Chapter 3.3 – Zeros of Polynomial Functions

Chapter 3.3 Zeros of Polynomial Functions

Recall Zero Factor Property from Chapter 1 that states that a product of factors is equal to 0 if and only if one or more of the factors is equal to zero. For example, if

0)52)(7)(1( =+−+ xxx then

25,7,1 −==−= xxx

Let’s extend this to help us find factors of polynomials of greater degree.

Sample Problem: Determine if 2+x is a factor of 604432)( 23 −−−= xxxxf Solution: If 2+x is a factor of )(xf , then 0)2( =−f . We use synthetic division by -2. -2 2 -3 -44 -60 -4 14 60

2 -7 -30 0 Remainder Since 0)2( =−f , then 2+x is a factor. Note that )(xf can now be written as a product of 2+x and the quotient, so that

)3072)(2()( 2 −−+= xxxxf Student Practice: Use the factor theorem and synthetic division to determine if the second polynomial is a factor of the first. 1. 135)( 23 ++−= xxxxf ; 1−x 2. 133852)( 234 ++−+= xxxxxf ; 1+x

Chapter 3.3 – Zeros of Polynomial Functions

Factoring Polynomials Sample Problem: Factor 604432)( 23 −−−= xxxxf into linear factors if -2 is a zero of )(xf . Solution: Since -2 is a zero of )(xf , we can use synthetic division to determine the quotient as we did before, knowing that 2+x is a factor. -2 2 -3 -44 -60 -4 14 60

2 -7 -30 0 Remainder Thus, )3072)(2()( 2 −−+= xxxxf . We can now fully factor the quadratic part to get

)6)(52)(2()( −++= xxxxf . Student Practice: Factor )(xf given that k is a zero of )(xf . 3. 314136)( 23 +−+= xxxxf ; 3−=k

The rational zeros theorem simply states that if a polynomial function has any rational roots, then they will be of the form p/q where p is a factor of the last term and q is a factor of the first term. This allows us to be able to list all the possible rational roots of a function.

Chapter 3.3 – Zeros of Polynomial Functions

Using the Rational Roots Theorem Sample Problem: For 10132)( 23 −−−= xxxxf a) list all possible rational zeros, b) find all rational zeros, c) factor f(x) Solution: a) For a rational number p/q to be a rational zero, p must be a factor of -10 and q must be a factor of 1. Factors of -10 are 10,5,2,1 ±±±± and factors of 1 are 1± . The possible rational zeros, p/q, are 10,5,2,1 ±±±± . b) We use synthetic division to show that -1 is a rational zero and thus 1+x is a factor. -1 1 -2 -13 -10 -1 3 10

1 -3 -10 0 Remainder Thus, )103)(1()( 2 −−+= xxxxf . We can now fully factor the quadratic part to get

)2)(5)(1()( +−+= xxxxf . So all rational solutions include 2,5,1 −==−= xxx (Note, if the quadratic part was not factorable, then we would use the quadratic formula to find the roots that would not be rational, but real) Student Practice: For each function, a) list all possible rational zeros, b) find all rational zeros, c) factor f(x) 4. 2425)( 23 −−+= xxxxf 5. 231276)( 234 +−−+= xxxxxf

Chapter 3.3 – Zeros of Polynomial Functions

Note that the number of zeros theorem only states that a polynomial MAY have up to n distinct zeros, but it may have less. For example, consider

22 )5(2510)( −=+−= xxxxf This function has only 1 zero which is 5 and occurs two times, once for each factor. The number of times a zero occurs is referred to as the multiplicity of the zero. Thus, 5 has a multiplicity of 2. Finding Polynomials Given Roots Suppose we knew the rational roots of a given polynomial, it would then be possible to determine the polynomial for which the roots occur. Sample Problem: Find a polynomial of least degree having only real coefficients that satisfies the given conditions: Zeros of -3,2, and 4 Solution: Since this polynomial has zeros of -3, 2, and 4, then )3( +x , )2( −x , and

)4( −x must all be factors of the polynomial; thus )4)(2)(3()( −−+= xxxxf or 24103)( 23 +−−= xxxxf

Student Practice: Find a polynomial of least degree having only real coefficients that satisfies the given conditions.

6. Zeros of 5, -2, 32 Tip: If

32 is a zero, then (3x-2) is a factor

Chapter 3.3 – Zeros of Polynomial Functions

Now let’s consider some polynomial with pure complex numbers as solutions.

Finding Zeros of a Polynomial Function Given Complex Root The conjugate zeros theorem can help determine how many real zeros (note they may not be rational) a polynomial function has. A polynomial function of odd degree must have at least 1 real zero since complex zeros come in pairs, while a polynomial function of even degree may have no real zeros. Sample Problem: Find all zeros of 1222187)( 234 +−+−= xxxxxf given that i−1 is a zero. FUN HUH !!! (Note that if the quadratic part was not factorable, we would use the quadratic formula to find solutions) Student Practice: 8. Find all zeros of 15177)( 23 −+−= xxxxf given that i−2 is a zero.

Chapter 3.3 – Zeros of Polynomial Functions

The following rule helps us determine the number of positive and negative real zeros of a polynomial function. By definition, a variation in sign is a change from positive to negative or negative to positive in success terms of a polynomial written in descending powers.

Sample Problem: Determine the number of possible positive and negative real zeros of the polynomial function 24103)( 23 +−−= xxxxf Solution: Since there are 2 variations in sign of )(xf , then there are either 2 positive real roots, or 2-2=0 positive real roots. Now consider )( xf − . 2410324)(10)(3)()( 2323 ++−−=+−−−−−=− xxxxxxxf Since there is only 1 variation of sign, there is either only 1 negative real zero, or none. (Note: The roots are shown above)