
Concept explainers
To calculate: The zeros of the function f(x)=48x4−52x3+13x−3 .

Answer to Problem 37PPS
The zeros of the function f(x)=48x4−52x3+13x−3 are x=−12 ,x=12,x=13 and x=34 .
Explanation of Solution
Given information:
The function f(x)=48x4−52x3+13x−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)=48x4−52x3+13x−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)=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.
So, count the number of times the sign changes between the coefficients of f(x)=48x4−52x3+13x−3
There are 3 or 1 positive real zeros.
Now,
f(−x)=48x4+52x3−13x−3
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 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)=48x4+52x3−13x−3 .
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 zeros3103+1+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 48 and constant term is 3 Therefore, p is a factor of 48 and q is a factor of 3.
The possible combinations of pq in simplest form are,
pq=±1,±3,±12,±32,±13,±14,±34,±16,±18,±38,±112,116,±316,±124,±148
Next, construct a table with help of synthetic substitution to compute the value of f(x) for real values of x .
x48−52013−3−148−100100−8784−348−196588−17515250−1248−3819−30−1348−6836894927−13027
As observed one zero is resulted at x=−12 .
Now, the depressed polynomial is obtained is 24x3−38x2+19x−3 . Again use synthetic substitution to factor the polynomial.
x24−3819−3124−14521212−1330138−1030146−8114−14346−510
The zeros are obtained at x=12,x=13 and x=34 .
Thus, the zeros of the function f(x)=48x4−52x3+13x−3 are x=−12 ,x=12,x=13 and x=34 .
Chapter 5 Solutions
Glencoe Algebra 2 Student Edition C2014
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