Download F7 - Sum of Cubes

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Solving Quadratic Equations I
Consider the following two problems.
Factor x2 − 3x + 2.
Solve x2 − 3x + 2 = 0.
The problem on the left is what we have been doing. The problem on the right
is what we will be doing. The difference between the problems is that the first
problem is factoring an expression. The problem on the right is solving an
equation. The connection between the problems is that being able to factor the
quadratic is the key to solving the problem. A warm-up is first in order.
The next few problems might seem simple, but do not overlook their significance.
You should try each problem and compare your answers.
Problems
1.
2·0
2.
0 · 11
3.
π·0
4.
0·0
5.
a·0
6.
0·b
7. (x + 1) · 0
8.
ab = 0
Solutions
2·0 =0
0 · 11 = 0
π·0 =0
0·0 =0
a·0 =0
0·b=0
(x + 1) · 0 = 0
a = 0 or, b = 0
The point is that if the product of any two numbers is zero, then either one or
both of the factors must be zero. In problem 1, the factors are 2 and zero. In
problem 2, the factors are 0 and 11, and so on. In problem 7 the factors are
(x + 1) and 0. Notice that (x + 1) is a factor. It is a number, and at this point
it can only be expressed as (x + 1). With this in mind, we can now solve the
following problems.
Example. Solve 2x = 0.
This simple linear equation could be solved by dividing both sides of the equation
by 2, then recognizing that the fraction 02 = 0. Or, based on what we learned
above, we know that at least one of the factors must be zero. Since 2 6= 0, then
x = 0.
Example. Solve 2(x + 1) = 0.
This problem is almost the same as above. Since one of the factors must be
1
zero and 2 6= 0, then x + 1 = 0. Set the factor to zero and solve for x.
x+1 = 0
−1 = −1
x = −1
Example. Solve (x − 5)(2x + 3) = 0.
Recall what was discovered in Problem 8: if ab = 0, then at least one of the
factors must equal zero. The factors are (x − 5) and (2x + 3). This implies that
either x − 5 = 0 or that 2x + 3 = 0. Solve each linear equation. Then,
x−5 = 0
+5 = +5
x = 5
2x + 3
−3
2x
x
= 0
= −3
= −3
= − 32
Thus x = 5, − 32 .
Example. Return to the first problem presented:
Solve x2 − 3x + 2 = 0.
This problem is a sum. Factoring is the process of writing a sum of terms as a
product of factors. Factor the left side of the equation, then apply the rule: if
ab = 0, then at least one factor must equal zero.
Begin the problem by playing the game and factoring x2 − 3x + 2.
Play the game
Solutions.
2
2
−2
−3
−1
−3
x2 − 3x + 2 = 0
x2 − 2x − 1x + 2 = 0
x(x − 2) − 1(x − 2) = 0
(x − 2)(x − 1) = 0 which implies,
2
x − 2 = 0 OR x − 1 = 0
+2 = +2
+1 = +1
x = 2
x = 1
Example. Solve x3 − 2x2 − x + 2 = 0.
Since the equation equals zero and is a polynomial of greater than the first
degree, the only way to solve is to write the sum as a product of factors. The
polynomial has four terms, so factor by grouping.
x3 − 2x2 − x + 2 = 0
x2 (x − 2) − 1(x − 2) = 0
(FT2)
(x − 2)(x2 − 1) = 0
(x − 2)(x + 1)(x − 1) = 0
(FT6)
OR x + 1 = 0 OR x − 1 = 0
−1 = −1
+1 = +1
x = −1
x = 1
x−2 = 0
+2 = +2
x = 2
Thus x = 2, 1, −1. Since the polynomial is of the third degree, it has three
solutions. To check that the solutions are correct. Substitute each solution into
the original equation, as you should try.
Example. Solve 8x2 − 16x + 6 = 0.
8x2 − 16x + 6 = 0
2(4x2 − 8x + 3) = 0
(FT1)
2(2x − 1)(2x − 3) = 0
(FT3)
2 6= 0 OR
2x − 1
+1
2x
x
= 0
= +1
=1
= 21
OR
2x − 3
+3
2x
x
= 0
= +3
= 3
= 23
Thus, x = 21 , 32 . This problem could also be solved by a slightly different approach. The greatest common factor is 2. Divide both sides of the equation by
2, then factor. This route is illustrated below.
3
8x2 − 16x + 6 = 0
0
8x2 − 16x + 6
=
2
2
16x 6
0
8x2
−
+ =
2
2
2
2
2
4x − 8x + 3 = 0
From this last step, proceed as in the examples above, i.e., factor the left side
of the equation, set each factor to zero and solve. You should try this.
Practice Problems Solve each of following equations presented in factored
form.
1. (x − 1)(x − 7) = 0
2. (x + 4)(x − 3) = 0
3. (x + 3)(x − 3) = 0
4. (x − 2)(x − 2) = 0
5. (2x + 5)(x − 4) = 0
6. (3x − 4)(2x + 1) = 0
7. (2x − 3)(x + 1)(x + 6) = 0
8. x(x − 5) = 0
Solve each of the following equations by factoring.
9. x2 + 7x + 12 = 0
10. x2 − 8x + 15 = 0
11. x2 + 3x − 10 = 0
12. 2x2 − 5x − 12 = 0
13. 3x2 − 48 = 0
14. 4x3 − 22x2 − 12x = 0
15. x3 + 3x2 − 4x − 12 = 0
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Solutions to Practice Problems
1. (x − 1)(x − 7) = 0 =⇒ x − 1 = 0 or x − 7 = 0 =⇒ x = 1, 7
2. (x + 4)(x − 3) = 0 =⇒ x + 4 = 0 or x − 3 = 0 =⇒ x = −4, 3
3. (x + 3)(x − 3) = 0 =⇒ x + 3 = 0 or x − 3 = 0 =⇒ x = −3, 3
4. (x − 2)(x − 2) = 0 =⇒ x − 2 = 0 or x − 2 = 0 =⇒ x = 2
5. (2x + 5)(x − 4) = 0 =⇒ 2x + 5 = 0 or x − 4 = 0 =⇒ x = − 25 , 4
6. (3x − 4)(2x + 1) = 0 =⇒ 3x − 4 = 0 or 2x + 1 = 0 =⇒ x = 43 , − 21
7. (2x − 3)(x + 1)(x + 6) = 0 =⇒ 2x − 3 = 0 or x + 1 = 0 or x + 6 = 0
=⇒ x = 32 , −1, −6
8. x(x − 5) = 0 =⇒ x = 0 or x − 5 = 0 =⇒ x = 0, 5
Note: x is a factor and can be thought of as (x + 0),
but for simplicity, it is written as x.
9. x2 + 7x + 12 = 0 =⇒ (x + 4)(x + 3) = 0 =⇒ x = −4, −3
10. x2 − 8x + 15 = 0 =⇒ (x − 3)(x − 5) = 0 =⇒ x = 3, 5
11. x2 + 3x − 10 = 0 =⇒ (x + 5)(x − 2) = 0 =⇒ x = −5, 2
12. 2x2 − 5x − 12 = 0 =⇒ (x − 4)(2x + 3) = 0 =⇒ x = 4, − 23
13. 3x2 − 48 = 0 =⇒ x2 − 16 = 0 =⇒ (x + 4)(x − 4) = 0
=⇒ x = −4, 4
14. 4x3 − 22x2 − 12x = 0 =⇒ 2x3 − 11x2 − 6x = 0
=⇒ x(x − 6)(2x + 1) = 0 =⇒ x = 0, 6, − 21
15. x3 + 3x2 − 4x − 12 = 0 =⇒ x2 (x + 3) − 4(x + 3) = 0
=⇒ (x + 3)(x + 2)(x − 2) = 0 =⇒ x = −3, −2, 2
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