THE FUNDAMENTAL THEOREM OF ALGEBRA Section 2 5

THE FUNDAMENTAL THEOREM OF ALGEBRA Section 2. 5

Objectives: � Use the Fundamental Theorem of Algebra to determine the number of zeros of a polynomial function. � Find all zeros of polynomial functions, including complex zeros. � Find conjugate pairs of complex zeros. � Find zeros of polynomials by factoring.

Fundamental Theorem of Algebra The fundamental theorem of algebra guarantees that every polynomial has a complete factorization, if we are allowed to use complex numbers.

The Linear Factorization Theorem � Using the Fundamental Theorem of Algebra and the equivalence of zeros and factors, we obtain the following theorem.

Example 1 All zeros for the given polynomial are distinct. Use the figure to determine graphically the number of real zeros and the number of imaginary zeros. f(x) = 3 x 3 – 3 x 2 – 3 x – 5

Example 1 Solution Real zeros correspond to x-intercepts; imaginary zeros do not. Graph has one xintercept so there is one real zero. Since f is degree 3 and all zeros are distinct, there are two imaginary numbers.

Polynomial Equations with Complex Solutions We can solve polynomial equations by writing them in their complete factored form and then use the fact that the solutions to the equation are the zeros of f(x).

Example 2 Confirm that the third-degree polynomial function f(x) = x 3 – x 2 + 4 x – 4 has exactly three zeros: x = 1, x = 2 i, and x = 2 i. Solution � Factor f(x) by grouping x 3 – x 2 + 4 x – 4 (x 3 – x 2) + (4 x – 4) x 2(x-1) + 4(x-1)(x 2+4) � Solve for the zeros (zero product property) x-1 = 0 x=1 x 2+4 = 0 x 2 = -4 x = ±√ 4 x = 2 i, -2 i

Your Turn: � Confirm that the third-degree polynomial function f(x) = x 3 + 4 x 2 + 9 x + 36 has exactly three zeros: x = -4, x = 3 i, and x = -3 i. Solution � Factors (x+4)(x-3 i)(x+3 i) � Zeros x = -4, 3 i, -3 i

Example 3 a: Find all zeros of f(x) = x 3 + 3 x 2 + 16 x + 48 (should be 3 total!) � p = ± 1 ± 2 ± 3 ± 4 ± 6 ± 8 ± 12 ± 16 ± 24 ± 48 � q ± 1 � Graph and you’ll see -3 is the only one on the graph you can see so synthetic divide with -3 � � -3 1 � 1 3 -3 16 0 0 16 48 -48 0 x 2 + 16 = 0 x 2 = -16 x = ±√-16 = ± 4 i The three zeros are -3, 4 i, -4 i

Example 3 b � Write f(x) = x 3 + 3 x 2 + 16 x + 48 as a product of linear factors. � Zeros are 3, 4 i, -4 i �Zero 3 ⇒ factor (x-3) �Zero 4 i ⇒ factor (x-4 i) �Zero -4 i ⇒ factor (x+4 i) � f(x) = (x-3)(x-4 i)(x+4 i)

Example 4 Solve x 3 - 3 x 2 7 x + 21. Solution Rewrite the equation as f(x) = 0, where f(x) = x 3 3 x 2 + 7 x 21. We can use factoring by grouping to find one real zero of f(x).

Example 3 Solution continued x 3 3 x 2 + 7 x 21 (x 3 – 3 x 2) + (7 x – 21) x 2(x – 3) + 7(x – 3)(x 2 + 7) x 3 3 x 2 + 7 x 21 = (x 3)(x 2 + 7) The solutions are 3 and

Your Turn: � Write f(x) = x 5 -3 x 4 -7 x 3 + 21 x 2 -18 x -54 as the product of linear factors, and list all the zeros of f. Solution � Factors: f(x) = (x+3)(x-3)(x-i√ 2)(x+i√ 2) � Zeros: x = 3, -3, i√ 2, -i√ 2

Conjugate Zeros Theorem If a polynomial f(x) has only real coefficients and if a + bi is a zero of f(x), then the conjugate a – bi is also a zero of f(x). Complex Zeros Occur in Conjugate Pairs

Zeros of a Polynomial Zeros (Solutions) Real Zeros Complex Zeros Rational Zeros Complex Number and its Conjugate

Summary of Theorems Fundamental Theorem of Algebra Every polynomial P(x) of degree n > 0 has at least one zero. Linear Factorization Theorem Every polynomial P(x) of degree n > 0 can be expressed as the product of n linear factors. Hence, P(x) has exactly n zeros—not necessarily distinct. Complex Conjugates Zeros Theorem Imaginary zeros of polynomials with real coefficients, if they exist, occur in conjugate pairs. Real Zeros and Odd-Degree Polynomials A polynomial of odd degree with real coefficients always has at least one real zero.

Example 4: Write a polynomial function that has the given zeros & has a leading coefficient of 1: 3, – 1, 2 i, – 2 i � Solution � f(x) = (x – 3)(x + 1)(x – 2 i)(x + 2 i) � Multiply these to get the polynomial � f(x) = (x 2 – 2 x – 3)(x 2 + 4) � f(x) = x 4 – 2 x 3 + x 2 – 8 x – 12

Example 5 Represent a polynomial f(x) of degree 4 with leading coefficient 2 and zeros of 3, 5, i and i in complete factored form and expanded form. Solution Let an = 2, x 1 = 3, x 2 = 5, x 3 = i, and x 4 = i. f(x) = 2(x + 3)(x 5)(x i)(x + i) Expanded: 2(x + 3)(x 5)(x i)(x + i) = 2(x + 3)(x 5)(x 2 + 1) = 2(x + 3)(x 3 5 x 2 + x 5) = 2(x 4 2 x 3 14 x 2 2 x 15) = 2 x 4 4 x 3 – 28 x 2 – 4 x – 30

Your Turn: � Find a fourth – degree polynomial function with real coefficients that has -2, and 2 i as zeros. Solution f(x) = x 4 + 4 x 3 + 8 x 2 + 16 x + 16

Example 6 Find the zeros of f(x) = x 4 + x 3 + 2 x 2 + x + 1 given one zero is i. Solution By the conjugate zeros theorem it follows that i must be a zero of f(x). (x + i) and (x i) are factors (x + i)(x i) = x 2 + 1, using long division we can find another quadratic factor of f(x).

Example 6 Solution continued Long division So x 4 + x 3 + 2 x 2 + x + 1 = (x 2 + 1)(x 2 + x + 1)

Example 6 Solution continued Use the quadratic formula to find the zeros of x 2 + x + 1. The four zeros are:

Example 7: Now write a polynomial function of least degree that has real coefficients, a leading coeff. of 1 and 1, -2+i, -2 -i as zeros. F(x)= (x-1)(x-(-2+i))(x-(-2 -i)) F(x)= (x-1)(x+2 -i)(x+2+i) f(x)= (x-1){(x+2)-i} {(x+2)+i} F(x)= (x-1){(x+2)2 -i 2} Foil F(x)=(x-1)(x 2 + 4 x + 4 –(-1)) Take care of i 2 � F(x)= (x-1)(x 2 + 4 x + 4 + 1) � F(x)= (x-1)(x 2 + 4 x + 5) Multiply � F(x)= x 3 + 4 x 2 + 5 x – x 2 – 4 x – 5 � f(x)= x 3 + 3 x 2 + x - 5 � � �

Example 8: Now write a polynomial function of least degree that has real coefficients, a leading coeff. of 1 and 4, 4, 2+i as zeros. Note: 2+i means 2 -i is also a zero � F(x)= (x-4)(x-(2+i))(x-(2 -i)) � F(x)= (x-4)(x-2 -i)(x-2+i) � F(x)= (x 2 – 8 x +16)((x-2)-i)((x-2)+i) � F(x)= (x 2 – 8 x +16)((x-2)2 -i 2) � F(x)= (x 2 – 8 x +16)(x 2 – 4 x + 4 –(-1)) � F(x)= (x 2 – 8 x +16)(x 2 - 4 x + 5) � F(x)= x 4– 4 x 3+5 x 2– 8 x 3+32 x 2 -40 x+16 x 2 -64 x+80 � F(x)= x 4 -12 x 3+53 x 2 -104 x+80 �

Example 9 Find all complex zeros of given that 1 – i is a zero. Solution

Example 9 Using the Conjugate Zeros Theorem, 1 + i is also a zero. The zeros of x 2 – 5 x + 6 are 2 and 3. Thus, and has four zeros: 1 – i, 1 + i, 2, and 3.

Your Turn: Write as a product of linear factors: f(x) = x 4 – 16 f(x) = (x-2)(x+2)(x-2 i)(x+2 i) � Find a third-degree polynomial with integer coefficients that has 2 and 3 -I as zeros. f(x) = x 3 – 8 x 2 + 22 x – 20 � Write the polynomial f(x) = x 4 – 4 x 3 + 5 x 2 – 2 x – 6 in completely factored form. (Hint: one factor is x 2 – 2 x – 2) �