6x 3 12x 2 7x 2 +16x 7x 2 +14x 2x 4

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1 2.3 Real Zeros of Polynomial Functions Name: Pre-calculus. Date: Block: 1. Long Division of Polynomials. We have factored polynomials of degree 2 and some specific types of polynomials of degree 3 using various methods. Long division is another method of factoring that is especially useful for polynomials of higher degree. Consider 6x 3 19x 2 +16x 4. Do we any algebraic method for factoring this polynomial? Suppose you are given that one of the is (x 2). This means that f(2) =, and that x =2 is a of the polynomial function. This also means there is an at (2,0). This means that there exists a second-degree polynomial q(x) such that f(x) =(x 2) q(x). To find q(x) you can use long division. Divide 6x 3 19x 2 +16x 4 by x 2. Use the result to factor the polynomial completely. 6x 3 19x 2 +16x 4 x 2 = First re-write the fraction in long division form as x 2) 6x 3 19x 2 +16x 4 The divisor isx 2, the dividend is 6x 3 19x 2 +16x 4. quotient divisor) dividend Notice that the terms are written so that the exponents are in descending order. Verify that all placeholders are written. 1. Divide the first term of the dividend by the first term of the divisor. The result is the first term of the quotient. 2. Multiply the first term of the quotient by the first term of the divisor. The result is the first term of the new dividend. 3. Multiply the first term of the quotient by the second term of the divisor. The result is the second term of the new dividend. 4. Subtract the original dividend from the new dividend. Bring down the next term from the original dividend. 5. Repeat #1-4 with the new dividend until you arrive at the remainder. 6x 2 7x +2 x 2 6x 3 19x 2 +16x 4 ) 6x 3 12x 2 7x 2 +16x 7x 2 +14x 2x 4 2x 4 The quotient is the result of dividing the dividend by the divisor. For this problem, the quotient is 6x 2 7x +2. Now you have the original factor (x 2) and another factor of degree two 6x 2 7x +2. Therefore we know that 6x3 19x 2 +16x 4 =6x 2 7x +2 x 2 Can you factor 6x 2 7x +2 further? 6x 3 19x 2 +16x 4 x 2 =(2x 1)(3x 2) Therefore the complete factorization is (x 2)(2x 1)(3x 2). Graph and verify. 0

2 In the previous example we used long division and the result had a remainder of zero. Often, long division will produce a nonzero remainder. Example 2: Divide x 2 +3x +5 by x+1. Label the dividend, divisor, quotient, and remainder. The result can be written as x2 +3x +5 x +1 =x Label the dividend, divisor, quotient, and remainder. x +1 This implies that x 2 +3x +5 =(x +1)(x +2) +3 which illustrates the Division Algorithm. The division algorithm is commonly written in two ways: f(x) =d(x)q(x) +r(x) and f(x) d(x) =q(x) +r(x) d(x) This means that if f(x) and d(x) are polynomials such that d(x) 0, and the degree of d(x) is less than or equal to the degree of f(x), there exist unique polynomials q(x) and r(x) where r(x) =0 or the degree of r(x) is less than the degree of d(x). If the remainder r(x) is zero, d(x) divides evenly into f(x) Example 3: Divide x 3 1 by x 1. Be careful to account for missing terms with zero coefficients or empty spaces. Answer: x3 1 x 1 =x2 +x+1, x 1 You can always check your answer by multiplying the resulting quotient by the divisor which will equal the original dividend. (x 2 +x +1)(x 1) =

3 Example 4: Divide 2x 4 +4x 3 5x 2 +3x 2 by x 2 +2x 3. Answer: 2x4 +4x 3 5x 2 +3x 2 x 2 +2x 3 =2x x +1 x 2 +2x 3 Individual Practice: try these on your own paper x + x + 10x x 2x 6x + 5x + 3x 5 3. x + 5 x 2 + 8x x 2 x + 3x x + 2 x + 5x + 7x x 3 x x x 36x + 13x 11x x + x 2 x + 2x 4x 5x x + 118x + 9x + 3x x + x x 8x + 2x 11. What does it mean if the remainder is zero? 12. If a 6 th degree polynomial is divided by a 3 rd degree polynomial, what is the degree of the quotient?

4 Recap. We already know if x =a is a zero, then is the corresponding factor of that zero. f( x) We also know that we can divide = q( x), where q (x) is the quotient. x a If (x a)is a factor and we divide f(x) by (x a), then the remainder, r(x) =. If the degree of f is n, what is the degree of q(x)? 2. Synthetic Division. Synthetic division can be used when dividing by a linear factor in the form. There is a pattern you must know when dividing a polynomial by a linear factor using synthetic division. The vertical pattern is addition; the diagonal pattern is multiplication by k. Example 1: Use synthetic division to divide x 4 10x 2 2x +4 by x+3. Write your answer in the form f(x) d(x) =q(x) +r(x) d(x) Individual practice: Use synthetic division to divide the following: Divide x 2x + 4x 2x + 5 by x (x ) (x +8) 3. 5x3 +6x +8 x Remainder & Factor Theorems. The Remainder Theorem: if a polynomial f(x) is divided by x k, the remainder is r = f(k). The Remainder Theorem tells you that synthetic division can be used to evaluate a polynomial function when x =k by dividing f(x) by x k. The remainder will be f(k). Example 1: Use the remainder theorem to evaluate the function f(x) =3x 3 +8x 2 +5x 7 at x = 2. Because the remainder r =, you can conclude that f( ) =. This also means that the ordered pair (, ) is a point on the graph of f. Example 2: Find the remainder if f(x) =x 3 4x 2 +2x 5 is divided by: a.(x 3) b. (x +2)

5 Individual practice: Use synthetic division to find each function value. Use a graphing calculator to verify your results. 1. f(x) =4x 3 13x p(x) =x 6 4x 4 +3x a. f(1) b. f( 2) c. f d. f(8) a. p(2) b. p( 4) c. p( 3) d. p( 1) 2 The Factor Theorem: a polynomial f(x) has a factor (x k) if and only if f(k) =0. In other words, to determine whether a polynomial has (x k) as a factor, evaluate the polynomial at x =k. If the result is 0, (x k) is a factor. Example 2: Show that (x 2) and (x +3) are factors of f(x) =2x 4 +7x 3 4x 2 27x 18. Find the remaining factors of f(x). Use synthetic division. Answer: f(x) =(x 2)(x +3)(2x +3)(x +1) Example 3: Is x+4 a factor of f(x) =4x 6 64x 4 +x 2 15?

6 Individual practice: Verify the given factors off. Find the remaining factors of f. Use your results to write the complete factorization of f. List all the real zeros of f. Confirm your results with a graphing calculator. 1. f(x) =2x 3 +x 2 5x +2, given the factors (x +2), (x 1) 2. f(x) =3x 3 +2x 2 19x +6, (x +3), (x 2) Using the remainder in synthetic division. In summary, the remainder r, obtained in the synthetic division of f(x) by x k, provides the following information. 1. The remainder r gives the value of f at x =k. That is, r = f(k). 2. If r =0, (x k) is a factor of f(x). 3. If r =0, (k,0) is an x-intercept of the graph of f. 4. The Rational Zero Test relates the possible rational zeros of a polynomial (having integer coefficients) to the leading coefficient and to the constant term of the polynomial. In other words, if a polynomial has integer coefficients, every rational zero of f has the form p q where p and q have no common factors other than 1, p is a factor of the constant term, and q is a factor of the leading coefficient. Possible rational zeros = ± From the list of possible rational zeros we can use trial-and-error to determine which (if any) are actual zeros of the polynomial. Example 1: Find the rational zeros of f(x) =x 3 +x+1 If there are no rational zeros, what must be true?

7 Example 2: Find the rational zeros of f(x) =2x 3 +3x 2 8x +3 using the rational zero test. Use synthetic division to verify whether the possible zero is an actual zero of the polynomial. Example 3: Find all the real zeros of f(x) =10x 3 15x 2 16x +12 using the rational zero test and synthetic division to verify whether the possible zeros are actual zeros of the polynomial. Individual practice: Use the rational zero test to list all possible rational zeros of f. Use a graphing calculator to verify that the zeros of f are contained in the list. 1. f(x) =x 3 4x 2 4x f(x) =4x 5 8x 4 5x 3 +10x 2 +x 2 3. f(x) =3x 5 +x 4 +4x 3 2x x 5 4. f ( x) = 2x + 11x 7x 6

8 How do we find the zeros? Rational Zero Test: gives us a list of possible rational zeros. To determine whether the possible zeros are actual zeros you may use synthetic or long division. The Remainder Theorem and Factor Theorem give us a quick way to determine if an input is a zero of the function. Synthetic division is a short cut for dividing linear factors. The degree of a polynomial gives the maximum number of zeros. Steps for finding the zeros of a polynomial: 1. Use the degree to determine the maximum number of zeros. 2. If the polynomial has integer coefficients, use the Rational Zero Test to identify potential rational zeros. 3. Test each zero using the Remainder Theorem (or Factor Theorem). 4. Once you find a zero, use synthetic division to break down the polynomial. Keep testing for zeros and use the depressed equation (quotient) to keep breaking down the polynomial until all zeros are found. * Factor out the GCF of a polynomial before you begin. ** If you are able to find all but two of the zeros, you can always use factoring or the quadratic formula to find the remaining two zeros. *** Note: Not all zeros are rational. Some may be irrational and some may be complex (imaginary) Letf ( x) = 2x + 3x 8x + 3. Find the real zeros and write the polynomial in factored form (as a product of linear factors.) Letg ( x) = 6x 4x + 3x 2. Find the real zeros and write the polynomial in factored form. 3. Let g(x) =x 5 5x 4 +12x 3 24x 2 +32x 16. Find the real zeros and write the polynomial in factored form (as a product of linear factors.)

More Polynomial Equations Section 6.4

MATH 11009: More Polynomial Equations Section 6.4 Dividend: The number or expression you are dividing into. Divisor: The number or expression you are dividing by. Synthetic division: Synthetic division