Date: 2.4 Real Zeros of Polynomial Functions Divide 4 – 5x – x2 + 6x3 by 3x – 2. Solution Write the divisor and dividend in descending powers of x. Next, consider how many times 3x divides into 6x3. 3x – 2 6x3 – x2 – 5x + 4 2x2 Divide: 6x3/3x = 2x2 Divide: 3x2/3x + x – 1 Multiply Multiply: 2x2(3x – 2) Divide: -3x/3x Multiply x(3x – 2) Subtract 6x3 – 4x2 from 6x3 – x2 – 5x and bring down –5x. + 4 and bring down 4. Multiply -1(3x – 2) 6x3 – 4x2 3x2 3x2 – 2x Subtract 3x2 – 2x from 3x2 – 5x -3x -3x + 2 Subtract -3x + 2 from -3x + 4, 2 leaving a remainder of 2.

The Division Algorithm If f (x) and d(x) are polynomials, with d(x) = 0, and the degree of d(x) is less than or equal to the degree of f (x), then there exist unique polynomials q(x) and r(x) such that f (x) = d(x) • q(x) + r(x).   The remainder, r(x), equals 0 or its is of degree less than the degree of d(x). If r(x) = 0, we say that d(x) divides evenly in to f (x) and that d(x) and q(x) are factors of f (x). Dividend Divisor Quotient Remainder

Example Divide:

Example cont.

Example cont. 3x

Example cont. 3x 3x3 + 9x2 + 9x

Example cont. 3x -(3x3 + 9x2 + 9x) -11x2 - 5x - 3

Example cont. 3x -11 -(3x3 + 9x2 + 9x) -11x2 - 5x - 3

Example cont. 3x -11 -(3x3 + 9x2 + 9x) -11x2 - 5x - 3 -11x2 - 33x - 33

-(3x3 + 9x2 + 9x) -11x2 - 5x - 3 -(-11x2 - 33x - 33) 28x+30 Example cont. 3x -11 -(3x3 + 9x2 + 9x) -11x2 - 5x - 3 -(-11x2 - 33x - 33) 28x+30

-(3x3 + 9x2 + 9x) -11x2 - 5x - 3 -(-11x2 - 33x - 33) 28x+30 Example cont. 3x -11 -(3x3 + 9x2 + 9x) -11x2 - 5x - 3 -(-11x2 - 33x - 33) 28x+30

Complete Student Checkpoint Divide

Complete Student Checkpoint Divide

Complete Student Checkpoint Divide x x2 + 9x

Complete Student Checkpoint Divide x -(x2 + 9x)

Complete Student Checkpoint Divide x -(x2 + 9x) 5x + 45

Complete Student Checkpoint Divide x + 5 -(x2 + 9x) 5x + 45

Complete Student Checkpoint Divide x + 5 -(x2 + 9x) 5x + 45 -(5x + 45)

Complete Student Checkpoint Divide x + 5 -(x2 + 9x) 5x + 45 -(5x + 45)

Complete Student Checkpoint Divide x + 5 -(x2 + 9x) 5x + 45 -(5x + 45)

Synthetic Division To divide a polynomial by x – c 3 1 4 -5 5 3 1 7 1. Arrange polynomials in descending powers, with a 0 coefficient for any missing terms. 3 1 4 -5 5 2. Write k for the divisor, x – k. To the right, write the coefficients of the dividend. 7 Add Multiply by 3 1 3 3. Write the leading coefficient of the dividend on the bottom row. 4. Multiply k (in this case, 3) times the value just written on the bottom row. Write the product in the next column in the 2nd row. 5. Add the values in this new column, writing the sum in the bottom row.

Synthetic Division 3 1 4 -5 5 3 1 7 Add. 53 21 48 16 1x2 + 7x + 16 + 3 1 4 -5 5 3 1 7 6. Repeat this series of multiplications and additions until all columns are filled in. Add. 53 Multiply by 3. 21 48 16 7. Use the numbers in the last row to write the quotient and remainder in fractional form. The degree of the first term of the quotient is one less than the degree of the first term of the dividend. The final value in the row is the remainder. 1x2 + 7x + 16 + 53 x – 3

Use the coefficients of the dividend in descending powers of x. Text Example Use synthetic division to divide 5x3 + 6x + 8 by x + 2. The divisor must be in the form x – k. Write x + 2 as x – (-2). This means that k = -2. Write a 0 coefficient for the missing x2-term in the dividend. -2 5 0 6 8 Use the coefficients of the dividend in descending powers of x. This is c in x-(-2). -10 26 -44 -52 20 -10 5 5x2 – 10x + 26 – x + 2 44 Thus, x + 2 5x3 + 0x2 + 6x + 8

Complete Student Checkpoint Use synthetic division to divide -2 1 0 -7 -6 -2 -3 6 4 -2 1 1x2 – 2x – 3 Thus, x + 2 x3 + 0x2 - 7x - 6

The Remainder Theorem The Factor Theorem If the polynomial f(x) is divided by x – k, then the remainder is f(k). The Factor Theorem Let f(x) be a polynomial. If f(k) = 0, then x – k is a factor of f(x). If x – k is a factor of f(x), then f(k) = 0.

Text Example Find f(-2) for f(x)= 5x3 + 6x + 8 f(-2)= 5(-2) 3 + 6(-2) + 8 -2 5 0 6 8 5 -10 20 26 -52 -44 Here is the previous problem. Notice the remainder: = 5(-8) + -12 + 8 = -40 + -12 + 8 = -52 + 8 = -44 Synthetic divisions remainder is an alternative to plugging in a x-value into the function. It is referred to in the remainder theorem.

Text Example Solve the equation 2x3 – 3x2 – 11x + 6 = 0 given that 3 is a zero of f(x) = 2x3 – 3x2 – 11x + 6. Given that f (3) = 0. The Factor Theorem tells us that x – 3 is a factor of f(x). Use synthetic division to divide f (x) by x – 3. 3 2 -3 -11 6 6 9 -6 2 3 -2 2x2 + 3x – 2 2x3 – 3x2 – 11x + 6 x – 3 Equivalently, 2x3 – 3x2 – 11x + 6 = (x – 3) (2x2 + 3x – 2)

Now we can solve the polynomial equation. Text Example cont. Equivalently, 2x3 – 3x2 – 11x + 6 = (x – 3) (2x2 + 3x – 2) Now we can solve the polynomial equation. 2x3 – 3x2 – 11x + 6 = 0 This is the given equation. (x – 3) (2x2 + 3x – 2) = 0 Factor using the result from the synthetic division. (x – 3)(2x – 1)(x + 2) = 0 Factor the trinomial. x – 3 = 0 or 2x – 1 = 0 or x + 2 = 0 Set each factor equal to 0. x = 3 x = 1/2 x = -2 Solve for x. The solution set is {-2, 1/2 , 3}.

Example Find a third-degree polynomial function f(x) with real coefficients that has -4, 3 and 5 as zeros and such that f(0)=180

Complete Student Checkpoint -4 3 4 -5 3 -8 27 -105 -108 32 -12 3

+ – 2 1 3x 5x 15x2 5x -6x -2 Complete Student Checkpoint -1 15 14 -3 -2 -30 -1 -2 2 1 -15 15 5 -6 -1 + – 2 1 3x 5x 15x2 5x -6x -2

Upper and Lower Bound Tests for Real Zeros DAY 2 Upper and Lower Bound Tests for Real Zeros If f (x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 be a polynomial of degree n > 1 with a positive leading coefficient. Suppose f (x) is divided by x - k using synthetic division: 1. If k > 0 and every number in the last line is non negative (positive or zero), then k is an upper bound for the real zeros of f. 2. If k > 0 and the numbers in the last line are alternating nonnegative and nonpositive, then k is a lower bound for the real zeros of f.

Prove that 5 is an upper bound (use synthetic division): Example Prove that all of the zeros must lie in the interval [-2, 5] for: Prove that 5 is an upper bound (use synthetic division): 2 -7 -8 14 8 5 10 15 35 -245 2 3 7 49 253 since the last line is all positive, 5 is an upper bound Prove that -2 is a lower bound: 2 -7 -8 14 8 -2 -4 22 -28 28 2 -11 14 -14 36 since the last line alternates signs, -2 is a lower bound Graph f(x) on a graphing calculator to view these upper and lower bounds for the zeros of the function. Use window [-2,5] by [-50,50].

The Rational Zero Theorem If f (x) = anxn + an-1xn-1 +…+ a1x + a0 has integer coefficients and p/q is a rational zero(or factor of f(x)), then p is a factor of the constant term a0 and q is a factor of the leading coefficient an.

Example Find all of the possible real, rational roots of f(x) = 2x3-3x2+5 p is a factor of 5 = +1, +5 q is a factor of 2 = +1, +2 Roots are: p/q = +1, +1/2, +5, +5/2

Find all zeros of f(x) = x3+12x2+21x+10 p/q = Example Find all zeros of f(x) = x3+12x2+21x+10 p/q = +1, +2, +5, +10 are possible zeros Start plugging them in to find the zeros or use synthetic division: f(1) = 1 12 21 10 1 1 13 34 1 13 34 44 44 f(-1) = 0 -1 is a zero -1 -1 -11 -10 1 11 10 0 Since f(-1)=0, x+1 is a factor, or a zero. Divide out -1 to get the other factor. Since we did synthetic division, we have already done that so: x3+12x2+21x+10 = 0 (x + 1)(x2+11x+10) = 0 (x + 1)(x + 1)(x+10) = 0 The solutions are -1, and -10

p is a factor of 3 = +1, +3 q is a factor of 4 = +1, +2 , +4 Complete Student Checkpoint List all possible rational zeros of p is a factor of 3 = +1, +3 q is a factor of 4 = +1, +2 , +4 Roots are: p/q = +1, +1/2, +1/4, +3, +3/2, +3/4

Real Zeros of Polynomial Functions