Polynomial equations
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Polynomial Equations. Whenever two polynomials are set equal to each other, the result is a polynomial equation. In this section we learn how to solve polynomial equations both graphically and algebraically by factoring. Solving Polynomial Equations Graphically. Example.

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Polynomial Equations

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Polynomial equations

Polynomial Equations

Whenever two polynomials are set equal to each other, the result is a polynomial equation.

In this section we learn how to solve polynomial equations both graphically and algebraically by factoring.

Solving Polynomial Equations Graphically


Example

Example

Find the zeros of the function given by f(x) = x3 – 2x2 – 5x + 6.

Solution

Graph the equation, choosing a window that shows the x-intercepts

of the graph. This may require several attempts.

To find the zeros use the ZERO option from the CALC menu.


Continued f x x 3 2 x 2 5 x 6

continued f(x) = x3 – 2x2 – 5x + 6

Use the same procedure for the other two zeros.

Hence, the zeros/roots are x = -2, x = 1, and x = 3.


Example1

Example

Solution


Example2

(4, 0)

(3, 0)

Solving Polynomial Equations Algebraically

Example

Solve: (x – 4)(x + 3) = 0.

Solution

According to the principle of zero products, at least one factor must be 0.

x – 4 = 0 or x + 3 = 0

x = 4 or x = 3

For 4:For 3:

(x – 4)(x + 3) = 0 (x – 4)(x + 3) = 0

(4 – 4)(4 + 3) = 0 (3 – 4)(3 + 3) = 0

0(7) = 0 0(7) = 0

0 = 0 TRUE 0 = 0 TRUE


Terms with common factors gcf

Terms with Common Factors (GCF)

When factoring a polynomial, we look for factors common to every term and then use the distributive law.

MultiplyFactor

4x(x2 + 3x 4) 4x3 + 12x2  16x

= 4xx2 + 4x3x  4x4= 4xx2 + 4x3x  4x4

= 4x3 + 12x2  16x= 4x(x2 + 3x 4)

Example

Factor: 28x6 + 32x3.

Solution

The prime factorization of 28x6 is

2  2  7  x  x  x  x  x  x

The prime factorization of 32x3 is

2  2  2  2  2  x  x  x

The greatest common factor is 2  2  x  x  x or 4x3.

28x6 + 32x3 = 4x3 7x3 + 4x3  8

= 4x3(7x3 + 8)


Polynomial equations

Example Factor: 12x5 21x4 + 24x3

Solution

The prime factorization of 12x5 is

2  2  3  x  x  x  x  x

The prime factorization of 21x4 is

3  7  x  x  x  x

The prime factorization of 24x3 is

2  2  2  3  x  x  x

The greatest common factor is 3  x  x  x or 3x3.

12x5 21x4 + 24x3 = 3x3 4x2  3x3  7x + 3x3  8

= 3x3(4x2 7x + 8)


Polynomial equations

Example

Solution


Polynomial equations

Factoring by Grouping

Example

Write an equivalent expression by factoring.

a) 3x3 + 9x2 + x + 3b) 9x4 + 6x 27x3  18

Solution

a) 3x3 + 9x2 + x + 3 = (3x3 + 9x2) + (x + 3)

= 3x2(x + 3) + 1(x + 3)

= (x + 3)(3x2 + 1)

Don’t forget to include the 1.

b) 9x4 + 6x 27x3  18

= (9x4 + 6x) + (27x3  18)

= 3x(3x3 + 2) + (9)(3x3 + 2)

= (3x3 + 2)(3x  9)

= (3x3 + 2)3(x  3)

= 3(3x3 + 2)(x  3)


Factoring and equations

Factoring and Equations

Example Solve: 7x2 = 35x.

Solution

Use the principle of zero products if there is a 0 on one side of the equation and the other side is in factored form.

7x2 = 35x

7x2 – 35x = 0 Subtracting 35x. One side is now 0.

7x(x – 5) = 0Factoring

x = 0 or x – 5 = 0 Use the principle of zero products

x = 0 or x = 5

To Use the Principle of Zero Products

1. Write an equivalent equation with 0 on one side, using the addition principle.

2. Factor the polynomial completely.

3. Set each factor that is not a constant equal to 0.

4. Solve the resulting equations.


Polynomial equations

To Use the Principle of Zero Products

1. Write an equivalent equation with 0 on one side, using the addition principle.

2. Factor the polynomial completely.

3. Set each factor that is not a constant equal to 0.

4. Solve the resulting equations.


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