__________
1.
Mathematics: Algebra 1
Solution:
18 = 2 x 3 x 3
12 = 2 x 2 x 3
27 = 3 x 3 x 3
What is the least common multiple of
18x and 24xy?
A. 27xy2
B. 144x2y
logx 2 = log3 x
C. 72xy *
D. 36xy
shift solve (x ? → 1)
x1 = 0.418
logx 2 = log3 x
shift solve (x ? → 999)
x 2 = 2.3932
LCM = 22 x 33 = 108
Answer: 108
x1 + x 2 = 0.418 + 2.3932 = 2.81
Solution:
18x = ( 2 )( 3 ) x
Another solution is by reverse. The choice must be
divisible by the given and has the least value.
108
108
108
=6 ;
=9;
=4
18
12
27
2
24xy = 23 ( 3 ) xy
( )
LCM = 23 32 xy = 72xy
2.
Suppose a number m is divisible by 6
and by 8. The number m must be
divisible by
A. 3
B. 4
7.
The number 10.097 has how many
significant digits?
A.
B.
C. 24
D. All of the above. *
Solution:
6 = 2 (3)
8.
8 = 23
1
3
A.
B.
C. 2
D. 5 *
3*
7
C. 5
D. 4
1. Solve for x if logx2 = log2 x .
A wife works three days then a day off
C. 1, 10
while his husband works five days then A. 1, 100 *
B. 10, 100
D. 2, 10
a day off. If the couple has a day-off
together today, how many days after will
Solution:
they have another day off together?
A. 15 days
B. 12 days *
C. 14 days
D. 11 days
Solution:
( ) − (log ( x ))
log x
2
CALC → substitute choices
A. 1, 100 → satisfy the equation
Solution:
Input in calcu:
3360 = 2  3  5  7
logC ( A ) xlogA (B ) xlogB ( C )
5
3960 = 23  32  5  11
GCD = 2  3  5 = 120
Or by reverse. The GCD is the choice with
greatest value and can divide evenly to
two given values.
x.
What is the greatest common factor of
40x3y5z7, 12x2y3z3, and 36x4yz6?
C. 4x2y2z
D. 4x2yz3 *
=
loga logb logc
logc
x =1
Or assign value for a, b and c.
3. If logx 2 = log3 x is satisfied by two values
of x, what is their sum?
1.73
2.35
C. 2.81 *
D. 3.14
Solution:
B. x = 5ay
D. x = ya5y *
Solution:
loga x = 5y + loga y
loga x − loga y = 5y
loga ( x / y ) = 5y
1
, then log10 10m2 =
2
A. 2 *
B. 2.5
1
2
SHIFT → SOVLE → =
X = 3.1622
Substitute :
log10 X =
(
*
C. 3
D. 10.5
Solution:
log10 10 ( 3.1622 )
Find LCM of 18, 12 and 27.
C. 108
D. 216
C. x = 5a y
7. If log10 m =
loga logb
A.
B.
A. x = a5y
x = ya5y
or LONG METHOD:
x = logc a loga b logb c
Answer: D. 4x2yz3
It’s obvious from the choices that 4 is
the GCF of 40, 12 and 36.
Now the variables, to know the GCF of
each variable, consider the variable with
the lowest power from the three given
expression (x2, y and z3)
A. 324
B. 240
6. Given: loga x = 5y + loga y. Find the value of
Assign value: A = 2, B = 3, C = 4
Re sult is 1.
If values are susbtituted to choices,
only A will satisfy.
3
6.
Solution:
mn = 2 x + y
Solution:
Get the prime factors of the numbers:
4x2yz2
4xyz
C. x + y
D. x - y
log2 mn = x + y
They will have same day-off every 12th day. 2. Find the value of x if
x = logc a loga b logb c
4. What is the GCD of 3360 and 3960?
A. 1* *
C. ab / c
A. 60
C. 110880
B. 120 *
D. 110800
B. bc / a
D. ca / b
A.
B.
A. 2 x + y
B. 2 x − y
log2 m + log2 n = x + y
LCM of 4 and 6 is 12.
5.
5. If log2 m = x and log2 n = y , then mn =
Substitute the choices. The value must be zero.
2
C. 7
D. 9
Substitute eqn2 to eqn1:
logy – 0.0579919 + logy = 1.748188
2logy = 1.8061799
logy = 0.90308995
applying anti-logarithm:
10log y = 100.90308995
y=8
Thus, m = 24 or its multiples.
Therefore, m is divisible by 3 or by 4
3.
8*
6
Answer: A. 8
Let x and y be the two numbers.
logx + logy = 1.748188 → eqn1
logx – logy = -0.0579919
→ logx = logy – 0.0579919 → eqn2
The number 7.93 x 10-2 has how many
significant digits?
A.
B.
LCM = 23 ( 3 ) = 24
4. The sum of the logarithm of two numbers
is 1.748188 and the difference of their
logarithms is – 0.0579919. What is one of
the numbers?
2
)2
8. Given the equation (ln x)2 – 2 lnx3 = - 5.
Solve for x.
GIVE YOUR BEST SHOT !
__________
Mathematics: Algebra 1
x = e and x = e3
x = e and x = e4
x = e and x = e5 *
x = e and x = e6
A.
B.
C.
D.
C.
3logb
x=
loga − 2logc − logb
D.
logb
x=
loga − 2logc − logb
Solution:
a
(
)
A.
B.
a
B.
14. Solve for x: x = log0.5 0 .
1
(y − x) *
2
1
D. x − y
2
1
xy
2
1
(x + y)
2
C.
A. 0
B. undefined
infinity
C. infinity *
D.negative
Solution:
let a = 10
log10 5 = 0.699
15. Solve for x: x = log ( 0 ) .
C. 0
D. undefined
infinity *
log10 7 = 0.845
log10 1.4 = 0.0731
substitute x and y to the choices :
1
1
C. ( y − x ) = (0.845 − 0.699) = 0.0731
2
2
10. Solve for x:
2log ( 3 − x ) = log2 + log ( 22 − 2x ).
D. 5x2 + 3x - 2 =
– 1 only
25 only
C. – 1 and 25 *
D. 1 and – 25
Solution:
From sum of roots formula:
 −k 
x1 + x 2 = −  
 2 
k
x1 + x 2 = → Eq.1
2
From the given that the difference
between roots is 5/2.
5
x1 − x 2 = → Eq.2
2
From product of roots formula:
3k
( x1 )( x2 ) = → Eq.3
2
value of loga 1.4 ?
A.
C. 2x2 – 5x - 12
22. Find the values of the constant k in the
equation 2x2 – kx + 3k = 0 if the
difference of the roots is 5/2.
Solution:
a x c −2 x = b3 x +1
Substitute the choices. The value must be zero.
a x c −2 xb−3x = b
2
ln ( x ) − 2ln x3 + 5
log a x c −2 xb−3x = logb
CALC → substitute choices
loga x + logc −2 x + logb−3x = logb
B. x = e, x = e5 → satisfy the equation
xloga − 2xlogc − 3xlogb = logb
logb
x=
9. If log 5 = x , log 7 = y , then what is the loga − 2 logc − 3logb
( )
A. 3x2 – 2x + 5 = 0
=0
B. 7x2 + 3x + 12 = 0
0*
C. infinity
D.
negative
Manipulating the 3 equations:
k = - 1, k = 25
23. Find the value of k if, in the equation 2x2
– kx + 4x + 5k = 0, one root is the
reciprocal of the other.
16. Find the value of k in the quadratic
equation 3x2 − kx + x − 7k = 0 if 3 is one
of the roots.
A. 3 * *
C. 6
B. 18
D. 12
A.
B.
2
1/2
C. 3
D. 2/5 *
Solution:
If one root is the reciprocal of the other,
A. -5 only * B. 7
5 and - 7
C. -5 and 7
D.
If 3 is a root, then it will satisfy the equation:
then the product of the roots must be 1.
3 ( 3 ) − k ( 3 ) + 3 − 7k = 0
2
C
=1
A
5k
=1
2
2
k=
5
k=3
11. If log4 7 = n , find log4
A.
B.
1
.
7
17. The sum of the zeros of y = 3x2 – 6x – 4
is
C. – n *
D. n
1/n
n2
A. 6
B. 2 *
C. 4/3
D. – 2
Solution:
log4 7 = n
1
log4 = log4 7−1
7
− log4 7 = −n
r1 + r2 = −
B
−6
=−
=2
A
3
24. If the equation x2 + 2(k + 2)x + 9k = 0 has
equal roots, find k.
A.
B.
1
3
C. 1 or 4 *
D. 1 or 3
18. x2 + 2x + 3 = 0 has
(
)
12. Solve the equation log3 x2 − 8x = 2 .
A.
B.
–1
9
C. 9 and – 1 *
D. 9 and 1
Solution:
log3 ( x 2 - 8x ) = 2
x 2 − 8x = 9
(x − 9)(x + 1) = 0
x = 9 l x = −1
13. Find x from the equation axc −2x = b3x +1 .
A.
x=
logb
*
loga − 2logc − 3logb
B.
x=
logb
loga − logc − 3logb
A.
B.
C.
D.
Solution:
two real rational roots
two real irrational roots
two equal real roots
two complex conjugate roots *
If the roots are equal, the discriminant
must be zero:
D = ( 2k + 4)2 − 4(9k) = 0
4k 2 + 16k + 16 − 36k = 0
19. Given the equation 2x2 + x -10 = 0. Find
the product of roots.
A. 5
B. -5 *
D. ½
C. -1/2
20. What is the discriminant of the equation
6x2 + 7x -13 = 0?
A. 412
B. 305
D. 361 *
4k 2 − 20k + 16 = 0
k = 1, k = 4
C. 388
25. If the product of the roots of 4x2 + (k + 5)x
– (k+13) = 0 is 15/4, find the one of the
roots.
A. 3/2
1.
21. Which of the following equations has
roots which are reciprocals of the roots of
the equation 2x2 – 3x – 5 = 0?
A. 24
GIVE YOUR BEST SHOT !
B. ¾ *
B. 30
C. 2/3
D.5/3
The roots of the quadratic
equation Ax2 + Bx + C = 0
are ¼ and -7/2. Determine
the value of B.
C. 7
D. 26 *
__________
2. Students A and B are requested to solve
the same quadratic equation. Student A
misread the linear term and gets the roots
3 and 8. Student B misreads the constant
term and gets the roots 1 and 5. Find the
equation.
Student A:
Sum of roots: 3+8  11
Product roots:3 ( 8 )  24
Equation read by A: x 2 − 11x + 24 = 0
but − 11x is a wrong term
Student B:
Sum of roots: 1+5  6
Product roots:1( 5 )  5
Equation read by B: x 2 − 6x + 5 = 0
but 5 is wrong.
 x 2 − 6x + 24 = 0 → ans
26. Find the equation whose roots are the
negatives of the roots of x2 + 7x – 2 = 0.
A.
B.
C.
D.
x2 + x – 2 = 0
x2 + 7x – 5 = 0
x2 – 7x – 2 = 0 *
x2 + 7x – 2 = 0
Solution:
Solving for the roots of the given
equation by Q.F., we get
x=
x=
−b  b2 − 4ac
2a
−7  7 2 − 4 (1)( −2 )
2(1)
−7 + 57
−7 − 57
x=
l x=
2
2
The new roots are:
7 − 57
7 + 57
x=
l x=
2
2
Thus, the required equation is:
x 2 − ( x1 + x 2 ) x + x1x 2 = 0
Mathematics: Algebra 1
Because the two are equal, therefore 5 is
another value of ‘x’.
28. Solve for x: 2 – 5ex = - 17
A.
B.
21
5
13
ln
5
19
5
Use calculator to solve this problem:
2 − 5e x = −17 → Shift → Solve → =
19
→ 1.335 or ln
5
x = 11 only
C. x =
−1  57
2
B. x = - 5, 3 *
D. x =
57
−1 
2
Answer: B. x = -5, 3
Use your calculator to solve this problem:
x +1 1
13
+ =
→ Shift → Solve →
x − 2 x x 2 − 2x
= →x=3
One value of ‘x’ is 3. To check if 5 is
another value of ‘x’:
x +1 1
13
+ :
→ CALC → x = −5
x − 2 x x 2 − 2x
x + 1 1 13
→
+ =
x − 2 x 35
13
13
→ 2
=
x − 2x 35
33. Solve ( 5x − 4 )
2x − y
2
A.
B.
= 10 . Find x.
0.8
2.06 *
= ( 2x + 1)
1/ 2
A.
B.
29. Given the equations 5x + y = 100 and
1/ 2
4 only *
4 or 8/9
( 5x − 4 )
12
+ 1.
C. 8/9 only
D. 4 and 8/9
= ( 2x + 1)
12
+1
5x − 4 = ( 2x + 1) + 2 2x + 1 + 1
C. 1.24
D. 3.12
5x − 2 x − 4 − 2 = 2 2 x + 1
3x − 6 = 2 2x + 1
Solution:
(x + y)log 5 = 2
2
x+y =
→ Eq.1
log 5
( 2x − y)log 2 = 1
1
2x − y =
→ Eq.2
log 2
9x 2 − 36x + 36 = 4 ( 2x + 1)
9x 2 − 44x + 32 = 0
8
x=4 l x=
9
34. Simplify:
Adding 1 and 2:
2
1
3x =
+
log 5 log 2
x = 2.06
30. Find the equation whose roots are 2, - 3
and 7/5.
A.
B.
C.
D.
5x3 – 2x2 – 37x + 42 = 0 *
5x3 + 2x2 – 37x + 42 = 0
5x3 – 2x2 – 37x – 42 = 0
5x3 – 2x2 + 37x + 42 = 0


 5x − 2x − 37 x + 42 = 0
A.
x(a2 + 4c 2 + 4ac) = 5a + 10c
5(a + 2c)
x= 2
(a + 4c 2 + 4ac)
5(a + 2c)
x=
(a + 2c)(a + 2c)
5
x=
(a + 2c)
19
*
5
17
D. ln
5
Answer: C. ln
3
x +1 1
13
, the solutions are
+ =
x − 2 x x 2 − 2x
a2 x + 4c 2 x + 4acx = 5a + 10c
C. ln
ln
 7 − 57 7 + 57 
 7 − 57  7 + 57 
x2 − 
+
x+
= 0 Solution:






2
2
2
2





7
( x − 2) ( x + 3)  x − 5 
x2 − 7x − 2 = 0
27. If
Solution:
a2 x + 4c 2 x − 10c = 5a − 4acx
2
31. Find the value of x of the following
system of equations:
x + 3y + 4z = 15
−2x + 4y + 5z = 12
3x + y + 6z = 29
A. 1
C. 3
B. 2 *
D. 4
 xy (a-b)2 


n-1
ab
xy
A.
a −b
C.
B.
xy *
D. ab (x-y)
Answer: B. xy
n
−1
2n −1
( xy ) ( a − b ) ( a − b )
 xy ( a − b )2 


n −1
( xy ) ( a − b )
n −1
2n − 2
( xy ) ( a − b )
n
=
2n − 2
= xy
Calculator:
Assume values for variables a, b, x and y
then use CALC function. (Same with the
previous problem)
35. Factor x4 + 5x2 − 36 .
A.
B.
C.
D.
Solution:
x + 3y + 4z = 15
−2 x + 4 y + 5z = 12
3x + y + 6z = 29
(xy)n (a − b)−1(a − b)2n - 1
( x − 4)( x + 9)
( x2 + 9) ( x − 2)( x + 2) *
( x + 4)( x − 9 )
( x − 2)( x + 2)( x − 3 )( x + 3 )
(
= (x
)(
)
+ 9) ( x − 2)( x + 2)
x 4 + 5x2 − 36 = x 2 + 9 x 2 − 4
2
Using calculator, we get: x=2
32. Solve the equation a2x + 4c2x – 10c = 5a
– 4acx for x.
A.
B.
5
a − 2c
5
2a + c
5
2a + 2c
5
D.
*
a + 2c
C.
GIVE YOUR BEST SHOT !
36. Solve for x: x + 1 + 2x + 3 − 8x + 1 = 0
A.
B.
– 3 and 1/17
3 and – 1/17
C. – 1/17
D. 3 *
__________
Mathematics: Algebra 1
x + 1 + 2x + 3 − 8x + 1 = 0
(
x + 1 = 8x + 1 − 2x + 3
)
B
4
sum of roots = − = −
A
3
2
x + 1 = ( 8x + 1) − 2 ( 8x + 1)( 2x + 3 ) + ( 2x + 3 )
( x + 1 − 8x − 1 − 2x − 3 )
=  −2 ( 8x + 1)( 2x + 3 ) 


2
3x 2
=
4y 2
41. If 3x = 4y, then
2
 −3 ( 3x + 1)  = 4 ( 8x + 1)( 2x + 3 )
3x 2
1
−
3
x
−1 .
x
−
1
48. Simplify
x2 − x
1− 2
x + x +1
2
A. ¾
9 ( 9x + 6x + 1) = 4 (16x + 26x + 3 )
2
B. 27/64
D. 9/16
2
81x 2 + 54x + 9 = 64x 2 + 104x + 12
37. If x, y and z are positive, with xy = 24, xz
= 48 and yz = 72, then what is the value
of x + y + z?
A. 22 *
B. 50
C. 36
(
equation x − 2
A.
(x − 2)2 = 0
1.9
A. k – 9
C. k – 14 *
D.
44. Solve for x:
x + x − 1− x = 1
A.
1/9
a5b3c 8 =
B. 9
C. 1/3
D. 3 *
9a3c 8
b−3
a2 = 9
a=3
46. If 2x + 4y = 7x - 6y, then
D. 6.8
B. 3:1
C. 1:2 *
1 1
: =
x y
H.
3x 4 + 4x 3 + x − 1 = 0
Descartes Rule :
no. of sign changes in f(x) is equal
to no. of positive real roots
no. of sign changes in f( −x) is equal
to no. of negative real roots
D. 1:3
47. Solve for x in the following equations:
x(x + y + z) = 2
y(x + y + z) = 6
z(x + y + z) = 8
A.
C. 6(2 – x) *
D. 3x
1/2 *
B. 2
C. -1/2
D. -2
(x + y + z)2 = 16
x+y+z=4
x ( 4) = 2
x = 0.5
Use calculator:
x 2 + x = 3 → x1 = 1.302 and x 2 = −2.302
so, x 4 + x = (1.302 ) + 1.302 = 4.175
4
1.
x 4 + x = ( −2.302 ) + −2.302 = 25.78
4
From the choices, find the expression
that will coincide to the ones
solved.
40. What is the sum of the roots of 3x3 + 4x2
– 4x = 0?
B. 4
G.
3x 4 − 4x 3 − x − 1 = 0 (1negative real root)
no. of complex roots
highest degree− 2 = 4 − 2 = 2
2 complex roots
x= 5
39. If x 2 + x = 3 , then x4 + x =
– 4/3 *
F.
Three positive real roots and one
negative real root
Three negative real roots and one
positive real root
One negative real root and three
complex roots
One positive real root, one negative
real root and two complex roots *
3x 4 + 4x 3 + x − 1 = 0 (1positive real root)
Add the three equations:
(x + y + z)(x + y + z) = 8 + 6 + 2
A.
C. 2
D. x
Use calculator:
Let x = 0.1
3x 2
1
−
x3 − 1 x − 1 → CALC → x = 0.1 → = →
x2 − x
1− 2
x + x +1
Answer: 1
E.
A. 13/25
B. 16/25 *
C. 17/25
D.
22/25
45. If a, b and c are real numbers and if
9a3c 8
a5b3c 8 = −3 , then a could equal
b
(x + 3) = 0 (x − 5) = 0
4(3 – x)
3(x – 1)
1*
–1
49. The nature of the roots of the equation
3x4 + 4x3 + x – 1 = 0 is
sum = 2 2 − 3 + 5 = 3.3
A.
B.
A.
B.
B. 25 – k C. 25 + 2k * D. 25 – 2k
A. 25k
2
x = 2 and 2 x = − 3
B. 2k – 9
k – 18
43. If x – y = 5 and xy = k, then x2 + y2 =
) ( x + 3 )( x − 5 ) = 0A.? 2:1
B. 3.3 * C. 2.2
*
equal to _____.
D. 62
xy = 24 yz = 72
yz 72
z
=
=3
xy 24
x
z
=x
3
xz = 48
xy = 24
xz 48
z
=
=2
xy 24
y
z
=y
2
z2
 z  z 
= xy
   = xy →
6
 3  2 
z2
xy = 24
= 24 z = 12
6
xz = 48 x(12) = 48 x = 4
xy = 24 4(y) = 24 y = 6
x + y + z = 4 + 6 + 12 = 22
38. What is the sum of the roots of the
4/3
( x − 2)( x + 3) = k, then ( x − 4)( x + 5) is
42. If
17x 2 − 50x − 3 = 0
1
x = 3, x = − → discarded
17
Substitute the two values in the equation given,
the value that will satisfy the equation is x =
3.
C.
C. – 3/4 D. 4/3
x+2
into
x 2 − 7x + 12
partial fractions.
Resolve
Solution:
50. How many possible rational roots are
there for 2x4 + 4x3 − 6x2 + 15x − 12 = 0 ?
B.
4
B. 12
C. 8
Rational root test:
factors of the
Possible
constant term
= 
rational roots
factors of the
leading coefficients
Possible
factors of 12
= 
rational roots
factors of 2
1,2,3,4,6,12
=
1,2
1,2,3,4,6,12
1,2,3,4,6,12
=
and 
1
2
So, the possible rational roots are:
1 3
1,2,3,4,6,12 and  ,
2 2
For a total of 16.
Use reverse by substituting value to x to
the choices and to the problem.
D.
6
5
−
x-4 x −3
GIVE YOUR BEST SHOT !
51. The solution set of
x −1
x
 2 is
D. 16 *
__________
x −1
x
x −1
Mathematics: Algebra 1
x + 48  16x
2
2
x 2 − 16x + 48  0
Equate left side to zero.
−20
x
x − 1 − 2x
Remainder,R:
R = f(1 3)
0
x
Equate the numerator and denominator
respectively to zero.
R = 3 (1 3 ) + 5 (1 3 ) − 5 (1 3 ) + 10 (1 3 ) − 1
4
x  12
4  x  12
x − 1 − 2x = 0
− ( x − 1) − 2x = 0
x − 1 − 2x = 0
4
x=0
A.
B.
Test the regions that if it satisfy
x 2 − 16x + 48  0
1
x  −1 −1  x  0 0  x 
Solution : 4  x  12
3 x  1/ 3
Integers:5, 6, 7, 8, 9, 10, 11
Test the regions if it satisfy
Answer: 7 integers
0
−1
1
x − 1 − 2x
3
0
x
1
Answer : 0  x 
3
55. Solve the equation: 9 − 3 x + 2 = 15 .
52. If logb (xy)  0 , which of the following
C. 2
C. 2 or – 2
D. – 2
D. No solution *
must be true? Note b < 1.
Follow these steps to solve an absolute value
equality which contains one absolute value:
1. Isolate the absolute value on one side of the
equation.
logb ( xy )  0, b  1
2. Is the number on the other side of the equation
assign b = 0.5 and equate the left side to 0. negative? If yes, then the equation has no
solution. If no, then go to step 3.
log0.5 ( xy ) = 0
3. Write two equations without absolute values.
0
xy = 0.5
The first equation will set the quantity inside
the side bars equal to the number on the
xy = 1
other side of the equal sign; the second
equation will set the quantity inside the bars
xy  1
( xy )  1
equal to the opposite of the number on the
other side.
I.
J.
xy < 0
xy < 1
C. xy > 1 *
D. xy > 0
56. How many integers are there in the
solution set of 3x – 2 < - 1?
1
Test the regions if it satisfy
log0.5 ( xy )  0
Answer : xy  1
53. |2x – 1| = 4x + 5 has how many numbers
in its solution set?
K.
L.
0
1*
C. 2
D. infinite
A. infinitely many * C. 3
B. none
D. only 1
3x − 2  −1
3x − 1  0
Equate left side to zero:
3x − 1 = 0
1
x=
3
2x − 1 = 4x + 5
x = −3
1/ 3
−2
3
The two values when tested to original
−2
equation, only x =
can satisfy.
3
Therefore, answer is 1.
x=
0
4
C. 7 *
D. infinite number
57. Find the remainder when 6x2 – 30 + 9x3
is divided by 3x – 4.
A.
B.
4
2*
2
1
C. 3
D. 6
Solution:
GIVE YOUR BEST SHOT !
C. - 2 *
D. - 1
Answer: C. - 2
Because x + 7 divides evenly with the
cubic polynomial, therefore when
we equate x + 7 to zero we can get
one root of the cubic polynomial.
x+7=0→x=-7
If we substitute x = - 7 to the cubic
polynomial, it will be equal to zero.
That is according to factor theorem.
x3 + ax 2 − 61x + 14 = 0
when x = −7
( −7 )
3
+ a ( −7 ) − 61( −7 ) + 14 = 0
2
49a + 98 = 0
a = −2
59. When the expression x4 + ax3 + 5x2 + bx
+ 6 is divided by (x – 2), the remainder is
16. When it is divided by (x + 1) the
remainder is 10. What is the value of the
constant a?
A.
B.
C. – 7
D. – 5 *
5
7
Answer: D. – 5
x–2=0→x=2
Substituting x = 2 to the polynomial, it’s
equal to 16 (remainder is the sum
according to the remainder
theorem)
x 4 + ax 3 + 5x 2 + bx + 6 = 16
when x = 2
( 2)
4
+ a ( 2 ) + 5 ( 2 ) + b ( 2 ) + 6 = 16
3
2
8a + 2b + 42 = 16
8a + 2b = −26
x+1=0→x=-1
Substituting x = - 1 to the polynomial, it’s
equal to 10 (remainder is the sum
according to the remainder
theorem)
x 4 + ax 3 + 5x 2 + bx + 6 = 10
when x = −1
( −1)
Test the regions if it satisfy
3x − 1  0
Solution : x  1/ 3
Therefore, infinitely many is the answer.
54. The number of integers that satisfy the
inequality x2 + 48 < 16x is
M.
N.
x  1/ 3
x  1/ 3
2x − 1 = 4x + 5
− ( 2x − 1) = 4x + 5
58. Calculate the value of a in the expression
x3 + ax2 – 61x + 14 if x + 7 divides evenly
into the cubic polynomial.
12
1
x=
3
x = −1
2
R=2
x 2 − 16x + 48 = 0
x = 12 and x = 4
x4
3
4
+ a ( −1) + 5 ( −1) + b ( −1) + 6 = 10
3
2
−a − b + 12 = 10
−a − b = −2
a+b = 2
Use your calculator:
Let a = x and b = y
Mode → 5EQN → 1an x + bn y = c n →
a b c 
8 2 −26  → = → x = −5 and y = 7


 1 1 2 
So, a = - 5 and b = 7
__________
Mathematics: Algebra 1
60. Given the equation x3 – 2x2 – 3k = 0. Find A. 2 *
B. 3
C. 1
D. -2
B. 45852x8
D. 46940x8
k so that 3 is one of the roots to this
equation.
n
8
66. Find the fourth term of the expansion of
r = + 1 = + 1 = 5th term
2
2
7
A. 2
C. 3 *
(a − 2x ) .
n−r +1
th
B. 4
D. 5
r term = nCr −1 ( x )
( y )r −1
A. 120a4 x3
C. −120a4 x3
8−5+1
Solution:
= 8C5−1 x 2
( −5 )5−1
B. −280a4 x3 *
D. 280a 4 x 3
Substitute x = 3 to the given equation.
4
4
3
2
= ( 8C4 ) x 2 ( −5 )
( 3) − 2 ( 3) − 3k = 0
( )
( )
k=3
67. Find
the
term
y5 in
involving
= 43750x8
the
expansion of ( 2x 2 + y ) .
10
61. Given f(x) = (x + 3)(x – 4) + 4. When f(x)
is divided by (x – k), the remainder is k.
Find k.
A.
B.
C. – 4
D. 4 & - 2 *
2
–2
A.
8064x10y5 *
C. 8564x10y5
B.
8464x10y5
D. 8264x10y5
(
A. 100,500x6 yz3
y
n − r +1
*
B. 500100x6 yz3
(y)
r −1
= 10C6-1 ( 2x 2 )
= (10C5 ) ( 2x
10
2
)
5
y
(y)
Solution:
Using long division:
r th term = nCr −1 ( x )
n − r +1
5
3x 2 + 6x + 8
x 3 − 2x 2 + 6 3x 5 + 0x 4 − 4x 3 + 2x 2 + 36x + 48
3x5 − 6x 4 + 0x 3 + 18x 2
+ 6x 4 − 4x3 − 16x 2 + 36x
0
+ 6x 4 − 12x3 + 0x 2 + 36x
8x3 − 16x 2 + 48
8x3 − 16x 2 + 48
0
63. Find the remainder if we divide
4y3 + 18y2 + 8y − 4 by ( 2y + 3 ) .
A. 11 *
B. 10
C. 672 *
)
7
Solve for a term with z 3
r th term = 7Cr −1 ( 3x 2 + 4y )
7 − r +1
r term = 7C4 −1 ( 3x 2 + 4y )
D.
th
( −2z )
r −1
7 − 4 +1
( −2z )
4
r term = nCr −1 ( x )
y
r th term = 4Cr −1 ( 3x 2 )
r −1
 1
= 9Cr −1 ( 2x )
x
 
Since the result is a constant term, x 0.
2
9 − r +1
4 −1
3
Solve for the term with x 6 y in ( 3x 2 + 4y )
r −1
4 − r +1
( 4y )
4
r −1
r = 2 → to produce y1
= ( 4C1) ( 3x 2 ) ( 4y )
3
1
3
1
3
Term : ( 7C3 ) ( 4C1) ( 3x 2 ) ( 4y )  ( −2z )


− 120960x 6 yz3 → Ans
r −1
 1
0
x = x
 
x 20 − 2r x −r +1 = x 0
20 − 2r − r + 1 = 0
r =7
Substitute :
73. Find
the
term
involving
x9
in
the
12
9 − 7 +1
 1
x
 
2

expansion of  x 2 +  .
x

7 −1
= 672
69. Find the 6th term of the expansion of
2y + 3 = 0
(
3x 2 + 4y as one term: ( 3x 2 + 4y ) − 2z
= ( 7C3 ) ( 3x 2 + 4y ) ( −2z )
n − r +1
9 − r +1
r −1
r = 4 → to produce z3
r th term = 9C7 −1 ( 2x 2 )
C. 13
D. 12
A.
B.
24339x9
25344x9 *
C. 23337x9
D. 22889x9
16
 1

 2a − 3  .


3
2
Substitute :
y=−
3
2
 3
 3
 3
R = 4  −  + 18  −  + 8  −  − 4
 2
 2
 2
R = 11
64. Find the remainder when 2x4 – 5x3 + 2x2
– 4x + 6 is divided by x – 2.
A.
9
1

of  2x 2 +  .
x

A. 664
B. 682
648
( x2 )
(y)
Consider the trinomial as binomial by assuming
5
th
3
6 −1
The term: 8064x y
62. Find the quotient of 3x5 – 4x3 + 2x2 + 36x
3
2
+ 48 divided by x – 2x + 6.
68. Find the constant term in the expansion
C. 3x2 – 6x - 8
D. 3x2 + 6x + 8
D.
−160920x yz
6
In order to have y , r should be 6.
10 − 6 +1
C.
−120960x6 yz3
5
0
)
3
n − r +1 r −1
=10Cr-1 ( 2x 2 )
3x2 – 4x – 8
3x2 + 4x + 8
expansion
of 3x + 4y − 2z , solve term involving
Substitute to the formula,
k 2 − 2k − 8 = 0
(k − 4)(k + 2) = 0
k = 4 l k = −2
binomial
7
x yz .
r term = nCr −1x
k 2 − 4k + 3k − 12 + 4 = k
A.
B.
the
2
6
th
Solution:
Solving for f(k)=k.
(k + 3)(k − 4) + 4 = k
72. In
B. – 4
4
C. 2
D. – 2 *
x−2=0
x=2
Substitute :
66939
256a11
66339
*
−
128a11
33939
256a11
66339
D. −
256a11
−
A.
B.
Answer: 25334x9
Use binomial expansion formula:
C. −
12
n
 2 2
 x + x  → (a + b)


rth term =n Cr −1an − r +1br −1
12
.
Find the sum of the coefficients.
A. 5094
C. 1245
B. 2942
D. 4095 **
Sum = ( 3 − 1)
12
− ( −1)
12
= 4095
R = 2 ( 2 ) − 5 ( 2 ) + 2 ( 2 ) − 4 ( 2 ) + 6 = −2
4
3
2
71. Find the middle term of the expansion of
(x
2
−5
)
65. If x – 3 is a factor of kx – 6x + 2kx – 12.
Solve for k.
A. 43750x8 **
2
2
x
 
rth term =12 Cr −1 ( 2 )
(x ) (x )
12 − r +1
70. From the given expression ( 3x − 1)
8
.
C. 44736x8
GIVE YOUR BEST SHOT !
r −1
rth term =12 Cr −1 ( x 2 )
r −1
rth term = 12 Cr −1 ( 2 )
2
13 − r
−1
r −1
(x) (x)
r −1
27 − 3r
rth term =12 Cr −1 ( 2 ) ( x )
r −1
26 − 2r
1− r
The term to find contains x9. For the rth
term to have x9, (x) 27-3r should be
equal to x9.
x 9 = x 27 − 3r
9 = 27 − 3r
r=6
__________
Mathematics: Algebra 1
So, r = 6 meaning, we are looking for the
6th term.
6th term =12 C6 −1 ( 2 )
6 −1
(x)
27 − 3 ( 6 )
6th term = 25344x9
GIVE YOUR BEST SHOT !