Rational Root TheoremPossible Rational Roots

What are the possible rational roots of

Factors of the constant term, 6, are

Factors of the lead coefficient, are

Possible rational roots are

±𝟏

¿±𝟏 , ±𝟐 ,±𝟑 , ±𝟔

Possible Rational Roots

±𝟐

¿±𝟏𝟐

, ±𝟏 , ±𝟐 , ±𝟒

What are the possible rational roots of

Factors of the constant term, 8, are

Possible rational roots are

Factors of the lead coefficient, 2, are

Always keep in mind the relationship among zeros, roots, and x-intercepts. The zeros of a function are the roots, or solutions of the equation . The real zeros, or real roots, are the x-intercepts of the graph of .

Finding Zeros of a Polynomial FunctionUse the Rational Zero Theorem to find all possible

rational zeros.Use Synthetic Division to try to find one rational

zero (the remainder will be zero).If “n” is a rational zero, factor the original

polynomial as (x – n)q(x).

Test remaining possible rational zeros in q(x). If one is found, then factor again as in the previous step.

Continue in this way until all rational zeros have been found.

See if additional irrational or non-real complex zeros can be found by solving a quadratic equation.

Finding Rational Zeros

So which one do you pick?

Pick any. Find one that is a zero using synthetic division...

Possible zeros are + 1, + 2, + 4, + 8

Find the rational zeros for

Find

Let’s try 1. Use synthetic division

1 1 1 –10 8 1 2 –8

1 2 –8 0 1 is a zero of the function

The depressed polynomial is x2 + 2x – 8Find the zeros of x2 + 2x – 8 by factoring or (by using the quadratic formula)…

(x + 4)(x – 2) = 0 x = –4, x = 2

The zeros of f(x) are 1, –4, and 2

Find all real zeros of

:isfactor another andfactor a is 1) (x zero, a is 1-

2-x-x-x-x 234

021111

21111

2322011

02322 235 xxxx

The possible rational zeros are

Use synthetic division

Find all possible rational zeros of:

Example ContinuedThis new factor has the

same possible rational zeros: Check to see if -1 is also a zero of this:

Conclusion:02121

2121

211111

2-x-x-x-x 234

2 and 1

:isfactor another andfactor a is 1) (x zero, a is 1-

2-x2x-x 23

Example ContinuedThis new factor has as possible

rational zeros: Check to see if -1 is also a zero of this:

Conclusion:6431

431

21211

2 and 1

:zero possibleanother try so zero, a NOT is 1- 1

2-x2x-x 23

Example ContinuedCheck to see if 1 is a zero:

Conclusion:

2011

011

21211

:zero possibleanother try so zero, a NOT is 1- 2

Example ContinuedCheck to see if 2 is a zero:

Conclusion:

0101

202

21212

:isfactor another andfactor a is 2) (x zero, a is 2

1x2

Example ContinuedSummary of work done:

012 x

2322 235 xxxxxf

121 22 xxxxf

ix

x

x

1

012

2

1 (double) 2Distinct zeros : - , , i, - i

is a zero of multiplicity two; 2 is a zero; and the other two zeros can be found by solving:

Using The Linear Factorization Theorem

Find a 4th degree polynomial function with real coefficients that has as zeros and such that .

Solution: Because is a zero , the conjugate, , must also be a zero.

We can now use the Linear Factorization Theorem for a fourth-degree polynomial.

𝑓 (𝑥 )=𝑎𝑛(𝑥−𝑐1)(𝑥−𝑐2)(𝑥−𝑐3)(𝑥−𝑐4)

¿𝑎𝑛(𝑥+2)(𝑥− 2)(𝑥−𝑖)(𝑥+𝑖)

¿𝑎𝑛(𝑥2− 4 )(𝑥2+1)

Using The Linear Factorization Theorem¿𝑎𝑛(𝑥

2− 4 )(𝑥2+1)

𝑓 (𝑥 )=𝑎𝑛(𝑥4 −3 𝑥2− 4)

𝑓 (3 )=𝑎𝑛 (34 −3 ∙ 32 − 4 )=−150

¿𝑎𝑛 (81−27− 4 )=−150

50𝑎𝑛=−150

𝑎𝑛=−3

Substituting for in the formula for , we obtain

𝑓 (𝑥 )=−3 (𝑥4 − 3 𝑥2 − 4)

𝑓 (𝑥 )=−3 𝑥4+9𝑥2+12

Descartes Rule of Signs is a method for determining the number of sign changes in a polynomial function.

Polynomial Function Sign Changes

Conclusion

\⁄ \⁄ \⁄

3 There are 3 positive real zeros, or positive real zero.

\⁄ \⁄

2 There are 2 positive real zeros, or positive real zeros.

\⁄1 There is one positive

real zero.

Descarte’s Rule of Signs and Positive Real Zeros

How do we determine the possible number of negative answers?

We substitute for every x-value in the equation.

Then we look for the sign changes.

Descarte’s Rule of SignsExample Determine the possible number of positive real zeros and negative real zeros of P(x) = x4 – 6x3 + 8x2 + 2x – 1.

We first consider the possible number of positive zeros by observing that P(x) has three variations in signs.

+ x4 – 6x3 + 8x2 + 2x – 1

Thus, by Descartes’ rule of signs, f has either 3 or 3 – 2 = 1 positive real zeros.

For negative zeros, consider the variations in signs for P(x).P(x) = (x)4 – 6(x)3 + 8(x)2 + 2(x) 1

= x4 + 6x3 + 8x2 – 2x – 1

Since there is only one variation in sign, P(x) has only one negative real root.

1 2 3

Total number of zeros 4Positive: 3 1Negative: 1 1Nonreal: 0 2