INTERMEDIATE ALGEBRA
math
hands
CH 01 SEC 08 SOLUTIONS
Solving
basic higher degree strategies
1. Solve by Grouping
− 3x3 + 6x2 + 5x − 10 = 0
Solution:
2
− 3x3 + 6x2 + 5x − 10 = 0
− 3x2 x − 2 + 5 x − 2 = 0
− 3x2 + 5 x − 2 = 0
− 3x + 5 = 0
OR
5
x2 =
3
r
5
x=±
3
(given)
(Grouping and Factor)
(factor using Dist. Law)
x−2=0
(ZFT)
OR
x=2
(algebra)
OR
x=2
(SRP)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra]
2. Solve by Grouping
3x3 + x2 + 12x + 4 = 0
Solution:
2
x
2
3x3 + x2 + 12x + 4 = 0
3x + 1 + 4 3x + 1 = 0
x2 + 4 3x + 1 = 0
x +4=0
OR
x2 = − 4
q
x=± −4
3x + 1 = 0
OR
OR
1
3
1
x= −
3
x= −
(given)
(Grouping and Factor)
(factor using Dist. Law)
(ZFT)
(algebra)
(SRP)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra]
3. Solve
x4 + 8x3 + 24x2 + 32x + 16 = 0
pg. 1
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
basic higher degree strategies
math
hands
CH 01 SEC 08 SOLUTIONS
Solution:
x4 + 8x3 + 24x2 + 32x + 16 = 0
4
3
2
2
3
(given)
4
(1x) + 4 (1x) (2) + 6 (1x) (2) + 4 (1x) (2) + (2) = 0
(recognize famous form a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 )
4
x + 2 = 0 (famous a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 = (a + b)4 )
x= −2
(ZFT, algebra, with multiplicity, 4)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra, counting multiplicity, ie each solutions gets counted each time
it occurs, even if repeated..]
4. Solve
2x3 − 5x2 − 4x + 3 = 0
Solution:
We first look at the constant term, 3, as well as the leading term, 2. From this we use the Rational Root
Theorem, to conclude that IF this equation has a rational root, it must be a fraction of the form
a
potential rational roots:
b
where the numerator, a, is an integer divisor of 3, and the denominator, b, is an integer divisor of 2. so
potential numerators, a, are:
± 1, or ± 3
while
potential denominators, b, are:
± 1, or ± 2
to check each potential root, a/b we check, the polynomial p(x) = 2x3 − 5x2 − 4x + 3 to see if p(a/b) = 0.
After, some trial and error, trying various combinations of the above numerators and denominators, we see that
p 3 =0
This implies that x + − 3 must be a factor of p(x), in other words p(x) must be of the form
p(x) = x + − 3 (some other polynomial)
to figure out what that other polynomial must be we divide p(x) by x + − 3
2x2 + x − 1
and we obtain:
x−3
2x3 − 5x2 − 4x + 3
− 2x3 + 6x2
x2 − 4x
− x2 + 3x
−x+3
x−3
0
pg. 2
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
basic higher degree strategies
math
hands
CH 01 SEC 08 SOLUTIONS
THUS:
p(x) = x + − 3
which you can further factor as
p(x) = x + − 3
which finally leads to the solution:
x = 3,
2x2 + x − 1
x+1
2x − 1
2x3 − 5x2 − 4x + 3 = 0
x + − 3 x + 1 2x − 1 = 0
1
x = − 1,
OR
x=
2
(given)
(factored, above)
(Zero Factor Theorem)
5. Solve by Grouping
− 3x4 + 6x3 + 5x2 − 10x = 0
Solution:
x=0
OR
− 3x4 + 6x3 + 5x2 − 10x = 0
− 3x3 x − 2 + 5x x − 2 = 0
− 3x3 + 5x x − 2 = 0
(x) − 3x2 + 5 x − 2 = 0
2
− 3x + 5 = 0
5
x=0
OR
x2 =
r3
5
x=0
OR
x=±
3
OR
(given)
(Grouping and Factor)
(factor using Dist. Law)
(factor using Dist. Law)
x−2=0
(ZFT)
OR
x=2
(algebra)
OR
x=2
(SRP)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra]
6. Solve by Grouping
− 3x4 + 2x3 + 9x2 − 6x = 0
pg. 3
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
basic higher degree strategies
math
hands
CH 01 SEC 08 SOLUTIONS
Solution:
x=0
− 3x4 + 2x3 + 9x2 − 6x = 0
− x3 3x − 2 + 3x 3x − 2 = 0
− x3 + 3x 3x − 2 = 0
(x) − x2 + 3 3x − 2 = 0
2
OR
−x +3=0
x=0
OR
x=0
OR
OR
x2 = 3
q
x=± 3
3x − 2 = 0
2
OR
x=
3
2
OR
x=
3
(given)
(Grouping and Factor)
(factor using Dist. Law)
(factor using Dist. Law)
(ZFT)
(algebra)
(SRP)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra]
7. Solve
16x4 + 96x3 + 216x2 + 216x + 81 = 0
Solution:
16x4 + 96x3 + 216x2 + 216x + 81 = 0
4
3
2
2
3
(given)
4
(2x) + 4 (2x) (3) + 6 (2x) (3) + 4 (2x) (3) + (3) = 0
(recognize famous form a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 )
2x + 3
4
(famous a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 = (a + b)4 )
3
x= −
(ZFT, algebra, with multiplicity, 4)
2
=0
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra, counting multiplicity, ie each solutions gets counted each time
it occurs, even if repeated..]
8. Solve
2x3 − 11x2 + 19x − 10 = 0
pg. 4
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
basic higher degree strategies
math
hands
CH 01 SEC 08 SOLUTIONS
Solution:
We first look at the constant term, -10, as well as the leading term, 2. From this we use the Rational Root
Theorem, to conclude that IF this equation has a rational root, it must be a fraction of the form
a
b
potential rational roots:
where the numerator, a, is an integer divisor of -10, and the denominator, b, is an integer divisor of 2. so
potential numerators, a, are:
± 1, ±2, ±5, or ± 10
while
potential denominators, b, are:
± 1, or ± 2
to check each potential root, a/b we check, the polynomial p(x) = 2x3 − 11x2 + 19x − 10 to see if p(a/b) = 0.
After, some trial and error, trying various combinations of the above numerators and denominators, we see that
p 1 =0
This implies that x + − 1 must be a factor of p(x), in other words p(x) must be of the form
p(x) = x + − 1 (some other polynomial)
to figure out what that other polynomial must be we divide p(x) by x + − 1
2x2 − 9x + 10
and we obtain:
x−1
2x3 − 11x2 + 19x − 10
− 2x3 + 2x2
− 9x2 + 19x
9x2 − 9x
10x − 10
− 10x + 10
0
THUS:
2x2 − 9x + 10
x−2
p(x) = x + − 1
which you can further factor as
p(x) = x + − 1
which finally leads to the solution:
x = 1,
2x − 5
2x3 − 11x2 + 19x − 10 = 0
x + − 1 x − 2 2x − 5 = 0
5
x = 2,
OR
x=
2
(given)
(factored, above)
(Zero Factor Theorem)
9. Solve
2x3 + 19x2 + 54x + 45 = 0
pg. 5
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
basic higher degree strategies
math
hands
CH 01 SEC 08 SOLUTIONS
Solution:
We first look at the constant term, 45, as well as the leading term, 2. From this we use the Rational Root
Theorem, to conclude that IF this equation has a rational root, it must be a fraction of the form
potential rational roots:
a
b
where the numerator, a, is an integer divisor of 45, and the denominator, b, is an integer divisor of 2. so
potential numerators, a, are:
± 1, ±3, ±5, ±9, ±15, or ± 45
while
potential denominators, b, are:
± 1, or ± 2
to check each potential root, a/b we check, the polynomial p(x) = 2x3 + 19x2 + 54x + 45 to see if p(a/b) = 0.
After, some trial and error, trying various combinations of the above numerators and denominators, we see that
p −5 =0
This implies that x + 5 must be a factor of p(x), in other words p(x) must be of the form
p(x) = x + 5 (some other polynomial)
to figure out what that other polynomial must be we divide p(x) by x + 5
2x2 + 9x + 9
and we obtain:
x+5
2x3 + 19x2 + 54x + 45
− 2x3 − 10x2
9x2 + 54x
− 9x2 − 45x
9x + 45
− 9x − 45
0
THUS:
p(x) = x + 5
which you can further factor as
p(x) = x + 5
which finally leads to the solution:
x = − 5,
2x2 + 9x + 9
x+3
2x + 3
2x3 + 19x2 + 54x + 45 = 0
x + 5 x + 3 2x + 3 = 0
3
x = − 3,
OR
x= −
2
(given)
(factored, above)
(Zero Factor Theorem)
10. Solve
pg. 6
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
basic higher degree strategies
math
hands
CH 01 SEC 08 SOLUTIONS
x3 + 6x2 + 12x + 8 = 0
Solution:
x3 + 6x2 + 12x + 8 = 0
3
2
2
3
(1x) + 3 (1x) (2) + 3 (1x) (2) + (2)
3
x+2 =0
(given)
(recognize famous form a3 + 3a2 b + 3ab2 + b3 )
(famous a3 + 3a2 b + 3ab2 + b3 = (a + b)3 )
x+2=0
OR
x+2 = 0
OR
x+2=0
x= −2
OR
x= −2
OR
x= −2
(ZFT)
(algebra)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra, counting multiplicity, ie each solutions gets counted each time
it occurs, even if repeated..]
11. Solve
x3 + 3x2 + 3x + 1 = 0
Solution:
x3 + 3x2 + 3x + 1 = 0
3
2
2
3
(1x) + 3 (1x) (1) + 3 (1x) (1) + (1)
3
x+1 =0
(given)
(recognize famous form a3 + 3a2 b + 3ab2 + b3 )
x+1=0
OR
x+1 = 0
OR
x+1=0
x= −1
OR
x= −1
OR
x= −1
(famous a3 + 3a2 b + 3ab2 + b3 = (a + b)3 )
(ZFT)
(algebra)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra, counting multiplicity, ie each solutions gets counted each time
it occurs, even if repeated..]
12. Solve by Grouping
− 9x3 − 6x2 + 12x + 8 = 0
pg. 7
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
basic higher degree strategies
math
hands
CH 01 SEC 08 SOLUTIONS
Solution:
− 9x3 − 6x2 + 12x + 8 = 0
− 3x2 3x + 2 + 4 3x + 2 = 0
− 3x2 + 4 3x + 2 = 0
2
OR
3x + 2 = 0
4
x2 =
OR
x= −
r3
4
x=±
OR
x= −
3
(given)
(Grouping and Factor)
(factor using Dist. Law)
− 3x + 4 = 0
(ZFT)
2
3
2
3
(algebra)
(SRP)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra]
13. Solve
1
2
=0
8x3 + 4x2 + x +
3
27
Solution:
2
1
8x3 + 4x2 + x +
=0
(given)
3
27
3
2
2
3
(2x) + 3 (2x) ((1/3)) + 3 (2x) ((1/3)) + ((1/3))
(recognize famous form a3 + 3a2 b + 3ab2 + b3 )
3
1
=0
(famous a3 + 3a2 b + 3ab2 + b3 = (a + b)3 )
2x +
3
1
1
1
2x + = 0
OR
2x + = 0
OR
2x + = 0
(ZFT)
3
3
3
1
1
1
x= −
OR
x= −
OR
x= −
(algebra)
6
6
6
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra, counting multiplicity, ie each solutions gets counted each time
it occurs, even if repeated..]
14. Solve
30x3 + 69x2 + 15x − 6 = 0
pg. 8
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
basic higher degree strategies
math
hands
CH 01 SEC 08 SOLUTIONS
Solution:
We first look at the constant term, -6, as well as the leading term, 30. From this we use the Rational Root
Theorem, to conclude that IF this equation has a rational root, it must be a fraction of the form
potential rational roots:
a
b
where the numerator, a, is an integer divisor of -6, and the denominator, b, is an integer divisor of 30. so
potential numerators, a, are:
± 1, ±2, ±3, or ± 6
while
potential denominators, b, are:
± 1, ±2, ±3, ±5, ±6, ±10, ±15, or ± 30
to check each potential root, a/b we check, the polynomial p(x) = 30x3 + 69x2 + 15x − 6 to see if p(a/b) = 0.
After, some trial and error, trying various combinations of the above numerators and denominators, we see that
1
=0
p −
2
This implies that x +
1
2
must be a factor of p(x), in other words p(x) must be of the form
1
(some other polynomial)
p(x) = x +
2
to figure out what that other polynomial must be we divide p(x) by x + 21
30x2 + 54x − 12
and we obtain:
x+
1
2
30x3 + 69x2 + 15x − 6
− 30x3 − 15x2
54x2 + 15x
− 54x2 − 27x
− 12x − 6
12x + 6
0
THUS:
1
30x2 + 54x − 12
p(x) = x +
2
which you can further factor as
1
p(x) = x +
6x + 12 30x − 6
2
which finally leads to the solution:
1
x= − ,
2
pg. 9
30x3 + 69x2 + 15x − 6 = 0
1
x+
6x + 12 30x − 6 = 0
2
1
x = − 2,
OR
x=
5
(given)
(factored, above)
(Zero Factor Theorem)
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
basic higher degree strategies
math
hands
CH 01 SEC 08 SOLUTIONS
15. Solve
8x3 + 12x2 + 6x + 1 = 0
Solution:
8x3 + 12x2 + 6x + 1 = 0
3
2
2
(given)
3
(2x) + 3 (2x) (1) + 3 (2x) (1) + (1)
(recognize famous form a + 3a b + 3ab2 + b3 )
3
2x + 1 = 0
(famous a3 + 3a2 b + 3ab2 + b3 = (a + b)3 )
2x + 1 = 0
OR
1
x= −
2
3
2
2x + 1 = 0
OR
OR
2x + 1 = 0
1
1
x= −
OR
x= −
2
2
(ZFT)
(algebra)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra, counting multiplicity, ie each solutions gets counted each time
it occurs, even if repeated..]
16. Solve
x3 − 3x2 + 3x − 1 = 0
Solution:
x3 − 3x2 + 3x − 1 = 0
3
2
2
3
(1x) + 3 (1x) (−1) + 3 (1x) (−1) + (−1)
3
x−1 =0
x−1=0
OR
x=1
x−1 = 0
OR
x=1
OR
OR
(given)
3
2
(recognize famous form a + 3a b + 3ab2 + b3 )
(famous a3 + 3a2 b + 3ab2 + b3 = (a + b)3 )
x−1=0
(ZFT)
x=1
(algebra)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra, counting multiplicity, ie each solutions gets counted each time
it occurs, even if repeated..]
17. Solve
x3 − 9x2 + 27x − 27 = 0
pg. 10
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
Solving
basic higher degree strategies
math
hands
CH 01 SEC 08 SOLUTIONS
Solution:
x3 − 9x2 + 27x − 27 = 0
3
2
2
(given)
3
3
(recognize famous form a + 3a b + 3ab2 + b3 )
(1x) + 3 (1x) (−3) + 3 (1x) (−3) + (−3)
3
x−3 =0
x−3=0
OR
x=3
x−3 = 0
OR
OR
x=3
2
(famous a3 + 3a2 b + 3ab2 + b3 = (a + b)3 )
x−3=0
(ZFT)
x=3
(algebra)
OR
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra, counting multiplicity, ie each solutions gets counted each time
it occurs, even if repeated..]
18. Solve
x4 + 4x3 + 6x2 + 4x + 1 = 0
Solution:
x4 + 4x3 + 6x2 + 4x + 1 = 0
4
3
2
2
3
(given)
4
(1x) + 4 (1x) (1) + 6 (1x) (1) + 4 (1x) (1) + (1) = 0
(recognize famous form a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 )
x+1
4
=0
(famous a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 = (a + b)4 )
x= −1
(ZFT, algebra, with multiplicity, 4)
[will let you should simplify.... if needed... or if you feel like it.. or if applicable..... also note 3 solutions.. as
promised by the fundamental theorem of algebra, counting multiplicity, ie each solutions gets counted each time
it occurs, even if repeated..]
19. Solve
1x3 + − 1 = 0
Solution:
x−1
1x3 + − 1 = 0
x2 + x + 1 = 0
(given)
(famous polynomial difference of cubes)
[from here, you may want to use ZFT, and the Quadratic Formula]
pg. 11
c
2007-2011
MathHands.com v.1010
INTERMEDIATE ALGEBRA
CH 01 SEC 08 SOLUTIONS
20. Solve
math
hands
Solving
basic higher degree strategies
8 3 1
x + =0
27
8
Solution:
8 3 1
x + =0
27
8
4 2 1
1
1
2
=0
x+
x − x+
3
2
9
3
4
(given)
(famous polynomial difference of cubes)
[from here, you may want to use ZFT, and the Quadratic Formula]
pg. 12
c
2007-2011
MathHands.com v.1010