Solving Polynomial Solving Polynomial EquationsEquations

PPT 5.3.2

Factor Polynomial Expressions

In the previous lesson, you factored various polynomial expressions.

Such as:x3 – 2x2 =x4 – x3 – 3x2 + 3x =

= =

Common Factor

x2(x – 2)

x[x2(x – 1) – 3(x – 1)]

x(x2 – 3)(x – 1)

x(x3 – x2 – 3x + 3)

Grouping – common factor the first two terms and then the last two terms.

Common Factor

Refer to 5.2.2 in Lesson 2 to review which strategy is required for each question.

Refer to 5.2.2 in Lesson 2 to review which strategy is required for each question.

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)

x2 = 0 or x – 2 = 0 x = 0 x = 2

Therefore, the roots are 0 and 2.

Let y = 0

Common factor

Separate the factors and set them equal to zero.

Solve for x

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)

x = 0 or x – 1 = 0 or x2 – 3 =

0 x = 0 x = 1 x =

Therefore, the roots are 0, 1 and ±1.73

Let y = 0

Common factor

Separate the factors and set them equal to zero.

Solve for x

3

Group

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

02

42

awherea

acbbx

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:

y = x3 + 5x2 – 9x

0 = x(x2 + 5x – 9)

x = 0 x2 + 5x – 9 = 0

We can, however, use the quadratic formula.

YES it can – YES it can – common factor.common factor.

Can this equation Can this equation be factored?be factored?

We still need to solve for x We still need to solve for x here. Can this equation be here. Can this equation be factored?factored?

No. There are no two No. There are no two integers that will multiply integers that will multiply to -9 and add to 5.to -9 and add to 5.a = 1

b = 5

c = -9

41.1,41.62

615

)1)(2(

)9)(1(4)5()5( 2

x

x

x

Therefore, the roots are 0, 6.41 and -1.41.

Remember, the root 0 came from an earlier step.