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if f(x)=6x^4-17x^3-57x^2+153x+27, and the graphing calculator tells you that x=3 and x=-3 are integer roots, algebraically determine when f(x)>0

if f(x)=6x^4-17x^3-57x^2+153x+27, and the graphing calculator tells you that x=3 and-example-1
User Dnozay
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2 Answers

13 votes

Answer:

Explanation:

if f(x)=6x^4-17x^3-57x^2+153x+27, and the graphing calculator tells you that x=3 and-example-1
User Amico
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4.3k points
6 votes

Answer:

The roots are x = 3, x = - 3 and x = - 1/6

Explanation:

The given polynomial is expressed as

f(x) = 6x^4 - 17x^3 - 57x^2 + 153x + 27

Since x = 3 and x = - 3 are roots of the polynomial, it means that the factors are

(x - 3) and (x + 3)

We would divide the polynomial by either of the factors. Let us divide it by x - 3 by using the long division method. The steps are shown below:

The quotient from the division is

6x^3 + x^2 - 54x - 9

We would divide 6x^3 + x^2 - 54x - 9 by (x - 3) again. The steps are shown below:

We would factor 6x^2 + 19x + 3

We have

6x^2 + 18x + x + 3

6x(x + 3) + 1(x + 3) = 0

(6x + 1)(x + 3) = 0

6x + 1 = 0 or x + 3 = 0

6x = - 1 or x = - 3

x = - 1/6 or x = 3

The roots are x = 3, x = - 3 and x = - 1/6



if f(x)=6x^4-17x^3-57x^2+153x+27, and the graphing calculator tells you that x=3 and-example-1
if f(x)=6x^4-17x^3-57x^2+153x+27, and the graphing calculator tells you that x=3 and-example-2
User Matthewsteele
by
4.7k points