PRE-CALC
CH 01 SEC 03 HW WORKSHEET
SOLVING
degree 3+
math
hands
1. Solve by Grouping
− 3x3 + 6x2 + 5x − 10 = 0
6. Solve by Grouping
− 3x4 + 2x3 + 9x2 − 6x = 0
2. Solve by Grouping
3x3 + x2 + 12x + 4 = 0
7. Solve
16x4 + 96x3 + 216x2 + 216x + 81 = 0
3. Solve
x4 + 8x3 + 24x2 + 32x + 16 = 0
8. Solve
2x3 − 11x2 + 19x − 10 = 0
4. Solve
2x3 − 5x2 − 4x + 3 = 0
9. Solve
2x3 + 19x2 + 54x + 45 = 0
5. Solve by Grouping
− 3x4 + 6x3 + 5x2 − 10x = 0
pg. 1
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2007-2011
MathHands.com v.1010
PRE-CALC
CH 01 SEC 03 HW WORKSHEET
SOLVING
degree 3+
math
hands
14. Solve
30x3 + 69x2 + 15x − 6 = 0
10. Solve
x3 + 6x2 + 12x + 8 = 0
15. Solve
11. Solve
8x3 + 12x2 + 6x + 1 = 0
x3 + 3x2 + 3x + 1 = 0
16. Solve
12. Solve by Grouping
x3 − 3x2 + 3x − 1 = 0
− 9x3 − 6x2 + 12x + 8 = 0
17. Solve
13. Solve
2
1
8x3 + 4x2 + x +
=0
3
27
pg. 2
x3 − 9x2 + 27x − 27 = 0
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2007-2011
MathHands.com v.1010
PRE-CALC
CH 01 SEC 03 HW WORKSHEET
18. Solve
SOLVING
degree 3+
math
hands
22. Solve
− 1x3 + 8 = 0
x4 + 4x3 + 6x2 + 4x + 1 = 0
23. Solve
2x3 − 5x2 − 9x + 18 = 0
19. Solve
1x3 + − 1 = 0
24. Solve by Grouping
20. Solve
8 3 1
x + =0
27
8
15x3 − 18x2 − 25x + 30 = 0
25. Solve by Grouping
21. Solve
8x3 + − 1 = 0
pg. 3
9
− 3x4 + 2x3 + x2 − 3x = 0
2
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2007-2011
MathHands.com v.1010
PRE-CALC
SOLVING
degree 3+
math
hands
CH 01 SEC 03 NOTES
THE IDEA HIGHER DEGREE EQUATIONS We take the opportunity here to re-emphazise the Fundamental Theorem
of Algebra which states that the number of solutions to a polynomial equation is equal to the degree of the
polynomial. The FTA tells you about the number of solutions, but unfortunately, it does not tell us how to
actually find them. In general, this can be quite challenging. However, there are several methods which can and
often do help. We survey some of these methods in this section.
HIGHER DEGREE EQs TOOLs
1.
2.
3.
4.
5.
FACTOR BY GROUPING
RECOGNIZE FAMOUS POLYNOMIALS
RECOGNIZE QUADRATIC-LIKE
RATIONAL ROOT THEOREM
WAIT FOR CALCULUS
SOLVE by GROUPING
EXAMPLE: Solve
x3 + 4x2 − 3x − 12 = 0
solution:
x3 + 4x2 − 3x − 12 = 0
(given)
2
x (x + 4) − 3(x + 4) = 0
(ALA,DL)
2
(x − 3)(x + 4) = 0
√
√
(x − 3)(x + 3)(x + 4) = 0
√
√
x= 3
x=− 3
x = −4
(DL)
(factor)
(ZFT)
Note, 3 solutions to the degree 3 equation, as expected.
SOLVE by RECOGNIZING FAMOUS POLYNOMIAL
EXAMPLE: Solve
x3 = 1
solution:
x3 − 1 = 0
(given)
(x − 1)(x + x + 1) = 0
(FAMOUS Poly)
3
2
x−1=0
or
3
x −1 =0
(BI)
2
x +x+1=0
x−1=0
or
x=1
or
(ZFT)
p
−1 ± 1 − 4(1)(1)
x=
√2 · 1
−1 ± i 3
x=
2
(QF)
(BI)
Note, 3 solutions to the degree 3 equation, as expected.
pg. 4
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2007-2011
MathHands.com v.1010
PRE-CALC
SOLVING
degree 3+
math
hands
CH 01 SEC 03 NOTES
SOLVE by RECOGNIZING FAMOUS POLYNOMIAL con’t...
Some FAMOUS POLYNOMIALs
1. a2 + 2ab + b2 = (a + b)2
2. a2 − 2ab + b2 = (a − b)2
3. a3 +3a2 b+3ab2 +b3 = (a+b)3
4. a3 −3a2 b+3ab2 −b3 = (a−b)3
5. x3 − x3 = (x− y)(x2 + xy + y 2)
6. x3 + y 3 = (x− y)(x2 − xy + y 2 )
7. x2 − y 2 = (x − y)(x + y)
8. x2 + 2xy + y 2 = (x + y)2
SOLVE QUADRATIC-LIKE
EXAMPLE: Solve
x6 + 4x3 + 3 = 0
solution:
x6 + 4x3 + 3 = 0
3
3
(x + 3)(x + 1) = 0
3
x + 3 = 0 or
(given)
(quadratic-like)
3
x +1=0
(ZFT)
Now solve each one of these separately. x3 + 1 is a famous polynomial, it factors as x3 + 1 = (x + 1)(x2 − x + 1),
then use ZFT to separate into x + 1 = 0, first solution, and x2 − x + 1 = 0, use QF on this equation to get two
more solutions. Then redo for other factor, x3 + 3 = 0 To solve this use the idea that
√
3
x3 + 3 = x3 + ( 3)3
Now, you can factor using the famous polynomial
x3 + y 3 = (x + y)(x2 − xy + y 2 )
. This will yield an additional 3 roots, for a total of 6 as expected.
SOLVE using Rational Root Theorem
EXAMPLE: Solve
p(x) = 5x3 + 36x2 + 67x + 12 = 0
The idea here is to hope for a rational root. We divide by 5 to clear the leading coefficient, to obtain.
x3 +
pg. 5
12
36 2 67
x + x+
=0
5
5
5
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2007-2011
MathHands.com v.1010
PRE-CALC
CH 01 SEC 03 NOTES
After some thinking, we can conclude that if such a
root exists, it must be a ’factor’ of 12
5 , Thus the numerator could be ±1, 2, 3, 4, 6, 12 and the denominator
could be ±1, 5 but that is it. This exhausts all rational possibilities. We try 1/5, -1/5, 2/5, -2/5, 2/1,
-2/1, 3/5, -3/5, 3/1, -3/1,4/5,-4/5, 4/1, -4/1 etc etc...
We try each one either by long division or by plugging
it into the polynomial. In other words, if p(1/5) = 0,
then (x − 1/5) must be a factor of p(x), the converse
is also true, thought a thorough explanation as to why
may be outside the scope of this class. Let us try
x = −4, it turns out p(−4) = 0, therefore, we know
(x + 4) is a factor, so we use long division to find the
pg. 6
SOLVING
degree 3+
math
hands
other factor.
5x2 + 16x + 3
x+4
3
5x + 36x2 + 67x + 12
− 5x3 − 20x2
16x2 + 67x
− 16x2 − 64x
3x + 12
− 3x − 12
0
Therefore, 5x3 + 36x2 + 67x + 12 = (x + 4)(5x2 + 16x +
3) = 0, and this is easier to solve using ZFT.
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2007-2011
MathHands.com v.1010