SOLUTION: Problem #1
For the function
h(x) = x³ - x² - 17x - 15,
Use long division to determine which of the following are factors of h(x).
a) x + 5 b) x + 1
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-> SOLUTION: Problem #1
For the function
h(x) = x³ - x² - 17x - 15,
Use long division to determine which of the following are factors of h(x).
a) x + 5 b) x + 1
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Question 108149: Problem #1
For the function
h(x) = x³ - x² - 17x - 15,
Use long division to determine which of the following are factors of h(x).
a) x + 5 b) x + 1 c) x + 3
Problem #2
Determine the oblique asymptote of the graph of the function.
For the function
h(x) = x³ - x² - 17x - 15,
Use long division to determine which of the following are factors of h(x).
a) x + 5 b) x + 1 c) x + 3
Divide x³ - x² - 17x - 15 by x + 5 to see if (a) is a factor of h(x)
x² - 6x + 13
--------------------
x + 5)x³ - x² - 17x - 15
x² + 5x²
--------
- 6x² - 17x
- 6x² - 30x
-----------
13x - 15
13x + 65
--------
- 80
No, (a) is not a factor because we get a remainder of -80,
not 0.
So, we divide x3 - x2 - 17x - 15 by x + 1 to see if (b) is a factor
of h(x)
x² - 2x - 15
--------------------
x + 1)x³ - x² - 17x - 15
x² + x²
--------
- 2x² - 17x
- 2x² - 2x
-----------
- 15x - 15
- 13x - 15
--------
0
Yes, (b) is a factor of h(x) because we get a remainder of 0.
Divide x3 - x2 - 17x - 15 by x + 5 to see if (c) is a factor
of h(x).
x² - 4x - 5
--------------------
x + 3)x³ - x² - 17x - 15
x² + 3x²
--------
- 4x² - 17x
- 4x² - 12x
-----------
- 5x - 15
- 5x - 15
--------
0
Yes, (c) is a factor of h(x) because we get a remainder of 0.
So (b) and (c) are factors and (a) is not a factor.
------------------------------------------------------------
Problem #2
Determine the oblique asymptote of the graph of the function.
g(x)= (x² + 4x - 1)/(x + 3)
We divide the expression on the right using long division:
x + 1
------------
x + 3)x² + 4x + 1
x² + 3x
-------
x + 1
x + 3
-----
- 2
Now use h(x) = = +
h(x) = x + 1 +
The fraction on the right becomes nearer and nearer to zero as x takes
on values with larger and larger absolute values. So therefore the graph
of h(x) becomes nearer and nearer to the graph of the right hand side
without the term, so the oblique (slanted)
asymptote has the equation:
y = x + 1
Here is the graph of h(x) and the green line is the oblique asymptote,
whose equation is y = x + 1
[The blue line is the vertical asymptote, whose equation is x = -3
gotten by setting the denominator x+3 equal to 0.]
Edwin