Solving Quadratic (and polynomial) Equations by Factoring

Solving Quadratic (and polynomial)

Equations by Factoring

For any real number a and b, if ab=0 then either a = 0, b = 0, or both equal 0.

Examples:

When solving polynomial equations, factor the expression and set each factor equal to zero.

Zero Product Property

4(0) 00(25) 0

2 358 (0) 0a b

( 3)(2 1) 0x x

In this example, the equation is already factored and is set equal to zero. To solve, apply the Zero Product Property by setting each factor equal to zero.

3 0x ( 3)(2 1) 0x x 2 1 0x If then and

Solve the factored equations1

2x 3x or

The solutions are x = -3 and x = ½ or {x | x = -3, ½}

Solving Quadratic (and polynomial) Equations

Subtract 3 Add 1, Divide by 2

29 4 0x Solve the following quadratic equation. Start by factoring the quadratic expression.

Factor using difference of squares

Set each factor equal to zero and solve each equation.

3 2x 3 2x

or

2

3x

2

3x or

-2 -2 +2 +2

(3 2)(3 2) 0x x

3 2 0x 3 2 0x


Solve the following quadratic equation. Notice this equations is not in standard form. Set the equation equal to zero by subtracting on both sides of this equation.

2 27 6x x 2 6 27 0x x

Factor the trinomial.

( 9)( 3) 0x x

9x 3x or


Where would the go so the

polynomial is ready to be factored?


Solve the following quadratic equations.

2

2

2

(3 5)(2 7) 0

12 0

2 8 0

3 15 0

4 25

x x

x x

x x

x

x

1)

2)

3)

4)

5)

5

34

0

5

5

2

x

x

x

x

x

7

23

4

5

2

x

x

x

x

and

and

and

and

Linear equations have only 1 solution

Let’s look at some of the graphs….

x2 + x -12 = 0 2x2 + 8x = 0 4x2 = 25

𝑥=− 4 ;3 𝑥=− 4 ; 0 𝑥=±52

Solve the following Cubic Equation

3 25 6 0x x x 2( 5 6) 0x x x

Factor out the GCF.

Factor the trinomial.

( 3)( 2) 0x x x

Don’t forget about your

0x 3 0x 2 0x

0x 3x 2x

oror

or or

Check out the graph…..3 25 6 0x x x

0x 3 0x 2 0x 0x 3x 2x

oror

or or

Solve the Quartic Equation

Put it in standard form

Factor – how low can you go?

We’re not done yet!!

4Solve: 4 64x

44 64 0x 4( 1 ) 04 6x 2 2(( 44)4 0)x x

2( 2)( 2 4 0)4 )(xx x

Set each factor equal to zero.

2( 2)( 2 4 0)4 )(xx x


( ) 02x ( ) 02x

2( ) 04x

2x 2x

2 4

4 2

x

x i

Why don’t we need to set the factor of 4

equal to zero?

Now solve each factor for x.

Solving by factoring, we found the zeros of 4x4 = 64 to be x = ±2, ±2i.

When factoring we can sometimes find imaginary solutions along with the real ones.

Notice – what degree was our polynomial?

How many solutions did we find?


4th

4!!

Roots of Equations In the last example we found two types

of roots – real and imaginary. Real roots can be seen on the graph –

the curve crosses the x-axis at those values.

Imaginary roots cannot been seen on the graph. Many functions have imaginary solutions.

For polynomial functions, the degree tells you the total number of roots – real and imaginary combined!

Roots of Equations

Thus a cubic equation should have how many roots?

A quintic equation should have how many roots?

Remember, all the roots may not be REAL, some could be imaginary.

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5

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Solving Quadratic (and polynomial) Equations by Factoring