<< Chapter < Page Chapter >> Page >

Let f be a polynomial function with real coefficients, and suppose a + b i b 0 , is a zero of f ( x ) . Then, by the Factor Theorem, x ( a + b i ) is a factor of f ( x ) . For f to have real coefficients, x ( a b i ) must also be a factor of f ( x ) . This is true because any factor other than x ( a b i ) , when multiplied by x ( a + b i ) , will leave imaginary components in the product. Only multiplication with conjugate pairs will eliminate the imaginary parts and result in real coefficients. In other words, if a polynomial function f with real coefficients has a complex zero a + b i , then the complex conjugate a b i must also be a zero of f ( x ) . This is called the Complex Conjugate Theorem .

Complex conjugate theorem

According to the Linear Factorization Theorem     , a polynomial function will have the same number of factors as its degree, and each factor will be in the form ( x c ) , where c is a complex number.

If the polynomial function f has real coefficients and a complex zero in the form a + b i , then the complex conjugate of the zero, a b i , is also a zero.

Given the zeros of a polynomial function f and a point ( c , f ( c )) on the graph of f , use the Linear Factorization Theorem to find the polynomial function.

  1. Use the zeros to construct the linear factors of the polynomial.
  2. Multiply the linear factors to expand the polynomial.
  3. Substitute ( c , f ( c ) ) into the function to determine the leading coefficient.
  4. Simplify.

Using the linear factorization theorem to find a polynomial with given zeros

Find a fourth degree polynomial with real coefficients that has zeros of –3, 2, i , such that f ( −2 ) = 100.

Because x = i is a zero, by the Complex Conjugate Theorem x = i is also a zero. The polynomial must have factors of ( x + 3 ) , ( x 2 ) , ( x i ) , and ( x + i ) . Since we are looking for a degree 4 polynomial, and now have four zeros, we have all four factors. Let’s begin by multiplying these factors.

f ( x ) = a ( x + 3 ) ( x 2 ) ( x i ) ( x + i ) f ( x ) = a ( x 2 + x 6 ) ( x 2 + 1 ) f ( x ) = a ( x 4 + x 3 5 x 2 + x 6 )

We need to find a to ensure f ( 2 ) = 100. Substitute x = 2   and f ( 2 ) = 100 into f ( x ) .

100 = a ( ( −2 ) 4 + ( −2 ) 3 5 ( −2 ) 2 + ( −2 ) 6 ) 100 = a ( −20 ) −5 = a

So the polynomial function is

f ( x ) = −5 ( x 4 + x 3 5 x 2 + x 6 )

or

f ( x ) = 5 x 4 5 x 3 + 25 x 2 5 x + 30
Got questions? Get instant answers now!
Got questions? Get instant answers now!

If 2 + 3 i were given as a zero of a polynomial with real coefficients, would 2 3 i also need to be a zero?

Yes. When any complex number with an imaginary component is given as a zero of a polynomial with real coefficients, the conjugate must also be a zero of the polynomial.

Find a third degree polynomial with real coefficients that has zeros of 5 and 2 i such that f ( 1 ) = 10.

f ( x ) = 1 2 x 3 + 5 2 x 2 2 x + 10

Got questions? Get instant answers now!

Using descartes’ rule of signs

There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order, Descartes’ Rule of Signs tells us of a relationship between the number of sign changes in f ( x ) and the number of positive real zeros. For example, the polynomial function below has one sign change.

The function, f(x)=x^4+x^3+x^2+x-1, has one sign change between x and -1.`

This tells us that the function must have 1 positive real zero.

Questions & Answers

write down the polynomial function with root 1/3,2,-3 with solution
Gift Reply
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
Pream Reply
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Oroke Reply
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
kiruba Reply
what is the answer to dividing negative index
Morosi Reply
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
Shivam Reply
give me the waec 2019 questions
Aaron Reply
the polar co-ordinate of the point (-1, -1)
Sumit Reply
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
Rockstar Reply
tanh`(x-iy) =A+iB, find A and B
Pankaj Reply
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
Branded Reply
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
archana Reply
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
Kc Reply
1+cos²A/cos²A=2cosec²A-1
Ramesh Reply
test for convergence the series 1+x/2+2!/9x3
success Reply
Practice Key Terms 6

Get the best Algebra and trigonometry course in your pocket!





Source:  OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Algebra and trigonometry' conversation and receive update notifications?

Ask