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Solving Polynomial Equations
Lesson 6-4
Additional Examples
Graph and solve x3 – 19x = –2x2 + 20.
Step 1: Graph y1 = x3 – 19x and y2 = –2x2 + 20
on a graphing calculator.
Step 2: Use the Intersect feature to find the
x values at the points of intersection.
The solutions are –5, –1, and 4.
Algebra 2
Solving Polynomial Equations
Lesson 6-4
Algebra 2
Additional Examples
(continued)
Check: Show that each solution makes the original equation a true statement.
x3 – 19x = –2x2 + 20
(–5)3 – 19(–5) –2(–5)2 + 20
–125 + 95 –50 + 20
–30 = –30
x3 – 19x = –2x2 + 20
(4)3 – 19(4) –2(4)2 + 20
64 – 76 –32 + 20
–12 = –12
x3 – 19x = –2x2 + 20
(–1)3 – 19(–1) –2(–1)2 + 20
–1 + 19
–2 + 20
18 = 18
Sum/Difference of Cubes
Algebra 2
Solving Polynomial Equations
Lesson 6-4
Algebra 2
Additional Examples
Factor x3 – 125.
x3 – 125 = (x)3 – (5)3
Rewrite the expression
as the difference of cubes.
= (x – 5)(x2 + 5x + (5)2)
Factor.
= (x – 5)(x2 + 5x + 25)
Simplify.
Example #1: x3 – 8
Example #2: x3 + 8
Example #3: 8x3 – 1
Example #4: 8x3 +1
Solving Polynomial Equations
Lesson 6-4
Algebra 2
Additional Examples
Solve 8x3 + 125 = 0. Find all complex roots.
8x3 + 125 = (2x)3 + (5)3
Rewrite the expression
as the difference of cubes.
= (2x + 5)((2x)2 – 10x + (5)2)
Factor.
= (2x + 5)(4x2 – 10x + 25)
Simplify.
Since 2x + 5 is a factor, x = –
5
is a root.
2
The quadratic expression 4x2 – 10x + 25 cannot be factored, so use the
Quadratic Formula to solve the related quadratic equation 4x2 – 10x + 25 = 0.
Solving Polynomial Equations
Lesson 6-4
Algebra 2
Additional Examples
(4x2 – 10x + 25)
(continued)
x=
–b ±
b2 – 4ac
2a
= –(–10) ±
=
– (–10) ±
8
=
10 ± 10i
8
=
5 ± 5i
4
Quadratic Formula
(–10)2 – 4(4)(25)
2(4)
Substitute 4 for a, –10 for b, and 25 for c.
–300
Use the Order of Operations.
3
–1 = 1
3
Simplify.
The solutions are – 5 and 5 ± 5i 3 .
2
4
Solving a Polynomial Equation
Algebra 2
Example #4: Solve 27x3 +1 = 0. Find all complex roots.
1.
2.
3.
4.
Rewrite equation as a sum of cubes
Factor
Simplify
Find complex roots
Solving Polynomial Equations
Lesson 6-4
Algebra 2
Additional Examples
Factor x4 – 6x2 – 27.
Step 1: Since x4 – 6x2 – 27 has the form of a quadratic expression, you can
factor it like one. Make a temporary substitution of variables.
x4 – 6x2 – 27 = (x2)2 – 6(x2) – 27
= a2 – 6a – 27
Rewrite in the form of a
quadratic expression.
Substitute a for x2.
Step 2: Factor a2 – 6a – 27.
a2 – 6a – 27 = (a + 3)(a – 9)
Step 3: Substitute back to the original variables.
(a + 3)(a – 9) = (x2 + 3)(x2 – 9)
= (x2 + 3)(x + 3)(x – 3)
Substitute x2 for a.
Factor completely.
The factored form of x4 – 6x2 – 27 is (x2 + 3)(x + 3)(x – 3).
Solving Polynomial Equations
Lesson 6-4
Algebra 2
Additional Examples
Solve x4 – 4x2 – 45 = 0.
x4 – 4x2 – 45 = 0
(x2)2 – 4(x2) – 45 = 0
Write in the form of a
quadratic expression.
Think of the expression as
a2 – 4a – 45, which factors
as (a – 9)(a + 5).
(x2 – 9)(x2 + 5) = 0
(x – 3)(x + 3)(x2 + 5) = 0
x = 3 or x = –3 or x2 = –5
Use the factor theorem.
x = ± 3 or x = ± i
Solve for x, and simplify.
5
Algebra 2
Homework
Page 324 #12-32 odd