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Siyavula textbooks: grade 12 Factorising cubic polynomials Factor theorem exercises, solving
Exercises - using factor theorem
Find the remainder when
4
x
3
-
4
x
2
+
x
-
5 is divided by
(
x
+
1
) .
Use the factor theorem to factorise
x
3
-
3
x
2
+
4 completely.
f
(
x
)
=
2
x
3
+
x
2
-
5
x
+
2
Find
f
(
1
) .
Factorise
f
(
x
) completely
Use the Factor Theorem to determine all the factors of the following expression:
x
3
+
x
2
-
17
x
+
15
Complete: If
f
(
x
) is a polynomial and
p is a number such that
f
(
p
)
=
0 , then
(
x
-
p
) is .....
Solving cubic equations
Once you know how to factorise cubic polynomials, it is also easy to solve cubic equations of the kind
a
x
3
+
b
x
2
+
c
x
+
d
=
0
Solve
6
x
3
-
5
x
2
-
17
x
+
6
=
0
.
Find one factor using the Factor Theorem
Try
f
(
1
)
=
6
(
1
)
3
-
5
(
1
)
2
-
17
(
1
)
+
6
=
6
-
5
-
17
+
6
=
-
10
Therefore
(
x
-
1
) is NOT a factor.
Try
f
(
2
)
=
6
(
2
)
3
-
5
(
2
)
2
-
17
(
2
)
+
6
=
48
-
20
-
34
+
6
=
0
Therefore
(
x
-
2
) IS a factor.
Division by inspection
6
x
3
-
5
x
2
-
17
x
+
6
=
(
x
-
2
)
(
)
The first term in the second bracket must be
6
x
2 to give
6
x
3 if one works backwards.
The last term in the second bracket must be
-
3 because
-
2
×
-
3
=
+
6 .
So we have
6
x
3
-
5
x
2
-
17
x
+
6
=
(
x
-
2
)
(
6
x
2
+
?
x
-
3
) .
Now, we must find the coefficient of the middle term (
x ).
(
-
2
)
(
6
x
2
) gives
-
12
x
2 . So, the coefficient of the
x -term must be 7.
So,
6
x
3
-
5
x
2
-
17
x
+
6
=
(
x
-
2
)
(
6
x
2
+
7
x
-
3
) .
Factorise fully
6
x
2
+
7
x
-
3 can be further factorised to
(
2
x
+
3
)
(
3
x
-
1
) ,
and we are now left with
6
x
3
-
5
x
2
-
17
x
+
6
=
(
x
-
2
)
(
2
x
+
3
)
(
3
x
-
1
)
Solve the equation
6
x
3
-
5
x
2
-
17
x
+
6
=
0
(
x
-
2
)
(
2
x
+
3
)
(
3
x
-
1
)
=
0
x
=
2
;
1
3
;
-
3
2
Sometimes it is not possible to factorise the trinomial ("second bracket"). This is when the quadratic formula
x
=
-
b
±
b
2
-
4
a
c
2
a
can be used to solve the cubic equation fully.
For example:
Solve for
x :
x
3
-
2
x
2
-
6
x
+
4
=
0
.
Find one factor using the Factor Theorem
Try
f
(
1
)
=
(
1
)
3
-
2
(
1
)
2
-
6
(
1
)
+
4
=
1
-
2
-
6
+
4
=
-
1
Therefore
(
x
-
1
) is NOT a factor.
Try
f
(
2
)
=
(
2
)
3
-
2
(
2
)
2
-
6
(
2
)
+
4
=
8
-
8
-
12
+
4
=
-
8
Therefore
(
x
-
2
) is NOT a factor.
f
(
-
2
)
=
(
-
2
)
3
-
2
(
-
2
)
2
-
6
(
-
2
)
+
4
=
-
8
-
8
+
12
+
4
=
0
Therefore
(
x
+
2
) IS a factor.
Division by inspection
x
3
-
2
x
2
-
6
x
+
4
=
(
x
+
2
)
(
)
The first term in the second bracket must be
x
2 to give
x
3 .
The last term in the second bracket must be 2 because
2
×
2
=
+
4 .
So we have
x
3
-
2
x
2
-
6
x
+
4
=
(
x
+
2
)
(
x
2
+
?
x
+
2
) .
Now, we must find the coefficient of the middle term (
x ).
(
2
)
(
x
2
) gives
2
x
2 . So, the coefficient of the
x -term must be
-
4 . (
2
x
2
-
4
x
2
=
-
2
x
2 )
So
x
3
-
2
x
2
-
6
x
+
4
=
(
x
+
2
)
(
x
2
-
4
x
+
2
) .
x
2
-
4
x
+
2 cannot be factorised any futher and we are now left with
(
x
+
2
)
(
x
2
-
4
x
+
2
)
=
0
Solve the equation
(
x
+
2
)
(
x
2
-
4
x
+
2
)
=
0
(
x
+
2
)
=
0
o
r
(
x
2
-
4
x
+
2
)
=
0
Apply the quadratic formula for the second bracket
Always write down the formula first and then substitute the values of
a
,
b and
c .
x
=
-
b
±
b
2
-
4
a
c
2
a
=
-
(
-
4
)
±
(
-
4
)
2
-
4
(
1
)
(
2
)
2
(
1
)
=
4
±
8
2
=
2
±
2
Final solutions
x
=
-
2
o
r
x
=
2
±
2
Exercises - solving of cubic equations
Solve for
x :
x
3
+
x
2
-
5
x
+
3
=
0
Solve for
y :
y
3
-
3
y
2
-
16
y
-
12
=
0
Solve for
m :
m
3
-
m
2
-
4
m
-
4
=
0
Solve for
x :
x
3
-
x
2
=
3
(
3
x
+
2
)
:
Remove brackets and write as an equation equal to zero.
Solve for
x if
2
x
3
-
3
x
2
-
8
x
=
3
End of chapter exercises
Solve for
x :
16
(
x
+
1
)
=
x
2
(
x
+
1
)
Show that
x
-
2 is a factor of
3
x
3
-
11
x
2
+
12
x
-
4
Hence, by factorising completely, solve the equation
3
x
3
-
11
x
2
+
12
x
-
4
=
0
2
x
3
-
x
2
-
2
x
+
2
=
Q
(
x
)
.
(
2
x
-
1
)
+
R for all values of
x . What is the value of
R ?
Use the factor theorem to solve the following equation for
m :
8
m
3
+
7
m
2
-
17
m
+
2
=
0
Hence, or otherwise, solve for
x :
2
3
x
+
3
+
7
·
2
2
x
+
2
=
17
·
2
x
A challenge :
Determine the values of
p
for which the function
f
(
x
)
=
3
p
3
-
(
3
p
-
7
)
x
2
+
5
x
-
3
leaves a remainder of 9 when it is divided by
(
x
-
p
) .
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Source:
OpenStax, Siyavula textbooks: grade 12 maths. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11242/1.2
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