The Rational Zero Theorem

The Rational Zero Theorem

Our goal in this section is to learn how we can ind the rational zeros o the polynomials. For example: x = x 4 + x x x + ( ) We could randomly try some actors and use synthetic division and know by the actor theorem that i the remainder is 0 then we have a actor. We might be trying things all day and not hit a actor so in this section we ll learn some techniques to help us narrow down the things to try.

Let ( x) = be a a n ( x) = x 4 + x x x + x polynomial unction o n + a n 1 x 1 We d need to try a lot o positive or negative numbers until we ound one that had 0 remainder. To help we have: The Rational Zeros Theorem n 1 + + a x + a where each coeicient is an integer. 1 degree1or higher o 0, p, in lowest terms, is a q rational zero o, then p must be a actor o a0, and q must be a actor o an. What this tells us is that we can get a list o the POSSIBLE rational zeros that might work by taking actors o the constant divided by actors o the leading coeicient. Factors o the constant Factors o the leading coeicient I a n 0, a 0 ± 0 1, 1 the orm Both positives and negatives would work or actors

REMEMBER! Descartes Rule o Signs Let denote a polynomial unction written in standard orm. The number o positive real zeros o either equals the number o sign changes o (x) or else equals that number less an even integer. The number o negative real zeros o either equals the number o sign changes o (-x) or else equals that number less an even integer. 1 starts Pos. changes Neg. changes Pos. ( x) = x 4 + x x x + There are sign changes so this means there could be or 0 positive real zeros to the polynomial.

Descartes Rule o Signs Let denote a polynomial unction written in standard orm. The number o positive real zeros o either equals the number o sign changes o (x) or else equals that number less an even integer. The number o negative real zeros o either equals the number o sign changes o (-x) or else equals that number less an even integer. 1 ( x) = x 4 + x x x + starts Pos. changes Neg. changes Pos. ( ) ( ) 4 ( ) ( ) ( x = ) x + x x ( x) + x = x 4 x x + x + There simpliy are (-x) sign changes so this means there could be or 0 negative real zeros to the polynomial.

± 1, So a list o possible things to try would be 1 any number rom the top divided by any rom the bottom with a + or - on it. In this case that just leaves us with ± 1 or ± ( x) = x 4 + x x x + Let s try 1 1 1 1 - -1 1-1 - 1-1 - 0 ( )( ) x 1 x + x x Since 1 is a zero, we can write the actor x - 1, and use the quotient to write the polynomial actored. YES! It is a zero since the remainder is 0 We ound a positive real zero so Descartes Rule tells us there is another one

± 1, 1 We could try, the other positive possible. IMPORTANT: Just because 1 worked doesn t mean it won t work again since it could have a multiplicity. ( ) ( )( ) x = x 1 x + x x 1 1-1 - 1 1 0 x Let s try 1 again, but we try it on the actored version or the remaining actor (once you have it partly actored use that to keep going--- don't start over with the original). YES! the remainder is 0 ( + )( + 1) + x + = x x Once you can get it down to numbers here, you can put the variables back in and actor or use the quadratic ormula, we are done with trial and error.

Let s take our polynomial then and write all o the actors we ound: = x 4 + x x x + ( x) = ( x 1) ( x + )( x + 1) There ended up being two positive real zeros, 1 and 1 and two negative real zeros, -, and -1. In this actored orm we can ind intercepts and let and right hand behavior and graph the polynomial Let & right hand behavior Rough graph Plot intercepts Touches at 1 crosses at -1 and -.

Let s try another one rom start to inish using the theorems and rules to help us. ( x) = x 4 + 1x + 9x + 7x + 9 Using the rational zeros theorem let's ind actors o the constant over actors o the leading coeicient to know what numbers to try. 1,, 9 ± actors o constant 1, So possible rational zeros are all possible combinations o numbers on top with numbers on bottom: 1 ± 1, ±, ±, ±, ± 9, ± actors o leading coeicient 9

starts Pos. Stays positive ( x) = x 4 + 1x + 9x + 7x + 9 Let s see i Descartes Rule helps us narrow down the choices. 1 ± 1, ±, ±, ±, ± 9, ± 9 No sign changes in (x) so no positive real zeros---we just ruled out hal the choices to try so that helps! 1 4 starts Pos. changes Neg. changes Pos. Changes Neg. Changes Pos. ( x) = x 4 1x + 9x 7x + 9 4 sign changes so 4 or or 0 negative real zeros.

( x) = x 4 + 1x + 9x + 7x + 9 Let s try -1-1 1 9 7 9 - -11-18 -9 11 18 9 0 Let s try -1 again -1 11 18 9 - - 9-9 9 9 0 x + 9x + 9 = x + x + ( )( ) 1 1,,,, 9, 9 Yes! We ound a zero. Let s work with reduced polynomial then. Yes! We ound 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 ound by setting these actors = 0 are -/ and -. Notice these were in our list o choices.

( x) = x 4 + 1x + 9x + 7x + 9 So our polynomial actored is: ( x) = ( x + 1) ( x + )( x + ) Its zeros are: -1,-1,-/,-.