I'm pretty positive it would be the second one from the top
Answer:
Step-by-step explanation:
First confirm that x = 1 is one of the zeros.
f(1) = 2(1)^3 - 14(1)^2 + 38(1) - 26
f(1) = 2 - 14 + 38 - 26
f(1) = -12 + 38 = + 26
f(1) = 26 - 26
f(1) = 0
=========================
next perform a long division
x -1 || 2x^3 - 14x^2 + 38x - 26 || 2x^2 - 12x + 26
2x^3 - 2x^2
===========
-12x^2 + 28x
-12x^2 +12x
==========
26x -26
26x - 26
========
0
Now you can factor 2x^2 - 12x + 26
2(x^2 - 6x + 13)
The discriminate of the quadratic is negative. (36 - 4*1*13) = - 16
So you are going to get a complex result.
x = -(-6) +/- sqrt(-16)
=============
2
x = 3 +/- 2i
f(x) = 2*(x - 1)*(x - 3 + 2i)*(x - 3 - 2i)
The zeros are
1
3 +/- 2i
Probably the subsitution method
y=1/2x
subsitute that fr y
2x+3(1/2x)=28
2x+3/2x=28
times 2 both sides
4x+3x=56
7x=56
divide by 7 both sides
x=8
sub back
y=1/2x
y=1/2(8)
y=4
(x,y)
(8,4)
That correct answer is 0.0625
D = 0.32
d - 8 / 5 = -4d
d - 1.6 = -4d
d + 4d = 1.6
5d = 1.6
d = 0.32
This is an exact question in my math textbook. It should be correct.
