All categories

2 years ago

PLEASE HELP ! Find the zeros of the given function.

Mathematics

2 answers:

photoshop1234 [79]2 years ago

4 0

Answer:

-1, 4+i, 4-i

Step-by-step explanation:

x^4- 6x^3 + 2x^2 + 26x + 17

Using the rational root theorem

we see if 1, -1, -17 or 17 are roots

Check and see if 1 is a root

1^4- 6(1^3) + 2(1^2) + 26(1) + 17=0

1-6+2+26+17 does not equal 0 1 is not a root

-1

1^4- 6(-1^3) + 2(1^2) + 26(-1) + 17=0

1 +6 +2 -26+17 = 0

-1 is a root

Factor out (x+1)

(x+1) ( x^3-7x^2+9x+17)

Using the rational root theorem again on x^3-7x^2+9x+17

Checking -1

-1 -7 -9 +17=0

-1 is a root

(x+1) (x+1) (x^2-8x+17)

Using the quadratic on the last

8 ±sqrt(8^2 - 4(1)17)

--------------------------------

gives imaginary roots

4±i

Dennis_Churaev [7]2 years ago

3 0

Answer:

x = -1, 4 + i, 4 - i

Step-by-step explanation:

Possible rational roots are:

+/- 1, +/- 17

Using trial method:

f(-1) = (-1)⁴ - 6(-1)³ + 2(-1)² + 26(-1) + 17

f(-1) = 0

f(1) = (1)⁴ - 6(1)³ + 2(1)² + 26(1) + 17

f(1) = 40

f(-17) = (-17)⁴ - 6(-17)³ + 2(-17)² + 26(-17) + 17

f(-17) = 113152

f(17) = (17)⁴ - 6(17)³ + 2(17)² + 26(17) + 17

f(17) = 55080

This implies that x = -1 is a root, with multiplicity of 2 or 4

Let the other quadratic factor be g(x)

g(x) = [x⁴ - 6x³ + 2x² + 26x + 17] ÷ (x + 1)²

g(x) = [x⁴ - 6x³ + 2x² + 26x + 17] ÷ (x² + 2x + 1)

g(x) = [x²(x² + 2x + 1) - 8x(x² + 2x + 1) + 17(x² + 2x + 1)] ÷ (x² + 2x + 1)

g(x) = x² - 8x + 17

g(x) = 0

x = [-(-8) +/- sqrt[(-8)² - 4(1)(17)]/2

x = [8 +/- 2i]/2

x = 4 +/- i

You might be interested in

For the function y=3x2: (a) Find the average rate of change of y with respect to x over the interval [3,6]. (b) Find the instant

nirvana33 [79]

Answer:

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

Step-by-step explanation:

a) Geometrically speaking, the average rate of change of $y$ with respect to $x$ over the interval by definition of secant line:

$r = \frac{y(b) -y(a)}{b-a}$ (1)

Where:

$a$ , $b$ - Lower and upper bounds of the interval.

$y(a)$ , $y(b)$ - Function exaluated at lower and upper bounds of the interval.

If we know that $y = 3\cdot x^{2}$ , $a = 3$ and $b = 6$ , then the average rate of change of $y$ with respect to $x$ over the interval is:

$r = \frac{3\cdot (6)^{2}-3\cdot (3)^{2}}{6-3}$

$r = 27$

The average rate of change of $y$ with respect to $x$ over the interval $[3,6]$ is 27.

b) The instantaneous rate of change can be determined by the following definition:

$y' = \lim_{h \to 0}\frac{y(x+h)-y(x)}{h}$ (2)

Where:

$h$ - Change rate.

$y(x)$ , $y(x+h)$ - Function evaluated at $x$ and $x+h$ .

If we know that $x = 3$ and $y = 3\cdot x^{2}$ , then the instantaneous rate of change of $y$ with respect to $x$ is:

$y' = \lim_{h \to 0} \frac{3\cdot (x+h)^{2}-3\cdot x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{(x+h)^{2}-x^{2}}{h}$

$y' = 3\cdot \lim_{h \to 0} \frac{2\cdot h\cdot x +h^{2}}{h}$

$y' = 6\cdot \lim_{h \to 0} x +3\cdot \lim_{h \to 0} h$

$y' = 6\cdot x$

$y' = 6\cdot (3)$

$y' = 18$

The instantaneous rate of change of $y$ with respect to $x$ at the value $x = 3$ is 18.

5 0

2 years ago

X - 4y = 28<br> 2x + y =2

Tresset [83]

X - 4y = 28
2x + y = 2

solve for x
y= -2x + 2
y= x/4 - 7

solve for system of equations
(4,-6)
x= 4, y= -6

7 0

2 years ago

Help me solve using either elimination or comparison.

Nadya [2.5K]

I hope this helps you

8 0

2 years ago

Write the formula, substitution, and solve. A parallelogram is being painted on the wall of a playroom. The parallelogram measur

Andreas93 [3]

Answer:

Area: 40

Step-by-step explanation:

To find the area of a parallelogram you multiply the base times the height, or the length times the height. This is why you multiply 5x8 to get 40. The substitution could be 5 x 8 = X

X is the area.

6 0

2 years ago

Which angles are adjacent angles to 3? Select all that apply

sattari [20]

Two Answers: Angle 1, Angle 4

Adjacent angles share a common segment, line, or ray. Think of two adjacent rooms sharing a common wall.

3 0

2 years ago