
The last sum is the remainder R, and R = P(r) (the remainder theorem).
•
R = 0 exactly when x − r is a factor (the factor theorem).
Step-by-step
WE KNOW
a = 2, b = −3, c = 1, d = −5, r = 2
FIND
quotient’s x² coefficient (q₂), quotient’s x coefficient (q₁), quotient’s constant (q₀), remainder, P(r) (R)
Step 1 · Find quotient’s x² coefficient (q₂)
Use q₂ = a
Bring down the leading coefficient.
q₂
=
2
Step 2 · Find quotient’s x coefficient (q₁)
Use q₁ = b + r × q₂
Multiply
by
r
and
add
to
the
next
coefficient.
q₁
=
−3
+
2
×
2
q₁
=
−3
+
4
q₁
=
1
Step 3 · Find quotient’s constant (q₀)
Use q₀ = c + r × q₁
Again:
multiply
by
r
and
add.
q₀
=
1
+
2
×
1
q₀
=
3
Step 4 · Find remainder, P(r): R
Use R = d + r × q₀
The
last
sum
is
the
remainder,
which
is
P(
r
).
R
=
−5
+
2
×
3
2
2
−3
1
−5
4
2
6
2
1
3
1
P(2) = 16 − 12 + 2 − 5 = 1
2x³ − 3x² + x − 5 = (x − 2)(2x² + x + 3) + 1
R
=
1
Check
Put the numbers back in. Both sides should match.
2 = 2 ✓
1 = −3 + 2 × 2 ✓
3 = 1 + 2 × 1 ✓
1 = −5 + 2 × 3 ✓
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Polynomial functions: graphs, end behavior, division and the remainder theorem
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Divide by a quadratic
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