Pc Polynomial Zeros

The Real Zeros of a Polynomial Function

REMAINDER THEOREM Let f be a polynomial function. If f (x) is divided by x – c, then the remainder is f (c). Let’s look at an example to see how this theorem is useful. 3 2

f ( x ) = 2 x − 3x + 2 x − 1

So the remainder we get in synthetic division is the same as the answer we’d get if we put -2 in the function. The root of x + 2 = 0 is x = -2 Find f(-2)

-2

2

-3 2 -1 -4 14 -32

2

-7

16 -33

using synthetic division let’s divide by x + 2 the remainder

f ( − 2 ) = 2( − 2 ) − 3( − 2 ) + 2( − 2 ) − 1 = −33 3

2

FACTOR THEOREM Let f be a polynomial function. Then x – c is a factor of f (x) if and only if f (c) = 0 If and only if means this will be true either way: 1. If f(c) = 0, then x - c is a factor of f(x) Try synthetic 2. If x - c is a factor of f(x) then f(c) = 0. division and see if the remainder 3 2 Is x + 3 a factor of − 4 x + 5 x + 8 ? is 0 Opposite -3 -4 5 0 8 NO it’s not a sign goes 12 -51 153 factor. In fact, here f(-3) = 161 -4 17 -51 161 We could have computed f(-3) at 3 2 f − 3 = −4 − 3 + 5 − 3 + 8 = 161 first to determine this. Not = 0 so not a factor

( )

( )

( )

Our goal in this section is to learn how we can factor higher degree polynomials. For example we want to factor: 4 3 2

f ( x ) = x + x − 3x − x + 2

We could randomly try some factors and use synthetic division and know by the factor theorem that if the remainder is 0 then we have a factor. We might be trying things all day and not hit a factor so in this section we’ll learn some techniques to help us narrow down the things to try. The first of these is called Descartes Rule of Signs named after a French mathematician that worked in the 1600’s. Rene Descartes 1596 - 1650

Descartes’ Rule of Signs Let f denote a polynomial function written in standard form. The number of positive real zeros of f either equals the number of sign changes of f (x) or else equals that number less an even integer. The number of negative real zeros of f either equals the number of sign changes of f (-x) or else equals that number less an even integer. 1 2 starts Pos.

changes Neg.

changes Pos.

f ( x ) = x + x − 3x − x + 2 4

3

2

There are 2 sign changes so this means there could be 2 or 0 positive real zeros to the polynomial.

Descartes’Rule of Signs Let f denote a polynomial function written in standard form. The number of positive real zeros of f either equals the number of sign changes of f (x) or else equals that number less an even integer. The number of negative real zeros of f either equals the number of sign changes of f (-x) or else equals that number less an even integer.

) = Pos. f ( xstarts x +changes x −Neg. 3x changes − x +Pos. 2 4

31

2

2

f ( − (x ) = ()− x ) 4+ ( − x3) − 3( −2 x ) − ( − x ) + 2 f − x = x − x −3 x + x + 2 4

3

2

There simplify are f(-x) 2 sign changes so this means there could be 2 or 0 negative real zeros to the polynomial.

Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros the polynomial may have. Counting multiplicities and complex (imaginary) zeros, the total 1 starts Neg. changes Pos. number of zeros will be the same as the degree of the polynomial.

f ( x ) = −3 x + 4 x + 2 5

4

There is one sign change so there is one positive real zero. starts Pos. Never changes

f ( − x ) = −3( − x ) + 4( − x ) + 2 = 3 x + 4 x + 2 5

4

5

4

There are no negative real zeros. Descartes rule says one positive and no negative real zeros so there must be 4 complex zeros for a total of 5. We’ll learn more complex zeros in Section 5.4

Back to our original polynomial we want to factor: 4 3 2

f ( x ) = 1x + x − 3x − x + 2

We’d need to try a lot of positive or negative numbers until we found one that had 0 remainder. To help we have:

The Rational Zeros Theorem Let f be a polynomial function of degree 1 or higher of the form f ( x ) = an x n + an −1 x n −1 +  + a1 x + a0 , an ≠ 0, a0 ≠ 0 p , in lowest terms, is a q rational zero of f , then p must be a factor of a0 , and q must be a factor of an . where each coefficient is an integer. If

What this tells us is that we can get a list of the POSSIBLE rational zeros that might work by taking factors of the constant divided by factors of the leading coefficient.

Factors of the constant Factors of the leading coefficient

± 1, 2 1

Both positives and negatives would work for factors

± 1, 2

So a list of possible things to try would be any number from the top divided by any from the bottom with a + or - on it. In this case that just leaves us with ± 1 or ± 2

1

f ( x ) = x + x − 3x − x + 2 1

4

3

2

1

1 1

-3 2

-1 -1

2 -2

1

2

-1

-2

0

( x − 1) ( x

3

+ 2x − x − 2 2

)

Since 1 is a zero, we can write the factor x - 1, and use the quotient to write the polynomial factored.

Let’s try 1 YES! It is a zero since the remainder is 0 We found a positive real zero so Descartes Rule tells us there is another one

± 1, 2 1

We could try 2, the other positive possible. IMPORTANT: Just because 1 worked doesn’t mean it won’t work again since it could have Let’s try 1 again, a multiplicity. but we try it on the factored version 3 2 for the remaining factor (once you have

(

f ( x ) = ( x − 1) x + 2 x − x − 2 1

1 1

2 1

-1 3

-2 2

3

2

0

)

it partly factored use that to keep going--don't start over with the original).

YES! the remainder is 0

x + 3 x + 2 = ( x + 2 )( x + 1) 2

Once you can get it down to 3 numbers here, you can put the variables back in and factor or use the quadratic formula, we are done with trial and error.

Let’s take our polynomial then and write all of the factors we found: There ended up 4 3 2 being two positive

f ( x ) = x + x − 3x − x + 2 2 = ( x − 1) ( x + 2 )( x + 1)

real zeros, 1 and 1 and two negative real zeros, -2, and -1.

In this factored form we can find intercepts and left and right hand behavior and graph the polynomial Left & right hand behavior

Plot intercepts Touches at 1 crosses at -1 and -2. “Rough” graph

Let’s try another one from start to finish using the theorems and rules to help us.

f ( x ) = 2 x + 13x + 29 x + 27 x + 9 4

3

2

Using the rational zeros 1, 3, 9 factors of constant theorem let's find factors of the constant over factors of leading 1, 2 factors of the leading coefficient coefficient to know what numbers to try. So possible rational zeros are all possible combinations of numbers on top with numbers on bottom:

±

1 3 9 ± 1, ± , ± 3, ± , ± 9, ± 2 2 2

starts Pos. Stays positive

f ( x ) = 2 x + 13x + 29 x + 27 x + 9 4

3

2

1 3 9 ± − 1, ± − ,± − 3, − ± ,− ± 9, − ± 2 2 2

Let’s see if Descartes Rule helps us narrow down the choices.

No sign changes in f(x) so no positive real zeros---we just ruled out half the choices to try so that helps! 1

2

3

4

starts Pos. changes Neg. changes Pos. Changes Neg. Changes Pos.

f ( − x ) = 2 x − 13 x + 29 x − 27 x + 9 4

3

2

4 sign changes so 4 or 2 or 0 negative real zeros.

f ( x ) = 2 x + 13x + 29 x + 27 x + 9 4

3

2

1 3 9 Let’s try -1 − 1, − , − 3, − , − 9, − 2 2 2 2 13 29 27 9 Yes! We found a zero. -2 -11 -18 -9 Let’s work with reduced polynomial 2 11 18 9 0 then.

-1

Let’s try -1 again

-1

2 2

11 -2 9

18 -9 9

9 -9 0

2 x + 9 x + 9 = ( 2 x + 3)( x + 3) 2

Yes! We found another one. We are done with trial and error since we can put variables back in and solve the remaining quadratic equation.

So remaining zeros found by setting these factors = 0 are -3/2 and -3. Notice these were in our list of choices.

f ( x ) = 2 x + 13x + 29 x + 27 x + 9 4

3

2

So our polynomial factored is:

f ( x ) = ( x + 1) ( 2 x + 3)( x + 3) 2