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Polynomial Functions and Zeros: Examples and Solutions

Learn how to find the degree, real and complex zeros, and factorize polynomial functions with step-by-step examples and solutions.

kennethsmith
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Polynomial Functions and Zeros: Examples and Solutions

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  1. _ + 1 i 2 1.What is the degree of f (x) = 8x6 – 4x5 + 3x2 + 2? ANSWER 6 2.Solve x2– 2x + 3 = 0 ANSWER

  2. 3. The function P given by x4+3x3 – 30x2 – 6x = 56 modelthe profit of a company. What are the real solutions of the function? -7.2, 4.5 ANSWER

  3. EXAMPLE 1 Find the number of solutions or zeros a. How many solutions does the equation x3 + 5x2 + 4x + 20 = 0have? SOLUTION Because x3 + 5x2 + 4x + 20 = 0 is a polynomial equation of degree 3,it has three solutions. (The solutions are – 5, – 2i, and 2i.)

  4. EXAMPLE 1 Find the number of solutions or zeros b. How many zeros does the function f (x) = x4 – 8x3 + 18x2 – 27have? SOLUTION Because f (x) = x4– 8x3 + 18x2– 27 is a polynomial function of degree 4, it has four zeros. (The zeros are – 1, 3, 3, and 3.)

  5. ANSWER 4 for Example 1 YOU TRY 1. How many solutions does the equation x4 + 5x2– 36 = 0 have?

  6. ANSWER 3 for Example 1 YOU TRY 2. How many zeros does the function f (x) = x3 + 7x2 + 8x – 16 have?

  7. STEP 1 Find the rational zeros of f. Because fis a polynomial function of degree 5, it has 5 zeros. The possible rational zeros are + 1, + 2, + 7, and +14. Using synthetic division, you can determine that – 1 is a zero repeated twice and 2 is also a zero. STEP 2 Write f (x) in factored form. Dividing f (x) by its known factors x + 1, x + 1, and x – 2 gives a quotient of x2– 4x + 7. Therefore: EXAMPLE 2 Find the zeros of a polynomial function Find all zeros of f (x) = x5 – 4x4 + 4x3 + 10x2 – 13x – 14. SOLUTION f (x) = (x + 1)2(x – 2)(x2 – 4x + 7)

  8. STEP 3 Find the complex zeros of f . Use the quadratic formula to factor the trinomial into linear factors. f(x) = (x + 1)2(x – 2) x – (2 + i 3 ) x – (2 – i 3 ) ANSWER The zeros of f are – 1, – 1, 2, 2 + i 3 , and2 – i 3. EXAMPLE 2 Find the zeros of a polynomial function

  9. Find the rational zero of f. because f is a polynomial function degree 3, it has 3 zero. The possible rational zeros are 1 , 3, using synthetic division, you can determine that 3 is a zero reputed twice and –3 is also a zero STEP 1 STEP 2 Writef (x) in factored form + + – – for Example 2 YOU TRY Find all zeros of the polynomial function. 3. f (x) = x3 + 7x2 + 15x + 9 SOLUTION Formula are (x +1)2 (x +3) f(x) = (x +1) (x +3)2 The zeros of f are – 1 and – 3

  10. Because the coefficients are rational and 2 + 5is a zero, 2–5 must also be a zero by the irrational conjugates theorem. Use the three zeros and the factor theorem to write f (x) as a product of three factors. EXAMPLE 3 Use zeros to write a polynomial function SOLUTION

  11. f (x) = (x – 3) [ x – (2 + √ 5 ) ] [ x – (2 – √ 5 ) ] = (x – 3) [ (x – 2) – √ 5 ] [ (x – 2) +√ 5 ] EXAMPLE 3 Use zeros to write a polynomial function Write f (x) in factored form. Regroup terms. = (x – 3)[(x – 2)2 – 5] Multiply. = (x – 3)[(x2 – 4x + 4) – 5] Expand binomial. = (x – 3)(x2 – 4x – 1) Simplify. = x3 – 4x2 – x – 3x2 + 12x + 3 Multiply. Combine like terms. = x3 – 7x2 + 11x + 3

  12. f(2 + √ 5 ) = (2 + √ 5 )3 – 7(2 + √ 5 )2 + 11( 2 + √ 5 ) + 3 = 38 + 17 √5– 63 – 28 √5+ 22 + 11√5+ 3 Sincef (2 + √5 ) = 0, by the irrational conjugates theorem f (2 –√ 5) = 0.  EXAMPLE 3 Use zeros to write a polynomial function CHECK You can check this result by evaluating f (x) at each of its three zeros. f(3) = 33 – 7(3)2 + 11(3) + 3 = 27 – 63 + 33 + 3 = 0  = 0 

  13. f (x) = (x – 4) [ x – (1 + √ 5 ) ] [ x – (1 – √ 5 ) ] Because the coefficients are rational and 1 + 5 is a zero, 1 – 5 must also be a zero by the irrational conjugates theorem. Use the three zeros and the factor theorem to write f(x) as a product of three factors = (x – 4) [ (x – 1) – √ 5 ] [ (x – 1) +√ 5 ] = (x – 4)[(x – 1)2 – ( 5)2] YOU TRY for Example 3 SOLUTION Write f (x) in factored form. Regroup terms. Multiply. = (x – 4)[(x2 – 2x + 1) – 5] Expand binomial.

  14. YOU TRY for Example 3 = (x – 4)(x2 – 2x – 4) Simplify. = x3 – 2x2 – 4x – 4x2 + 8x + 16 Multiply. = x3 – 6x2 + 4x +16 Combine like terms.

  15. √6 Because the coefficients are rational and 2i is a zero, –2i must also be a zero by the complex conjugates theorem. 4 + 6 is also a zero by the irrational conjugate theorem. Use the five zeros and the factor theorem to write f(x) as a product of five factors. f (x) = (x–2) (x +2i)(x-2i)[(x –(4–√6 )][x –(4+√6) ] = (x – 2) [ (x2–(2i)2][x2–4)+√6][(x– 4) – √6 ] = (x – 2)[(x2+ 4)[(x– 4)2 – ( 6 )2] YOU TRY for Example 3 SOLUTION Write f (x) in factored form. Regroup terms. Multiply. = (x – 2)(x2 + 4)(x2– 8x+16 – 6) Expand binomial.

  16. YOU TRY for Example 3 = (x – 2)(x2 + 4)(x2– 8x + 10) Simplify. = (x–2) (x4– 8x2 +10x2 +4x2 –3x +40) Multiply. = (x–2) (x4 – 8x3+14x2 –32x + 40) Combine like terms. = x5– 8x4 +14x3 –32x2 +40x– 2x4 +16x3 –28x2 + 64x – 80 Multiply. = x5–10x4 + 30x3 – 60x2 +104x– 80 Combine like terms.

  17. TACHOMETER A tachometer measures the speed (in revolutions per minute, or RPMs) at which an engine shaft rotates. For a certain boat, the speed xof the engine shaft (in 100s of RPMs) and the speed sof the boat (in miles per hour) are modeled by s (x) = 0.00547x3 – 0.225x2 + 3.62x – 11.0 What is the tachometer reading when the boat travels 15miles per hour? EXAMPLE 4 Approximate real zeros of a polynomial model

  18. ANSWER The tachometer reading is about 1990RPMs. EXAMPLE 4 Approximate real zeros of a polynomial model SOLUTION Substitute 15 for s(x) in the given function. You can rewrite the resulting equation as: 0 = 0.00547x3– 0.225x2 + 3.62x – 26.0 Then, use a graphing calculator to approximate the real zeros of f (x) = 0.00547x3– 0.225x2 + 3.62x – 26.0. From the graph, there is one real zero:x ≈ 19.9.

  19. ANSWER The zeros are x ≈– 2.2,x ≈ – 0.3, and x ≈ 1.1. for Examples 4 YOU TRY 6. Approximate the real zeros of f (x) = 3x5 + 2x4 – 8x3 + 4x2 – x – 1.

  20. ANSWER about 70,000 envelopes YOU TRY 7. The profit P for printing envelopes is modeled by P = x – 0.001x3 – 0.06x2 + 30.5x, where x is the number of envelopes printed in thousands. What is the least number of envelopes that can be printed for a profit of $1500?

  21. ANSWER +√5, + 2i ANSWER x3 + x2 + 44x + 150. KEEP GOING 1. Find all the zeros of f(x) = x4 – x2 – 20. 2. Write a polynomial function of least degree that has rational coefficients, a leading coefficient of 1, and – 3 and 1 – 7i

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