George C.
Jul 24, 2018
(
x
+
2
)
(
x
+
6
)
2
=
0
Explanation:
Given:
x
3
+
14
x
2
+
60
x
+
72
=
0
By the rational roots theorem, any rational zeros of the given cubic are expressible in the form
p
q
for integers
p
,
q
with
p
a divisor of the constant term
72
and
q
a divisor of the coefficient
1
of the leading term.
That means that the only possible rational zeros are:
±
1
,
±
2
,
±
3
,
±
4
,
±
6
,
±
8
,
±
9
,
±
12
,
±
18
,
±
24
,
±
36
,
±
72
In addition, note that all of the coefficients are positive and the constant term is non-zero. As a result, any real zero (rational or otherwise) of this cubic must be negative.
So that leaves rational possibilities:
−
1
,
−
2
,
−
3
,
−
4
,
−
6
,
−
8
,
−
9
,
−
12
,
−
18
,
−
24
,
−
36
,
−
72
We find:
(
−
2
)
3
+
14
(
−
2
)
2
+
60
(
−
2
)
+
72
=
−
8
+
56
−
120
+
72
=
0
So
x
=
−
2
is a zero and
(
x
+
2
)
a factor:
x
3
+
14
x
2
+
60
+
72
=
(
x
+
2
)
(
x
2
+
12
x
+
36
)
Without trying any more of our "possible" zeros, we can recognise the remaining quadratic factor as a perfect square trinomial:
x
2
+
12
x
+
36
=
x
2
+
2
(
x
)
(
6
)
+
6
2
=
(
x
+
6
)
2
So the factored form of the given cubic equation can be written:
(
x
+
2
)
(
x
+
6
)
2
=
0
Answer:
A
<em>can i have brainliest lol</em>
Step-by-step explanation:
when x = 3, the volume is 36.
4 times 9 = 36, so you multiply f(x) by g(x) to get 36
Answer:
The coordinates of the fourth vertex of the rectangle is (3,-8).
Step-by-step explanation:
It is given that a rectangle has vertices at these coordinates. (1, 2) , (3, 2) , (1, −8).
Let the coordinates of the fourth vertex of the rectangle be (x,y).
The diagonals of rectangle intersect each other at their midpoint.
Plot these points on a coordinate plane. From the graph it is noticed that the end point of one diagonal are (3, 2) and (1, −8).
The end point of second diagonal are (1, 2) and (x,y).
Since the diagonals of rectangle intersect each other at their midpoint, therefore
On comparing both the sides,
Therefore the coordinates of the fourth vertex of the rectangle is (3,-8).
<span>The answer to this question is c, two. The top arc of the circle intersects the y=x^2 as each 'limb' extends into infinity. The part of the circle below the x-axis does not intersect the circle. The y=x^2 curve does not dip below the x-axis, but does extend into infinity on both the negative x-axis and positive x-axis.</span>
