Review 4.1-4.4Pre-Calculus

Factors, Roots, Zeros

y x2 2x 15For our Polynomial Function:

The Factors are: (x + 5) & (x - 3)

The Roots/Solutions are: x = -5 and 3

The Zeros are at: (-5, 0) and (3, 0)

Solving a Polynomial Equation

The only way that x2 +2x - 15 can = 0 is if x = -5 or x = 3

Rearrange the terms to have zero on one side:

x2 2x 15 x2 2x 150 Factor:

(x 5)(x 3) 0 Set each factor equal to zero

and solve: (x 5) 0 and (x 3)0 x 5 x 3

y x2 2x 15

x-Intercepts of a Polynomial

The points where y = 0 are called the x-intercepts of the graph.

The x-intercepts for our graph are the points...

and(-5, 0) (3, 0)

Real/Imaginary Roots

If a polynomial has ‘n’ complex roots will its graph have ‘n’ x-intercepts?

In this example, the degree n = 3, and if we factor the polynomial, the roots are x = -2, 0, 2. We can also see from the graph that there

are 3 x-intercepts.

y x3 4x

Find Roots/Zeros of a Polynomial

We can find the Roots or Zeros of a polynomial by setting the polynomial equal to 0 and factoring.

Some are easier to factor than others!

f (x) x3 4x

x(x2 4)x(x 2)(x 2)

The roots are: 0, -2, 2

Real/Imaginary Roots

Just because a polynomial has ‘n’ complex roots doesn’t mean that they are all Real!

y x3 2x2 x 4In this example, however, the degree is still n = 3, but there is only one Real x-intercept or root at x = -1, the other 2 roots must have imaginary components.

Find the zeros of 2f x x 2 x 3 x 1 x 4

2x 2 0 x 3 0 x 1 0 x 4 0

2, -3 (d.r), 1, -4

EXAMPLE 1

Solve by factoring.

Answer: The solution set is {0, –4}.

Original equation

Add 4x to each side.

Factor the binomial.

Solve the second equation.

Zero Product Propertyor

Solve by factoring.

Original equation

Subtract 5x and 2 from each side.Factor the trinomial.Zero Product PropertyorSolve each equation.

Answer: The solution set is Check each solution.

Solve each equation by factoring. a.

b.

Answer: {0, 3}

Answer:

Answer: The solution set is {3}.

Solve by factoring.

Original equation

Add 9 to each side.

Factor.

Zero Product Propertyor

Solve each equation.

CheckThe graph of the related function, intersects the x-axis only once. Since the zero of the function is 3, the solution of the related equation is 3.

Answer: {–5}

Solve by factoring.

Factor Theorem

1. If f(c)=0, that is c is a zero of f, then x - c is a factor of f(x).

2. Conversely if x - c is a factor of f(x), then f(c)=0.

Example – Find the roots/zeros for this polynomial 3 210 9 19 6x x x

Given -2 is a zero, use synthetic division to find the remaining roots.

-2 10 9 -19 6Don’t forget your remainder should be zero

-20 22 -6

10 -11 3 0

The new, smaller polynomial is:

210 11 3x x

This quadratic can be factored into: (5x – 3)(2x – 1)

Therefore, the zeros to the problem

3 12, ,5 2

x

Example: Find all the zeros of each

polynomial function3 210 9 19 6x x x If we were to graph

this equation to check you could see the zero

From looking at the graph you can see that there is a zero at -2

ZERO

ZEROZERO

We have already known that the possible rational zeros are:

Find All the Rational Zeros

EXAMPLE 3

So 1, 2, -2, and -3 are rational roots

Example:The zeros of a third-degree polynomial are 2, 2, and -5. Write a polynomial.

(x – 2)(x – 2)(x+5)First, write the zeros in factored form

Second, multiply the factors out to find your polynomial

3 2 16 20x x x ANSWER

Given 1 ± 3i and 2 are roots, find the remaining roots.

ixixx 31312 Multiply the last two factors together. All i terms should disappear when simplified. 22 9333132 iixiixxixxx

-1 1022 2 xxx Now multiply the x – 2 through

20144 23 xxxHere is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i

If x = the root then x - the root is the factor

form. ixixx 31312

EXAM

PLE

5

f(x)= (x-1)(x-(-2+i))(x-(-2-i)) f(x)= (x-1)(x+2 - i)(x+2+ i) f(x)= (x-1)[(x+2) - i] [(x+2)+i] f(x)= (x-1)[(x+2)2 - i2] Foil f(x)=(x-1)(x2 + 4x + 4 – (-1)) Take care of

i2 f(x)= (x-1)(x2 + 4x + 4 + 1) f(x)= (x-1)(x2 + 4x + 5) Multiply f(x)= x3 + 4x2 + 5x – x2 – 4x – 5 f(x)= x3 + 3x2 + x - 5

Now write a polynomial function of least degree that has real coefficients, a leading coeff. of 1 and 1, -2+i, -2-i as zeros.

Note: 2+i means 2 – i is also a zero F(x)= (x-4)(x-4)(x-(2+i))(x-(2-i)) F(x)= (x-4)(x-4)(x-2-i)(x-2+i) F(x)= (x2 – 8x +16)[(x-2) – i][(x-2)+i] F(x)= (x2 – 8x +16)[(x-2)2 – i2] F(x)= (x2 – 8x +16)(x2 – 4x + 4 – (– 1)) F(x)= (x2 – 8x +16)(x2 – 4x + 5) F(x)= x4– 4x3+5x2 – 8x3+32x2 – 40x+16x2 –

64x+80 F(x)= x4-12x3+53x2-104x+80

Now write a polynomial function of least degree that has real coefficients, a leading coeff. of 1 and 4, 4, 2+i as zeros.

Conjugate Pairs Complex Zeros Occur in Conjugate Pairs = If a +

bi is a zero of the function, the conjugate a – bi is also a zero of the function (the polynomial function must have real coefficients)

EXAMPLES: Find a polynomial with the given zeros

-1, -1, 3i, -3i

2, 4 + i, 4 – i

Find Roots/Zeros of a Polynomial

If we cannot factor the polynomial, but know one of the roots, we can divide that factor into the polynomial. The resulting polynomial has a lower degree and might be easier to factor or solve with the quadratic formula.

We can solve the resulting polynomial to get the other 2 roots:

f (x) x3 5x2 2x 10one root is x 5

x 5 x3 5x2 2x 10x3 5x2

2x 10 2x 10 0

x2 2

(x - 5) is a factor

x 2, 2

EXAMPLE: Solving a Polynomial Equation

Solve: x4 6x2 8x + 24 0.Solution Now we can solve the original equation as follows.

x4 6x2 8x + 24 0 This is the given equation.

(x – 2)(x – 2)(x2 4x 6) 0

This was obtained from the second synthetic division.

x – 2 0 or x – 2 0 or x2 4x 6 0 Set each factor equal to

zero. x 2 x 2 x2 4x 6 0 Solve.

The solution set of the original equation is {2, 2 – i, 2 i }.

2,i 2i

Ise Quadratic Formula

FIND ALL THE ZEROS

f (x) x 4 3x 3 6x 2 2x 60

(Given that 1 + 3i is a zero of f)

f (x) x 3 7x 2 x 87

(Given that 5 + 2i is a zero of f)

More Finding of Zeros

f (x) x 5 x 3 2x 2 12x 8

f (x) 3x 3 4x 2 8x 8

FIND ALL RATIONAL ROOTS:

EXAMPLE: Using the Rational Zero TheoremList all possible rational zeros of f (x) 15x3 14x2 3x – 2.Solution The constant term is –2 and the leading coefficient is 15.

1 2 1 2 1 25 53 3 15 15

Factors of the constant term, 2Possible rational zeros Factors of the leading coefficient, 151, 2

1, 3, 5, 15

1, 2, , , , , ,

Divide 1 and 2 by 1.

Divide 1 and 2 by 3.

Divide 1 and 2 by 5.

Divide 1 and 2 by 15.

There are 16 possible rational zeros. The actual solution set to f (x) 15x3 14x2 3x – 2 = 0 is {-1, 1/3, 2/5}, which contains 3 of the 16 possible solutions.

EXAMPLE: Solving a Polynomial EquationSolve: x4 6x2 8x + 24 0.Solution Recall that we refer to the zeros of a polynomial function and the roots of a polynomial equation. Because we are given an equation, we will use the word "roots," rather than "zeros," in the solution process. We begin by listing all possible rational roots.

Factors of the constant term, 24Possible rational zeros Factors of the leading coefficient, 11, 2 3, 4, 6, 8, 12, 24

11, 2 3, 4, 6, 8, 12, 24

FIND ALL POSSIBLE RATIONAL ROOTS

Descartes’ Rule of Signs

Arrange the terms of the polynomial P(x) in descending degree:

• The number of times the coefficients of the terms of P(x) change sign = the number of Positive Real Roots (or less by any even number)

• The number of times the coefficients of the terms of P(-x) change sign = the number of Negative Real Roots (or less by any even number)

In the examples that follow, use Descartes’ Rule of Signs to predict the number of + and - Real Roots!

EXAMPLE: Using Descartes’ Rule of SignsDetermine the possible number of positive and negative real zeros of f (x) x3 2x2 5x + 4.Solution1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f (x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f (x). We obtain this equation by replacing x with x in the given function.

f (x) (x)3 2(x)2 5x 4

f (x) x3 2x2 5x + 4 This is the given polynomial function.

Replace x with x.

x3 2x2 5x + 4

3.4: Zeros of Polynomial Functions

EXAMPLE: Using Descartes’ Rule of SignsDetermine the possible number of positive and negative real zeros of f (x) x3 2x2 5x + 4.SolutionNow count the sign changes.

f (x) x3 2x2 5x + 4

1 2 3

There are 0 positive roots, and 3 negative roots

Find the Roots of a Polynomial

For higher degree polynomials, finding the complex roots (real and imaginary) is easier if we know one of the roots.

Descartes’ Rule of Signs can help get you started. Complete the table below:

List the Possible Rational Roots

For the polynomial:

All possible values of:

All possible Rational Roots of the form p/q:

Narrow the List of Possible Roots

For the polynomial:

Descartes’ Rule:

All possible Rational Roots of the form p/q:

Find a Root That Works

For the polynomial:

Substitute each of our possible rational roots into f(x). If a value, a, is a root, then f(a) = 0. (Roots are solutions to an equation set equal to zero!)

Descartes’s Rule of Signs

EXAMPLES: describe the possible real zeros

f (x) 3x 3 5x 2 6x 4

f (x) 3x 3 2x 2 x 3

Find the zeros of 3 2f x 2x x 13x 6 Hint: 2 is a zero

X

2 2 1 -13 6

2

4

5

10

-3

-6

0

22x2 3 0x 5x

2x 1 x 3 0x 2

1x 2, , 32

Find the zeros of 3f x x 11x 20 Hint: 4 is a zero

X

4 1 0 -11 -20

1

4

4

16

5

20

0

2x 4x 5 0x 4

x 4, 2 i, 2 i

4 16 4 1 52

4 42

2 i, 2 i

Find the zeros of 3 2f x x 2x 3x 6 Hint: 2 is a zero

X

2 1 -2 -3 6

1

2

0

0

-3

-6

0

2 32 xx 0

2x 3

x 2, 3, 3

x 3, 3

Find the zeros of 3 2f x x 5x 2x 24 Hint: 2 is a zero

X

2 1 5 -2 -24

1

2

7

14

12

24

0

2x 7 12 0x x2

x 4 x 3 0x 2

x 2, 4, 3

No Calculator

Given –2 is a zero of 3 2f x x 2x 5x 6, find ALL the zeros of the function.

-2 1 -2 -5 6

1

-2

-4

8

3

-6

0

2x 4x 3 0x 2

x 1 x 3 0x 2

x 2,1, 3

No Calculator

Given 5 is a zero of 4 3 2f x x 5x 4x 20x, find ALL the zeros of the function.

5 1 -5 -4 20

1

5

0

0

-4

-20

0

2x 45 xx 0

x 2 x 2x 0x 5

x 5,0, 2, 2

No constant, so 0 is a zero: 3 2f x x x 5x 4x 20

No Calculator

Given -1 and 3 are zeros of 4 3 2f x x 9x 23x 3x 36, find ALL the zeros of the function.

3

1

3

-7

-21

12

36

0 2x 1 x 3 x 7x 12 0

x 4 x 3x 1 x 3 0

x 1, 3 d.r. , 4

-1 1 -9 23 -3 -36

1

-1

-10

10

33

-33

-36

36

0

No Calculator

Given is a zero of 3 2f x 2x 9x 13x 6, find ALL the zeros of the function.

32

2

3

-6

-9

4

6

0

2x 03 22x 3x

x 1 x 2x 02 3

3x ,1, 22

/2 2 -9 13 -63

1 -3 2

No Calculator

Given is a zero of 3 2f x 3x 8x 5x 6, find ALL the zeros of the function.

23

3

2

-6

-4

-9

-6

0

2x 02 33x 2x

x 3 x 1x 03 2

2x , 3, 13

/3 3 -8 -5 62

1 -2 -3

No Calculator

Given 2 is a zero of 3 2f x x 6x 13x 10, find ALL the zeros of the function.

2 1 -6 13 -10

1

2

-4 5

10

0

2x 4x 5 0x 2

x 2, 2 i, 2 i

4 16 4 1 52

4 42

2 i, 2 i

-8

No Calculator

Given –3 is a zero of 3 2f x x 3x x 3, find ALL the zeros of the function.

-3 1 3 1 3

1

-3

0

0

1

-3

0

2 13 xx 0

2x 1

x 3, i, i

x i, i

No Calculator

Find a polynomial function with real coefficients which has zeros of 1, -2, and 3.

f x x 1 x 2 x 3

2xf x xx 1 6

2 2x x 6f x x 1 x x 6

3 2 2f x x x 6x x x 6

3 2f x x 2x 5x 6

No Calculator

Find a polynomial function with real coefficients which has zeros of 0, 2, -2, and 5.

f x x x 2 x 2 x 5

2xf x x 0x 2 1x 3

2 2x 3x 10 x 3x 1x 0f x x 2

3 2 2f x x 3x 10x 2x 6x 20x

4 3 2f x x 5x 4x 20x

3 2f x x 5x 20x 4x

No Calculator

Find a polynomial function with real coefficients which has zeros of 3/2, 2, and 1.

f x 2x 3 x 2 x 1

2xf x 3 22 3 xx

2 2x 3f x 2x x 2 3x 23 x

3 2 2f x 2x 6x 4x 3x 9x 6

3 2f x 2x 9x 13x 6

No Calculator

Find a polynomial function with real coefficients which has zeros of 2 and i.

f x x 2 x i x i

2f x x 2 x 1

2 2x 1f x x 2 x 1

3 2f x x x 2x 2

3 2f x x 2x x 2

If i is a root, then –i is a root as well

No Calculator

Find a polynomial function with real coefficients which has zeros of 1 and 2 + i.

f x x 1 x 2 i x 2 i

2 2x 2f x x 1 i

2f x x 1 x 4x 4 1

2 2x 4x 5 xx x 5f 4x 1

3 2f x x 5x 9x 5

If 2 + i is a root, then 2 – i is a root as well

2xf x 4x 1 x 5

3 2 2f x x 4x 5x x 4x 5