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4
3
2
1
Outside the high school program
/ Algebra
32 exercises (100% corrected)
a
Area of resolution, definition
(5 exercices)
Handling frames, inequalities
(1 exercice)
Development
(1 exercice)
Factoring
(4 exercices)
Factoring by a common factor
(6 exercices)
Factorisation by a common factor and equation
(7 exercices)
Factorization: remarkable identity
(2 exercices)
Factoring by common factor and remarkable identity
(3 exercices)
Equations
(1 exercice)
Square roots
(1 exercice)
Euclidean division
(1 exercice)
image PNG : /home/_math/_exercice/d1/1343/metapost/exo_faux.png
(1
x
2
1
x
(1
x
2
1
x
1
x
1
x
1
x
1
x
0
5(
x
2
x
3(
x
2
5
x
10
2
x
3
x
15
2
7
x
10
3
x
17
7
x
3
x
17
−
10
7
x
3
x
7
7
x
−
3
x
7
4
x
7
x
7
4
1.
A
r
e
a
of
r
esolution,
definition
E.1343
Here
is
the
w
ork
of
a
studen
t
trying
to
solv
e
the
equation
:
(1+
x
)
2
=
1+
x
1
a
V
erify
that
the
v
alue
−
1
is
a
solution
of
this
equa-
tion.
b
What
can
b
e
said
ab
out
the
studen
t’s
prop
osed
solu-
tion?
2
The
diagram
b
elo
w
represen
ts
the
differen
t
steps
p
er-
formed
b
y
this
studen
t.
Complete
the
ˇ
lab
els
ı
b
y
in-
dicating
the
action
p
erformed
b
y
the
studen
t
:
3
a
F
or
what
v
alues
of
x
are
defined
the
t
w
o
expressions
:
(1
+
x
)
2
;
1
+
x
b
F
or
what
v
alues
of
x
are
defined
the
t
w
o
expressions
:
(1
+
x
)
2
1
+
x
;
1
+
x
1
+
x
c
Explain
where
the
error
made
b
y
the
studen
t
comes
from.
E.4389
W
e
wish
to
solv
e
the
equation
:
(
E
)
:
(
x
−
2)
x
x
+1
=
2
x
2
+1
x
+1
1
What
is
the
solving
range
of
this
equation?
2
F
or
x
=
−
1
,
establish
the
follo
wing
equalit
y
:
2
x
2
+
1
x
+
1
−
(
x
−
2)
x
x
+
1
=
x
+
1
3
What
can
b
e
said
ab
out
the
set
of
solutions
of
(
E
)
?
E.4402
The
diagram
b
elo
w
sho
ws
the
steps
in-
v
olv
ed
in
solving
an
algebraic
equation.
Briefly
describ
e
eac
h
of
these
steps,
indicating
in
the
la-
b
els
the
mathematical
action
p
erformed
(factoring,
exp
an-
sion,
subtr
action.
.
.
)
.
E.4544
Consider
the
function
f
defined
b
y
the
relation
:
f
(
x
)
=
−
2
x
+
3
1
Justify
that
the
defining
set
of
the
function
f
is
R
+
.
2
Establish
that
the
function
f
is
strictly
increasing
on
R
+
.
E.454
Solv
e
the
follo
wing
equation
:
2
x
−
2
x
−
1
=
3
x
+3
2
x
+1
2.
Development
E.6994
Giv
e
the
reduced
dev
elop
ed
form
of
the
follo
wing
expressions
:
a
2
x
+
1
x
+
2
b
x
−
1
(2
x
+
2
c
−
2
−
x
3
x
+
1
d
2
x
+
3
2
3.
F
actoring
E.4424
F
actor
the
follo
wing
expressions
:
a
(3
x
−
1)(2
x
+
1)
+
(5
−
x
)(2
x
+
1)
b
x
(2
−
x
)
+
(3
x
+
1)(2
−
x
)
c
(
x
+
1)(
x
−
1)
−
(2
x
+
3)(
x
−
1)
h
ttps:/
/c
hingmath.fr
chapExoCorrec/1343
sacados/1343
(1
x
2
1
x
(1
x
2
1
x
1
x
1
x
1
x
1
x
0
chapExoCorrec/4389
sacados/4389
chapExoCorrec/4402
sacados/4402
5(
x
2
x
3(
x
2
5
x
10
2
x
3
x
15
2
7
x
10
3
x
17
7
x
3
x
17
−
10
7
x
3
x
7
7
x
−
3
x
7
4
x
7
x
7
4
chapExoCorrec/4544
sacados/4544
chapExoCorrec/454
sacados/454
chapExoCorrec/6994
sacados/6994
chapExoCorrec/4424
sacados/4424
E.4403
Solv
e
the
follo
wing
equations
:
a
(4
x
+
3)(2
−
3
x
)
+
(2
−
6
x
)(2
−
3
x
)
=
0
b
2
·
(
x
−
2)(3
x
+
2)
=
3(2
x
+
3)(
x
−
2)
c
(
x
+
1)(3
x
−
2)
=
(
x
+
1)
2
E.9423
F
actor
the
follo
wing
expressions
:
a
(3
x
+
4)(2
x
−
1)
+
4(3
x
+
4)
b
(2
x
+
4)(3
−
3
x
)
+
(2
x
+
4)
c
(
x
+
1)(3
−
2
x
)
+
(3
−
2
x
)
2
E.9426
Consider
equation
:
(
F
)
:
3
−
3
x
2
x
+
2
+
3
x
−
2
2
x
+
3
=
0
1
Establish
the
follo
wing
equalit
y
:
3
−
3
x
2
x
+
2
+
3
x
−
2
2
x
+
3
=
5
−
x
(2
x
+
2)(2
x
+
3)
2
Solv
e
equation
(
F
)
.
4.
F
actoring
by
a
c
ommon
factor
E.9420
F
actor
the
follo
wing
expressions
:
a
(
x
+
2)(
x
−
3)
+
(3
x
+
6)(2
x
−
1)
b
(3
x
+
6)(
x
−
4)
−
(2
x
−
8)(
x
−
2)
c
(2
x
+
7)(27
x
+
18)
−
(5
−
x
)(3
x
+
2)
E.9422
F
actor
the
follo
wing
expressions
:
a
(3
x
+
2)(5
−
x
)
+
(6
x
+
4)
b
(5
−
2
x
)(
x
+
1)
−
3
x
(2
x
−
5)
c
(5
−
2
x
)
+
(4
x
−
10)(7
−
2
x
)
E.9418
F
actor
the
follo
wing
expressions
:
a
(3
x
+
2)(4
x
+
2)
−
(5
x
+
1)(2
x
+
1)
b
(2
x
−
1)(9
x
+
3)
+
(3
x
+
1)
2
E.4613
Reminders
:
The
opp
osite
of
the
expression
3
x
−
4
can
b
e
expressed
with
the
follo
wing
t
w
o
forms
:
−
3
x
+4
;
4
−
3
x
The
opp
osite
of
the
pro
duct
a
×
b
can
b
e
expressed
with
the
t
w
o
forms
:
(
−
a
)
×
b
;
a
×
(
−
b
)
A
tten
tion,
the
pro
duct
(
−
a
)
×
(
−
b
)
is
égal
to
the
pro
duct
a
×
b
.
Applications
:
In
the
expression
(3
x
−
4)(5
x
+1)+(4
−
3
x
)(2
−
2
x
)
,
the
common
factor
is
3
x
−
4
.
W
e
ha
v
e
the
follo
wing
transformations
:
(2
−
x
)(
x
−
4)
=
−
(
x
−
2)
(
x
−
4)
=
(
x
−
2)(4
−
x
)
F
actor
the
follo
wing
expressions
:
a
(3
x
−
4)(5
x
+
1)
+
(4
−
3
x
)(2
−
2
x
)
b
(
x
−
2)(2
x
+
3)
+
(2
−
x
)(
x
−
4)
E.6995
Reminder
:
when
the
co
efficien
ts
of
a
p
olynomial
are
m
ul-
tiples
of
the
same
n
um
b
er,
a
factorization
is
p
ossible.
Example
:
2
x
+
2
=
2
·
x
+
1
6
x
−
3
=
3
·
2
x
−
1
F
actor
the
follo
wing
expressions
:
a
x
+
1
2
x
−
1
+
2
x
+
2
3
x
+
2
b
6
x
−
3
x
+
1
+
2
x
−
1
x
+
2
c
9
x
−
3
(3
x
+
1)
−
x
−
2
3
x
−
1
d
2
x
+
2
2
+
x
+
1
E.7219
F
actor
the
follo
wing
expressions
:
a
x
+
3
2
x
−
1
−
3
x
+
1
x
+
3
b
6
−
3
x
x
+
2
−
2
−
x
2
5.
F
actorisation
by
a
c
ommon
factor
and
e
quation
E.9414
Solv
e
the
follo
wing
equations
:
a
(4
x
+
2)(3
x
−
1)
+
x
(6
x
+
3)
=
0
b
(3
x
+
9)
2
=
2(2
x
+
6)(3
x
−
2)
h
ttps:/
/c
hingmath.fr
chapExoCorrec/4403
sacados/4403
chapExoCorrec/9423
sacados/9423
chapExoCorrec/9426
sacados/9426
chapExoCorrec/9420
sacados/9420
chapExoCorrec/9422
sacados/9422
chapExoCorrec/9418
sacados/9418
chapExoCorrec/4613
sacados/4613
chapExoCorrec/6995
sacados/6995
chapExoCorrec/7219
sacados/7219
chapExoCorrec/9414
sacados/9414
E.2858
Solv
e
the
follo
wing
equations
:
a
(3
x
+
1)(1
−
3
x
)
+
(6
x
+
2)(3
x
−
1)
=
0
E.2113
Solv
e
the
follo
wing
equations
using
the
metho
d
of
y
our
c
hoice
:
a
(2
x
−
3)(4
x
−
2)
+
(3
−
6
x
)(
x
−
3)
=
0
b
(2
−
6
x
)(3
x
+
3)
=
(
−
9
x
+
3)(2
x
+
1)
E.9416
Solv
e
the
follo
wing
equations
:
a
(5
x
+
2)(4
x
−
3)
=
(2
x
−
1)(3
−
4
x
)
b
(2
−
3
x
)(2
x
−
4)
+
(10
−
5
x
)(3
x
−
1)
=
0
E.9419
Solv
e
the
follo
wing
equation
:
(5
x
+
1)(3
x
−
1)
−
2(
x
+
1)(1
−
3
x
)
=
0
E.4566
Solv
e
the
follo
wing
equations
:
a
(
x
+
2)(3
−
x
)
+
2(
x
−
3)(2
x
−
5)
=
0
b
(6
−
2
x
)(3
x
+
2)
=
(3
x
−
9)(
x
+
2)
c
(9
−
3
x
)(5
+
2
x
)
+
(
x
−
3)(5
+
x
)
=
0
E.4425
F
actor
the
follo
wing
expressions
:
a
(
x
−
1)(2
x
+
1)
−
(2
x
−
2)(5
−
2
x
)
b
3(4
+
2
x
)
−
(3
+
x
)(10
+
5
x
)
c
(2
−
x
)(3
x
−
4)
+
2
−
3
2
x
(2
x
+
3)
6.
F
actorization
:
r
emarkable
identity
E.9421
F
actor
the
follo
wing
expressions
:
a
(2
x
+
1)
2
−
4(2
−
3
x
)
2
b
18
x
2
−
24
x
+
8
+
(3
x
−
2)(2
−
x
)
E.4528
F
actor
the
follo
wing
expressions
:
a
−
5(3
x
+
2)
−
5
x
(2
x
+
1)
7.
F
actoring
by
c
ommon
factor
and
r
emarkable
identity
E.4551
F
actor
the
follo
wing
expressions
:
a
(3
x
+
2)(5
x
−
1)
−
(2
−
10
x
)(2
−
4
x
)
b
(
x
+
3)(3
x
+
6)
+
(4
x
+
8)
2
c
(2
x
+
3)(5
x
−
4)
−
(2
x
−
2)(6
−
x
)
d
(
x
−
3)(
−
3
x
−
3)
−
(2
x
+
2)(3
−
3
x
)
E.9425
Solv
e
the
follo
wing
equations
:
a
(1
−
5
x
)(2
x
+
1)
=
(1
−
3
x
)(3
x
+
2)
b
(6
x
+
1)(3
x
+
1)
+
(2
x
+
1)(2
−
9
x
)
=
0
c
(
−
5
x
−
4)(2
x
+
1)
=
(
−
4
x
−
3)(3
x
+
2)
E.4547
Solv
e
the
follo
wing
equations
:
a
(4
x
+
2)(3
x
−
1)
+
x
(6
x
+
3)
=
0
b
(
x
+
3)
2
=
2(2
x
+
6)(3
x
−
2)
c
(
−
5
x
−
4)(2
x
+
1)
=
(
−
4
x
−
3)(3
x
+
2)
d
(1
−
5
x
)(2
x
+
1)
=
(1
−
3
x
)(3
x
+
2)
8.
Equations
E.1940
Consider
t
w
o
real
n
um
b
ers
x
and
y
v
er-
ifying
the
follo
wing
frames
:
−
4
x
−
1
;
2
y
5
1
Giv
e
the
frames
of
the
follo
wing
n
um
b
ers
:
a
3
×
(1
−
x
)
+
1
b
1
−
3
−
x
4
2
Giv
e
the
frames
of
the
follo
wing
n
um
b
ers
:
a
x
2
b
x
×
y
c
2
x
−
3
y
9.
Squar
e
r
o
ots
h
ttps:/
/c
hingmath.fr
chapExoCorrec/2858
sacados/2858
chapExoCorrec/2113
sacados/2113
chapExoCorrec/9416
sacados/9416
chapExoCorrec/9419
sacados/9419
chapExoCorrec/4566
sacados/4566
chapExoCorrec/4425
sacados/4425
chapExoCorrec/9421
sacados/9421
chapExoCorrec/4528
sacados/4528
chapExoCorrec/4551
sacados/4551
chapExoCorrec/9425
sacados/9425
chapExoCorrec/4547
sacados/4547
chapExoCorrec/1940
sacados/1940
E.296
1
Let
a
and
b
b
e
t
w
o
n
um
b
ers
suc
h
that
a
0
and
b
0
.
Compare
n
um
b
ers
:
a
+
b
;
a
+
b
2
What
conditions
on
a
and
b
m
ust
there
b
e
for
the
inequal-
it
y
to
b
e
strict?
10.
Euclide
an
division
E.2299
Here
are
t
w
o
Euclidean
divisions
of
p
oly-
nomials
:
2
x
3
+
2
x
2
−
16
x
−
12
2
x
3
+
6
x
2
−
4
x
2
−
16
x
−
4
x
2
−
12
x
−
4
X
−
12
−
4
X
−
12
0
x
+
3
2
x
2
−
4
x
−
4
3
x
3
−
5
x
2
+
x
−
7
3
x
3
−
6
x
2
x
2
+
x
x
2
−
2
x
3
X
−
7
3
X
−
6
−
1
x
−
2
3
x
2
+
x
+
3
They
pro
duce
the
follo
wing
m
ultiplicativ
e
en
tries
:
2
x
3
+
2
x
2
−
16
x
−
12
=
(
x
+
3)(2
x
2
−
4
x
−
4)
This
equalit
y
yields
the
three
ro
ots
of
this
second-degree
p
olynomial.
3
x
3
−
5
x
2
+
x
−
7
=
(
x
−
2)(3
x
2
+
x
+
3)
−
1
This
equalit
y
yields
the
follo
wing
iden
tit
y
called
ˇ
simple
elemen
t
decomp
osition
of
a
rational
fraction
ı
:
3
x
3
−
5
x
2
+
x
−
7
x
−
2
=
3
x
2
+
x
+
3
−
1
x
−
2
Using
Euclidean
division
of
p
olynomials,
answ
er
the
follo
wing
questions
:
1
Noting
that
−
3
is
a
ro
ot
of
x
3
+
3
x
2
−
2
x
+
3
,
determine
that
this
p
olynomial
admits
no
other
ro
ots.
2
Sho
w
that
:
2
x
3
−
5
x
2
+
x
−
4
2
x
+
1
=
x
2
−
3
x
+
2
−
−
6
2
x
+
1
11.
Unclassifie
d
financial
ye
ars
E.7332
Consider
the
t
w
o
functions
f
and
g
de-
fined
on
R
+
b
y
the
relations
:
f
(
x
)
=
2
·
x
−
1
·
x
;
g
(
x
)
=
3
·
x
2
−
9
·
x
+
6
Using
the
calculator,
conjecture
the
relativ
e
p
osition
on
R
+
of
the
curv
es
C
f
and
C
g
represen
tativ
e
of
the
functions
f
and
g
resp
ectiv
ely
.
Calculator
results
rounded
to
the
nearest
h
undredth
will
b
e
used.
h
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/c
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chapExoCorrec/296
sacados/296
chapExoCorrec/2299
sacados/2299
chapExoCorrec/7332
sacados/7332
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