Answer The rational zeros of the function f ( x ) = 20 x 4 − 16 x 3 + 11 x 2 − 12 x − 3 are 1 and − 1 5 . Explanation Given information: The function f ( x ) = 20 x 4 − 16 x 3 + 11 x 2 − 12 x − 3 . 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 ) = 20 x 4 − 16 x 3 + 11 x 2 − 12 x − 3 . Observe that degree of polynomial is 4, so the functions has 4 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 ) = 20 x 4 − 16 x 3 + 11 x 2 − 12 x − 3 There are 3 sign changes, so there are 3 or 1 positive real zeros. Now, f ( − x ) = 20 ( − x ) 4 − 16 ( − x ) 3 + 11 ( − x ) 2 − 12 ( − x ) − 3 f ( − x ) = 20 x 4 + 16 x 3 + 11 x 2 + 12 x − 3 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 ) = 20 x 4 + 16 x 3 + 11 x 2 + 12 x − 3 . There is 1 sign change, so there is 1 negative real zero. 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 3 1 0 1 + 3 + 0 = 4 1 1 2 1 + 1 + 2 = 4 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 20 and constant term is − 3 . Therefore, p is a factor of 3 and q is a factor of 20. Factors of 3 are ± 1 and ± 3 . Factors of 20 are ± 1 , ± 2 , ± 4 , ± 5 , ± 10 and ± 20 The possible combinations of p q in simplest form are, p q = ± 1 , ± 1 2 , ± 1 4 , ± 1 5 , ± 1 10 , ± 1 20 , ± 3 , ± 3 2 , ± 3 4 , ± 3 5 , ± 3 10 , ± 3 20 Next, construct a table with help of synthetic substitution to compute the value of f ( x ) for real values of x . p q 20 − 16 11 − 12 − 3 − 3 20 − 76 239 − 729 2184 − 1 20 − 36 47 − 59 56 1 20 4 15 3 0 3 20 44 143 417 1248 As observed one zero is resulted at x = 1 . Since, only one real positive zero is obtained so use the depressed polynomial to obtain remaining zeroes. Now, the depressed polynomial is obtained is f ( x ) = 20 x 3 + 4 x 2 + 15 x + 3 . Group the terms together and factor the polynomial. 20 x 3 + 4 x 2 + 15 x + 3 = 0 4 x 2 ( 5 x + 1 ) + 3 ( 5 x + 1 ) = 0 ( 5 x + 1 ) ( 4 x 2 + 3 ) = 0 Apply the zero product property. Either ( 5 x + 1 ) = 0 or ( 4 x 2 + 3 ) = 0 That is, x = − 1 5 , x = ± i 3 2 . Thus, the rational zeros of the function f ( x ) = 20 x 4 − 16 x 3 + 11 x 2 − 12 x − 3 are 1 and − 1 5 .

1st Edition

ISBN: 9780078884825

Author: McGraw-Hill/Glencoe

Publisher: Glencoe/McGraw-Hill School Pub Co

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Question

Chapter EP, Problem 6.8.6EP

To determine

To calculate: The rational zeros of the function $f(x)=20x4−16x3+11x2−12x−3$ .

Expert Solution & Answer

Answer to Problem 6.8.6EP

The rational zeros of the function $f(x)=20x4−16x3+11x2−12x−3$ are $1 and −15$ .

Explanation of Solution

Given information:

The function $f(x)=20x4−16x3+11x2−12x−3$ .

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 $f(x)=20x4−16x3+11x2−12x−3$ .

Observe that degree of polynomial is 4, so the functions has 4 zeros which can be either real or imaginary.

So, count the number of times the sign changes between the coefficients of $f(x)=20x4−16x3+11x2−12x−3$

There are 3 sign changes, so there are 3 or 1 positive real zeros.

Now,

$f(−x)=20(−x)4−16(−x)3+11(−x)2−12(−x)−3f(−x)=20x4+16x3+11x2+12x−3$

So, count the number of times the sign changes between the coefficients of $f(−x)=20x4+16x3+11x2+12x−3$ .

There is 1 sign change, so there is 1 negative real zero.

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

$Number of PositiveReal zerosNumber of NegativeReal zerosNumber ofImaginary zerosTotal Numberof zeros3101+3+0=41121+1+2=4$

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 20 and constant term is $−3$ . Therefore, p is a factor of $3$ and q is a factor of 20.

Factors of $3$ are $±1 and ±3$ .

Factors of 20 are $±1,±2,±4,±5,±10 and ±20$

The possible combinations of $pq$ in simplest form are,

$pq=±1,±12,±14,±15,±110,±120,±3,±32,±34,±35,±310,±320$

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

$pq20−1611−12−3−320−76239−7292184−120−3647−595612041530320441434171248$

As observed one zero is resulted at $x=1$ . Since, only one real positive zero is obtained so use the depressed polynomial to obtain remaining zeroes.

Now, the depressed polynomial is obtained is $f(x)=20x3+4x2+15x+3$ .

Group the terms together and factor the polynomial.

$20x3+4x2+15x+3=04x2(5x+1)+3(5x+1)=0(5x+1)(4x2+3)=0$

Apply the zero product property.

Either $(5x+1)=0 or (4x2+3)=0$

That is, $x=−15,x=±i32$ .

Thus, the rational zeros of the function $f(x)=20x4−16x3+11x2−12x−3$ are $1 and −15$ .

Chapter EP, Problem 6.8.5EP

Chapter EP, Problem 6.8.7EP

Chapter EP Solutions

Algebra 2

Ch. EP - Prob. 1.1.1EP Ch. EP - Prob. 1.1.2EP Ch. EP - Prob. 1.1.3EP Ch. EP - Prob. 1.1.4EP Ch. EP - Prob. 1.1.5EP Ch. EP - Prob. 1.1.6EP Ch. EP - Prob. 1.1.7EP Ch. EP - Prob. 1.1.8EP Ch. EP - Prob. 1.1.9EP Ch. EP - Prob. 1.1.10EP

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The rational zeros of the function f ( x ) = 20 x 4 − 16 x 3 + 11 x 2 − 12 x − 3 .