Explanation Given: The polynomial is P ( x ) = x 5 − x 4 + 5 x 3 − 5 x 2 + 4 x − 4 and one zero of P ( x ) is i . Theorems used: Conjugate zeros theorem: If a polynomial function P ( x ) has a zero a + b i , then the conjugate a − b i is also a zero of the polynomial P ( x ) . Number of zeros theorem : The function defined by a polynomial of degree n can have at most n distinct zeros. Factor theorem: If a polynomial P ( x ) has a zero k then the factor of that polynomial is ( x − k ) . Calculation: The polynomial P ( x ) has a zero i , so by the conjugate zero theorem − i is also a zero of the polynomial and as the polynomial has a degree of 5 so it must have five zeros by number of zeros theorem. The polynomial P ( x ) have zeros i and − i then by the factor theorem the linear factors of the polynomial are ( x − i ) and ( x + i ) . Use the synthetic division method by dividing the polynomial by ( x − i ) to get the remaining zeros. In the synthetic division method, take the coefficients of the polynomial in the descending order of powers of variable and if any term is missing, use a zero in that place to act as a placeholder. Since there is no term missing, take the dividend expression same as x 5 − x 4 + 5 x 3 − 5 x 2 + 4 x − 4 . Arrange the coefficients in descending order and write the values in the form as shown below. i 1 − 1 5 − 5 4 − 4 Bring down the first coefficient of expression and then multiply it with the given number to get i ( 1 ) = i . i 1 − 1 5 − 5 4 − 4 ↓ i _ 1 Now, add − 1 and i to obtain − 1 + i . i 1 − 1 5 − 5 4 − 4 ↓ i _ 1 − 1 + i The obtained value is again multiplied with the given number to get i ( − 1 + i ) = − i − 1 . i 1 − 1 5 − 5 4 − 4 ↓ i − i − 1 _ 1 − 1 + i Add 5 and − i − 1 to obtain 4 − i . i 1 − 1 5 − 5 4 − 4 ↓ i − i − 1 _ 1 − 1 + i 4 − i The obtained value is again multiplied with the given number to get i ( 4 − i ) = 4 i + 1 . i 1 − 1 5 − 5 4 − 4 ↓ i − i − 1 4 i + 1 _ 1 − 1 + i 4 − i Add − 5 and 4 i + 1 to obtain 4 i − 4 . i 1 − 1 5 − 5 4 − 4 ↓ i − i − 1 4 i + 1 _ 1 − 1 + i 4 − i 4 i − 4 The obtained value is again multiplied with the given number to get i ( 4 i − 4 ) = − 4 − 4 i . i 1 − 1 5 − 5 4 − 4 ↓ i − i − 1 4 i + 1 − 4 − 4 i _ 1 − 1 + i 4 − i 4 i − 4 Add 4 and − 4 − 4 i to obtain − 4 i . i 1 − 1 5 − 5 4 − 4 ↓ i − i − 1 4 i + 1 − 4 − 4 i _ 1 − 1 + i 4 − i 4 i − 4 − 4 i The obtained value is again multiplied with the given number to get i ( − 4 i ) = 4 . i 1 − 1 5 − 5 4 − 4 ↓ i − i − 1 4 i + 1 − 4 − 4 i 4 _ 1 − 1 + i 4 − i 4 i − 4 − 4 i Add − 4 and 4 to obtain 0

A Graphical Approach to College Algebra, Books a la Carte Edition (6th Edition)

6th Edition

ISBN: 9780321900715

Author: John Hornsby, Margaret L. Lial

Publisher: PEARSON

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Chapter 3.7, Problem 71E

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To factor: The polynomial P(x)=x5−x4+5x3−5x2+4x−4 completely by using the zero i.

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Chapter 3.7, Problem 70E

Chapter 3.7, Problem 72E

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A Graphical Approach to College Algebra, Books a la Carte Edition (6th Edition)

Ch. 3.1 - For each complex number, (a) state the real part,...Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - For each complex number, (a) state the real part,...Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E

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To factor: The polynomial P ( x ) = x 5 − x 4 + 5 x 3 − 5 x 2 + 4 x − 4 completely by using the zero i .

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To factor: The polynomial P(x)=x5−x4+5x3−5x2+4x−4 completely by using the zero i.

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Chapter 3 Solutions