Verifying Upper & Lower Bounds

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Section 2.5
Question 88

Please do 88(a & b) as a guide for me to do the rest? Thank you.
20210919_212059.jpg
 
88.

f(x)=x^3-4x^2+1

Find all of the possible roots, ±p/q, then apply the synthetic division to find the upper and lower bounds.
p=±1
q=±1

p/q= 1/1, 1/-1,-1/-1,-1/1 These are the possible roots of the polynomial function

If you divide a polynomial function f (x) by (x - c), where c > 0, using synthetic division and this yields all positive numbers, then c is an upper bound.

If you divide a polynomial function f(x) by (x - c), where c < 0, using synthetic division and this yields alternating signs, then c is a lower bound to the real roots of the equation f(x) = 0. Special note that zeros can be either positive or negative. Note that two things must occur for c to be a lower bound.

x^3-4x^2+1
1| 1 -4 1
..|
..| 1 -3
-----------------------
..| 1 -3 -2 ->

x^2-5x+6
(1)^2-5(1)+6=1-5+6-> c > 0 or positive, the other is that all the coefficients of the quotient as well as the remainder are not positive =>No Upper Bounds

-1| 1 -4 1
..|
..| -1 5
-----------------------
..| 1 -5 6

x^2-5x+6
(-1)^2-5(-1)+6=1+5+6-> successive coefficients of the quotient and the remainder have alternating signs=>Lower Bound: -1
 
88.

f(x)=x^3-4x^2+1

Find all of the possible roots, ±p/q, then apply the synthetic division to find the upper and lower bounds.
p=±1
q=±1

p/q= 1/1, 1/-1,-1/-1,-1/1 These are the possible roots of the polynomial function

If you divide a polynomial function f (x) by (x - c), where c > 0, using synthetic division and this yields all positive numbers, then c is an upper bound.

If you divide a polynomial function f(x) by (x - c), where c < 0, using synthetic division and this yields alternating signs, then c is a lower bound to the real roots of the equation f(x) = 0. Special note that zeros can be either positive or negative. Note that two things must occur for c to be a lower bound.

x^3-4x^2+1
1| 1 -4 1
..|
..| 1 -3
-----------------------
..| 1 -3 -2 ->

x^2-5x+6
(1)^2-5(1)+6=1-5+6-> c > 0 or positive, the other is that all the coefficients of the quotient as well as the remainder are not positive =>No Upper Bounds

-1| 1 -4 1
..|
..| -1 5
-----------------------
..| 1 -5 6

x^2-5x+6
(-1)^2-5(-1)+6=1+5+6-> successive coefficients of the quotient and the remainder have alternating signs=>Lower Bound: -1

This concept is new to me. Thanks. I will play with a few questions on paper. We move on to Rational Functions or Section 2.6.
 

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