Answer The zeros of the function f ( x ) = x 5 − 2 x 4 − 12 x 3 − 12 x 2 − 13 x − 10 are x = − 1 , x = − 2 , x = i , x = − i and x = 5 . Explanation Given information: The function f ( x ) = x 5 − 2 x 4 − 12 x 3 − 12 x 2 − 13 x − 10 . Formula used: A polynomial of n degree has n zeros, which can be either real or imaginary. Descartes’ rule of signs states that consider a polynomial P ( x ) = a n x n + ⋯ + a 1 x + a 0 with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of P ( x ) or is less than this by an even number, the number of times the sign between coefficients changes is equal to count of negative real zeroes of P ( − x ) or is less than this by an even number. Calculation: Consider the function f ( x ) = x 5 − 2 x 4 − 12 x 3 − 12 x 2 − 13 x − 10 . Observe that degree of polynomial is 5, so the functions has 5 zeros which can be either real or imaginary. Descartes’ rule of signs states that consider a polynomial P ( x ) = a n x n + ⋯ + a 1 x + a 0 with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of P ( x ) or is less than this by an even number. So, count the number of times the sign changes between the coefficients of f ( x ) = x 5 − 2 x 4 − 12 x 3 − 12 x 2 − 13 x − 10 There is 1 positive real zero. Now, f ( − x ) = − x 5 − 2 x 4 + 12 x 3 − 12 x 2 + 13 x − 10 Descartes’ rule of signs states that consider a polynomial P ( x ) = a n x n + ⋯ + a 1 x + a 0 with real coefficients then, the number of times the sign between coefficients changes is equal to count of negative real zeroes of P ( − x ) or is less than this by an even number. So, count the number of times the sign changes between the coefficients of f ( − x ) = − x 5 − 2 x 4 + 12 x 3 − 12 x 2 + 13 x − 10 . There are 4, 2 or 0 negative real zeros. Next, construct a table with possible combinations of real and imaginary zeros. Number of Positive Real zeros Number of Negative Real zeros Number of Imaginary zeros Total Number of zeros 1 4 0 5 1 2 2 5 1 0 4 5 Recall that the Rational zero theorem states that provided a polynomial P ( x ) with integral coefficients then every rational zero is of the form p q that is a rational number in simplest form. Here, p is a factor of the constant term and q is a factor of leading coefficient. For the provided function leading coefficient is 1 and constant term is 10 Therefore, p is a factor of 10 and q is a factor of 1. The possible combinations of p q in simplest form are, p q = ± 1 , ± 2 , ± 5 , ± 10 Next, construct a table with help of synthetic substitution to compute the value of f ( x ) for real values of x . x 1 − 2 − 12 − 12 − 13 − 10 1 1 − 1 − 13 − 25 − 38 − 48 − 1 1 − 3 − 9 − 3 − 10 0 2 1 0 − 12 − 36 − 85 − 180 − 2 1 − 4 − 4 − 4 − 5 0 5 1 3 3 3 2 0 − 5 1 − 7 23 − 127 622 − 3120 As observed three zeros are resulted at x = − 1 , x = − 2 and x = 5 . Multiply the factors together, ( x + 1 ) ( x + 2 ) ( x − 5 ) = x 3 − 2 x 2 − 13 x − 10 Now divide the polynomial f ( x ) = x 5 − 2 x 4 − 12 x 3 − 12 x 2 − 13 x − 10 by ( x 3 − 2 x 2 − 13 x − 10 ) . ( x 3 − 2 x 2 − 13 x − 10 ) x 2 + 1 x 5 − 2 x 4 − 12 x 3 − 12 x 2 − 13 x − 10 − x 5 + 2 x 4 + 13 x 3 + 10 x 2 x 3 − 2 x 2 − 13 x − 10 − x 3 + 2 x 2 + 13 x + 10 0 Now, the depressed polynomial is obtained is x 2 + 1 . Factor the polynomial. x 2 + 1 = 0 x = ± i The zeros are obtained at ± i . Thus, the zeros of the function f ( x ) = 48 x 4 − 52 x 3 + 13 x − 3 are x = − 1 , x = − 2 , x = i , x = − i and x = 5 .

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Author: McGraw-Hill Glencoe

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Chapter 5.8, Problem 36PPS

To determine

To calculate: The zeros of the function f(x)=x5−2x4−12x3−12x2−13x−10 .

Expert Solution & Answer

Answer to Problem 36PPS

The zeros of the function f(x)=x5−2x4−12x3−12x2−13x−10 are x=−1,x=−2,x=i,x=−i and x=5 .

Explanation of Solution

Given information:

The function f(x)=x5−2x4−12x3−12x2−13x−10 .

Formula used:

A polynomial of n degree has n zeros, which can be either real or imaginary.

Descartes’ rule of signs states that consider a polynomial P(x)=anxn+⋯+a1x+a0 with real coefficients then, the number of times the sign between coefficients changes is equal to count of positive real zeroes of P(x) or is less than this by an even number, the number of times the sign between coefficients changes is equal to count of negative real zeroes of P(−x) or is less than this by an even number.

Calculation:

Consider the function f(x)=x5−2x4−12x3−12x2−13x−10 .

Observe that degree of polynomial is 5, so the functions has 5 zeros which can be either real or imaginary.

So, count the number of times the sign changes between the coefficients of f(x)=x5−2x4−12x3−12x2−13x−10

There is 1 positive real zero.

Now,

f(−x)=−x5−2x4+12x3−12x2+13x−10

So, count the number of times the sign changes between the coefficients of f(−x)=−x5−2x4+12x3−12x2+13x−10 .

There are 4, 2 or 0 negative real zeros.

Next, construct a table with possible combinations of real and imaginary zeros.

Number of PositiveReal zerosNumber of NegativeReal zerosNumber ofImaginary zerosTotal Numberof zeros140512251045

Recall that the Rational zero theorem states that provided a polynomial P(x) with integral coefficients then every rational zero is of the form pq that is a rational number in simplest form. Here, p is a factor of the constant term and q is a factor of leading coefficient.

For the provided function leading coefficient is 1 and constant term is 10 Therefore, p is a factor of 10 and q is a factor of 1.

The possible combinations of pq in simplest form are,

pq=±1,±2,±5,±10

Next, construct a table with help of synthetic substitution to compute the value of f(x) for real values of x .

x1−2−12−12−13−1011−1−13−25−38−48−11−3−9−3−100210−12−36−85−180−21−4−4−4−505133320−51−723−127622−3120

As observed three zeros are resulted at x=−1,x=−2 and x=5 .

Multiply the factors together,

(x+1)(x+2)(x−5)=x3−2x2−13x−10

Now divide the polynomial f(x)=x5−2x4−12x3−12x2−13x−10 by (x3−2x2−13x−10) .

(x3−2x2−13x−10)x2+1x5−2x4−12x3−12x2−13x−10−x5+2x4+13x3+10x2x3−2x2−13x−10−x3+2x2+13x+100

Now, the depressed polynomial is obtained is x2+1 .

Factor the polynomial.

x2+1=0x=±i

The zeros are obtained at ±i .

Thus, the zeros of the function f(x)=48x4−52x3+13x−3 are x=−1,x=−2,x=i,x=−i and x=5 .

Chapter 5.8, Problem 35PPS

Chapter 5.8, Problem 37PPS

Chapter 5 Solutions

Glencoe Algebra 2 Student Edition C2014

Ch. 5.1 - Prob. 1AGP Ch. 5.1 - Prob. 1BGP Ch. 5.1 - Prob. 1CGP Ch. 5.1 - Prob. 1DGP Ch. 5.1 - Prob. 2AGP Ch. 5.1 - Prob. 2BGP Ch. 5.1 - Prob. 3AGP Ch. 5.1 - Prob. 3BGP Ch. 5.1 - Prob. 4AGP Ch. 5.1 - Prob. 4BGP

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The zeros of the function f ( x ) = x 5 − 2 x 4 − 12 x 3 − 12 x 2 − 13 x − 10 .