Transcript Slide 1

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Precalculus Review
Copyright © Cengage Learning. All rights reserved.
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Solving Polynomial Equations
Copyright © Cengage Learning. All rights reserved.
Solving Polynomial Equations
Polynomial Equation
A polynomial equation in one unknown is an equation that
can be written in the form
axn + bxn – 1 + ··· + rx + s = 0
where a, b, . . . , r, and s are constants.
We call the largest exponent of x appearing in a nonzero
term of a polynomial the degree of that polynomial.
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Solving Polynomial Equations
Quick Examples
1. 3x + 1 = 0 has degree 1 because the largest power of x
that occurs is x = x1. Degree 1 equations are called
linear equations.
2. x2 – x – 1 = 0 has degree 2 because the largest power
of x that occurs is x2. Degree 2 equations are also
called quadratic equations, or just quadratics.
3. x3 = 2x2 + 1 is a degree 3 polynomial (or cubic) in
disguise. It can be rewritten as x3 – 2x2 – 1 = 0, which is
in the standard form for a degree 3 equation.
4. x4 – x = 0 has degree 4. It is called a quartic.
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Solution of Linear Equations
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Solution of Linear Equations
By definition, a linear equation can be written in the form
ax + b = 0.
a and b are fixed numbers with a  0.
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Solution of Quadratic Equations
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Solution of Quadratic Equations
By definition, a quadratic equation has the form
ax2 + bx + c = 0.
a, b, and c are fixed numbers with a  0.
The solutions of this equation are also called the roots of
ax2 + bx + c.
We’re assuming that you saw quadratic equations
somewhere in high school but may be a little hazy about
the details of their solution.
There are two ways of solving these equations—one works
sometimes, and the other works every time.
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Solution of Quadratic Equations
Solving Quadratic Equations by Factoring
(works sometimes)
If we can factor a quadratic equation ax2 + bx + c = 0, we
can solve the equation by setting each factor equal to zero.
Quick Example
x2 + 7x + 10 = 0
(x + 5)(x + 2) = 0
Factor the left-hand side.
x + 5 = 0 or x + 2 = 0
If a product is zero, one or both factors is zero.
Solutions: x = –5 and x = –2
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Solution of Quadratic Equations
Test for Factoring
The quadratic ax2 + bx + c, with a, b, and c being integers
(whole numbers), factors into an expression of the form
(rx + s)(tx + u) with r, s, t, and u integers precisely when the
quantity b2 – 4ac is a perfect square. (That is, it is the
square of an integer.)
If this happens, we say that the quadratic factors over the
integers.
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Solution of Quadratic Equations
Quick Example
x2 + x + 1 has a = 1, b = 1, and c = 1, so b2 – 4ac = –3,
which is not a perfect square. Therefore, this quadratic
does not factor over the integers.
Solving Quadratic Equations with the Quadratic Formula
(works every time)
The solutions of the general quadratic ax2 + bx + c = 0
(a ≠ 0) are given by
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Solution of Quadratic Equations
We call the quantity  = b2 – 4ac the discriminant of the
quadratic ( is the Greek letter delta), and we have the
following general rules:
• If  is positive, there are two distinct real solutions.
• If  is zero, there is only one real solution:
(Why?)
• If  is negative, there are no real solutions.
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Solution of Quadratic Equations
Quick Example
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Solution of Cubic Equations
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Solution of Cubic Equations
By definition, a cubic equation can be written in the form
ax3 + bx2 + cx + d = 0.
a, b, c, and d are fixed numbers and a  0.
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Solution of Cubic Equations
Solving Cubics by Finding One Factor
Start with a given cubic equation ax3 + bx2 + cx + d = 0.
Step 1 By trial and error, find one solution x = s. If
a, b, c, and d are integers, the only possible rational
solutions are those of the form s = ±(factor of d)/(factor of a).
Step 2 It will now be possible to factor the cubic as
ax3 + bx2 + cx + d = (x – s)(ax2 + ex + f ) = 0
To find ax2 + ex + f, divide the cubic by x – s, using long
division.
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Solution of Cubic Equations
Step 3 The factored equation says that either x – s = 0 or
ax2 + ex + f = 0.
We already know that s is a solution, and now we see that
the other solutions are the roots of the quadratic. Note that
this quadratic may or may not have any real solutions, as
usual.
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Solution of Cubic Equations
Quick Example
To solve the cubic x3 – x2 + x – 1 = 0, we first find a single
solution.
Here, a = 1 and d = –1. Because the only factors of ±1 are
±1, the only possible rational solutions are x = ±1. By
substitution, we see that x = 1 is a solution.
Thus, (x – 1) is a factor. Dividing by (x – 1) yields the
quotient (x2 + 1). Thus,
x3 – x2 + x – 1 = (x – 1)(x2 + 1) = 0
so that either x – 1 = 0 or x2 + 1 = 0.
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Solution of Cubic Equations
Because the discriminant of the quadratic x2 + 1 is
negative, we don’t get any real solutions from x2 + 1 = 0,
so the only real solution is x = 1.
Possible Outcomes When Solving a Cubic Equation
If you consider all the cases, there are three possible
outcomes when solving a cubic equation:
1. One real solution (as in the Quick Example above)
2. Two real solutions (try, for example, x3 + x2 – x – 1 = 0)
3. Three real solutions (see the next example)
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Example 1 – Solving a Cubic
Solve the cubic 2x3 – 3x2 – 17x + 30 = 0.
Solution:
First we look for a single solution. Here, a = 2 and d = 30.
The factors of a are 1 and 2, and the factors of d are
1, 2, 3, 5, 6, 10, 15, and 30.
This gives us a large number of possible ratios: 1, 2, 3,
5, 6, 10, 15, 30, 1/2, 3/2, 5/2, 15/2. Undaunted,
we first try x = 1 and x = –1, getting nowhere.
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Example 1 – Solution
cont’d
So we move on to x = 2, and we hit the jackpot, because
substituting x = 2 gives 16 – 12 – 34 + 30 = 0. Thus, (x – 2)
is a factor.
Dividing yields the quotient 2x2 + x – 15. Here is the
calculation:
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Example 1 – Solution
cont’d
Thus, 2x3 – 3x2 – 17x + 30 = (x – 2)(2x2 + x – 15) = 0.
Setting the factors equal to zero gives either x – 2 = 0 or
2x2 + x – 15 = 0.
We could solve the quadratic using the quadratic formula,
but, luckily, we notice that it factors as
2x2 + x – 15 = (x + 3)(2x – 5).
Thus, the solutions are x = 2, x = –3 and x = 5/2.
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Solution of Higher-Order Polynomial
Equations
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Solution of Higher-Order Polynomial Equations
Logically speaking, our next step should be a discussion of
quartics, then quintics (fifth degree equations), and so on
forever. Well, we’ve got to stop somewhere, and cubics
may be as good a place as any.
On the other hand, since we’ve gotten so far, we ought to
at least tell you what is known about higher order
polynomials.
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Solution of Higher-Order Polynomial Equations
Quartics
Just as in the case of cubics, there is a formula to find the
solutions of quartics.
Quintics and Beyond
All good things must come to an end, we’re afraid. It turns
out that there is no “quintic formula.”
In other words, there is no single algebraic formula or
collection of algebraic formulas that gives the solutions to
all quintics.
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