PRECAL PROBELMS HELP ASAP?

Question

Consider the following.
g(x) = 6x4 − 19x3 − 86x2 + 304x − 160
(a) Use the zero or root feature of a graphing utility to approximate the zeros of the function accurate to three decimal places. (Enter your answers as a comma-separated list.)
x =


(b) Determine the exact value of one of the zeros (use synthetic division to verify your result).
x =


(c) Factor the polynomial completely.
g(x) =

 

Answer 1

a) The polynomial function g(x) = 6x⁴ - 19x³- 86x² + 304x - 160

y = 6x⁴ - 19x³- 86x² + 304x - 160

Test points

Make the table of values to for the polynomial.

Choose random values for x and find the corresponding values for y.

x	y = 6x4 - 19x3- 86x2 + 304x - 160	(x, y)
- 3	y = 6(-3)4 - 19(-3)3- 86(-3)2 + 304(-3) - 160	(-3, -847)
-2	y = 6(-2)4 - 19(-2)3- 86(-2)2 + 304(-2) - 160	(- 2, -864)
-1	y = 6(-1)4 - 19(-1)3- 86(-1)2 + 304(-1) - 160	(- 1, - 525)
0	y = 6(0)4 - 19(0)3- 86(0)2 + 304(0) - 160	(0, -160)
1	y = 6(1)4 - 19(1)3- 86(1)2 + 304(1) - 160	(1, 45)
2	y = 6(2)4 - 19(2)3- 86(2)2 + 304(2) - 160	(2, 48)
3	y = 6(3)4 - 19(3)3- 86(3)2 + 304(3) - 160	(3, - 49)

End behavior g(x) = 6x⁴ - 19x³- 86x² + 304x - 160

Degree of polynomial is 4 and leading coefficient 6.

The graph of a polynomial function is always a smooth curve; that is, it has no breaks or corners.

All even degree polynomials behave on their ends like quadratics.

All even degree polynomials are either up on both ends and or down on both ends.depending on whether the polynomial has, respectively, a positive or negative leading coefficient.

The above polynomial even degree  polynomial with a positive leading coefficient .

So the graph up on both ends.

Comments

Contd....

Graph

1.Draw a coordinate plane.

2.Plot the coordinate points found in the table.

3.Then sketch the graph, connecting the points with a smooth curve.

The points where it crosses the x axis will give solutions to the polynomial function g(x).

The graph crosses the x - axis at x = -4, 2/3, 5/2, 4

From the graph the zeros are x = {-4, 0.666, 2.5, 4}.

Answer 2

b) g(x) = 6x⁴ - 19x³ - 86x² + 304x - 160

Rational Root Theorem, if a rational number in simplest form p/q is a root of the polynomial equation a_nxⁿ + a_n – 1x^{n – 1} + ... + a₁x + a₀ = 0, then p is a factor of a₀ and q is a factor if a_n.

If p/q is a rational zero, then p is a factor of 6 and q is a factor of 1.

The possible values of p are   ± 1,  ± 2, ± 3,± 4, ± 8, ± 10, ± 20, ± 40, ± 80 and ± 160.

The possible values for q are ± 1, ± 2, ± 3 and ± 6.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1,  ± 2, ± 3,± 4, ± 8, ± 10, ± 20, ± 40, ± 80 , ± 60,±1/2, ±3/2, ±5, ±1/3, ±2/3, ±4/3, ±8/3, ±10/3, ±20/3, ±40/3, ±80/3, ±160/3, ± 1/6, ±5/3.

Make a table for the synthetic division and test possible real zeros.

p/q	6	-19	-86	304	-160
1	6	-13	-99	205	45
2	6	-7	-100	104	48
3	6	-1	-89	37	- 49
4	6	5	-66	40	0

Since f(4) = 0,  x = 4 is a zero. The depressed polynomial is  6x³ + 5x² - 66x + 40 = 0.

If p/q is a rational zero, then p is a factor of 40 and q is a factor of 6.

Comments

Contd..

The possible values of p are   ± 1,  ± 2, ± 4, ± 5, ± 8, ± 10, ± 20 and ± 40.

The possible values of q are   ± 1,  ± 2, ± 3, and ± 6.

By the Rational Roots Theorem, the only possible rational roots are, p/q = ± 1,  ± 2, ± 4, ± 5, ± 8, ± 10, ± 20, ± 40, ±1/2, ±5/2, ±1/3, ±2/3, ±4/3, ±5/3, ±8/3, ±10/3, ± 20/3 , ±40/3, ±1/6,±5/6.

Make a table for the synthetic division and test possible real zeros.

p/q	6	5	-66	40
1	6	11	-55	-15
2	6	17	-32	-24
- 4	6	-19	10	0

Since f(- 4) = 0,  x = - 4 is a zero. The depressed polynomial is  6x² - 19x + 10 = 0

6x² - 15x - 4x + 10 = 0

3x(2x - 5) - 2(2x - 5) = 0

(2x - 5)(3x - 2) = 0

x = 5/2, x = 2/3.

Zeros are x = 5/2, 2/3, 4 and - 4.

Answer 3

c) The polynomial function g(x) = 6x⁴ - 19x³- 86x² + 304x - 160

From Factor theorem,

When x - c is a factor of the polynomial then f(c) = 0.

Observe the solution b,

Real zeros are 4, - 4, 5/2 and 2/3.

Factoring of 6x⁴ - 19x³- 86x² + 304x - 160 = (x - 4)(x + 4)(2x - 5)(3x - 2).