Transcript Solving Polynomial Equations

Solving Polynomial
Equations
PPT 5.3.2
Factor Polynomial Expressions
In the previous lesson, you factored
various polynomial expressions. Refer to 5.2.2
Common
Factor
in Lesson 2 to
review which
strategy is
required for
each question.
Such as:
2(x – 2)
3
2
x
x – 2x =
3 – x2 – 3x + 3)
4
3
2
x(x
x – x – 3x + 3x =
2(x – 1) – 3(x – 1)]
x[x
=
2 – 3)(x – 1)
x(x
=
Solving Polynomial Equations
The expressions on the previous slide are
now equations:
y = x3 – 2x2 and y = x4 – x3 – 3x2 +3x
To solve these equations, we will be
solving for x when y = 0.
Solve
y = x3 – 2x2
0 = x3 – 2x2
0 = x2(x – 2)
Let y = 0
Common factor
Separate the factors and
set them equal to zero.
x2 = 0 or x – 2 = 0
x=0 x=2
Solve for x
Therefore, the roots are 0 and 2.
Solve
y = x4 – x3 – 3x2 + 3x
0 = x4 – x3 – 3x2 + 3x
0 = x(x3 – x2 – 3x + 3)
0 =x[x2(x – 1) – 3(x – 1)]
0 = x(x – 1)(x2 – 3)
Let y = 0
Common factor
Group
Separate the factors and set them
equal to zero.
x = 0 or x – 1 = 0 or x2 – 3 = 0
x=0
x=1
x= 3
Therefore, the roots are 0, 1
and ±1.73
Solve for x
What are you solving for?
In the last two slides we solved for x when
y = 0, which we call the roots. But what
are roots?
If you have a graphing calculator follow
along with the next few slides to discover
what the roots of an equation represent.
What are roots?
Press the Y= button on your calculator.
Type x3 – 2x2
Press the GRAPH button.
Look at where the graph is crossing the x-axis.
The x-intercepts are 0 and 2.
If you recall, when we solved for the roots of the
equation y = x3 – 2x2, we found them to be 0 and
2. Don’t forget, we also put 0 in for y, so it
makes sense that the roots would be the
x-intercepts.
Use your graphing calculator to graph the
other equation we solved,
y = x4 – x3 – 3x2 + 3x
As you would now expect, the roots that
we found earlier, 0, 1 and ±1.73, are in fact
the x-intercepts of the graph.
The Quadratic Formula
 b  b 2  4ac
x
2a
where a  0
For equations in quadratic form: ax2 + bx + c = 0, we can
use the quadratic formula to solve for the roots of the
equation.
This equation is normally used when factoring is not an
option.
Using the Quadratic Formula
Solve the following cubic equation:
Can thisWe
equation
still need to solve for x
y = x3 + 5x2 – 9x
be factored?
here. Can this equation be
2
0=
x(xit can
+ 5x– – 9)
YES
factored?
factor.
x =common
0
x2 + 5x – 9 = 0
No. There
are nouse
twothe quadratic formula.
We can,
however,
integers that will multiply
a=1
 (
5)  to( 5.
5)2  4(1)(9)
to -9xand
add

b=5
c = -9
(2)(1)
5  61
2
x  6.41,  1.41
x
Remember, the
root 0 came from
an earlier step.
Therefore, the roots are 0, 6.41
and -1.41.