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Answer:
The quotient when is divided by
is
with no remainder, so x+2 is a factor of p(x).
Step-by-step explanation:
As the question suggests, there are two ways to solve this problem: long division (which can always be used to divide polynomials), and synthetic division (which can only be used when you are dividing by something of the form . In this case, we may use synthetic division, since
is equal to
. I will use synthetic division here as it is slightly faster than long division.
The -2 on the left represents that we are dividing by x-(-2), and the rest of the numbers in the top row are the coefficients of p(x): 1, -4, -8, and 8.
The first step in synthetic division is to being down the leading coefficient of the polynomial: in this case, the 1 (as indicated in red). Now, we multiply the 1 by -2. We get -2, which we place directly below the -4.
Next, we add directly down the column, (-4 + (-2) is equal to -6), and this answer is placed in the box below. We can continue this process, getting the coefficients 1, -6, 4, and 0 in the bottom row.
This is the answer: the quotient is , and the remainder is zero (which indicates that x+2 is a factor of p(x)).
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Edit: long division
We may also solve this problem using long division (see the second image). The first step is to look at the leading coefficients: since is equal to
, the first term in the quotient will be
. Since
is equal to
, we must now subtract that from
.
We repeat the same process, as shown in the image. Since we eventually get to zero, the remainder is zero, and the polynomial at the top (x^2-6x+4) is the quotient.
<em>I know this might be difficult to follow, so please comment if you have any questions.</em>
Answer:
Given the data in the question;
Samples of size 100: 6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.
a)
For a p chart ( control chart for fraction nonconforming), the center line and upper and lower control limits are;
UCL = p" + 3√[ (p"(1-P")) / n ]
CL = p"
LCL = p" - 3√[ (p"(1-P")) / n ]
here, p" is the average fraction defective
Now, with the 30 samples of size 100
p" = [∑(6, 7, 3, 9, 6, 9, 4, 14, 3, 5, 6, 9, 6, 10, 9, 2, 8, 4, 8, 10, 10, 8, 7, 7, 7, 6, 14, 18, 13, 6.)] / [ 30 × 100 ]
p" = 234 / 3000
p" = 0.078
so the trial control limits for the fraction-defective control chart are;
UCL = p" + 3√[ (p"(1-P")) / n ]
UCL = 0.078 + 3√[ (0.078(1-0.078)) / 100 ]
UCL = 0.078 + ( 3 × 0.026817 )
UCL = 0.078 + 0.080451
UCL = 0.1585
LCL = p" - 3√[ (p"(1-P")) / n ]
LCL = 0.078 - 3√[ (0.078(1-0.078)) / 100 ]
LCL = 0.078 - ( 3 × 0.026817 )
LCL = 0.078 - 0.080451
LCL = 0 ( SET TO ZERO )
Diagram of the Chart uploaded below
b)
from the p chart for a) below, sample 28 violated the first western electric rule,
summary report from Minitab;
TEST 1. One point more than 3.00 standard deviations from the center line.
Test failed at points: 28
Hence, we conclude that the process is out of statistical control
Lets Assume that assignable causes can be found to eliminate out of control points.
Since 28 is out of control, we should eliminate this sample and recalculate the trial control limits for the P chart.
so
p" = 0.0745
UCL = p" + 3√[ (p"(1-P")) / n ]
UCL = 0.0745 + 3√[ (0.0745(1-0.0745)) / 100 ]
UCL = 0.0745 + ( 3 × 0.026258 )
UCL = 0.0745 + 0.078774
UCL = 0.1532
LCL = p" - 3√[ (p"(1-P")) / n ]
LCL = 0.0745 - 3√[ (0.0745(1-0.0745)) / 100 ]
LCL = 0.0745 - ( 3 × 0.026258 )
LCL = 0.0745 - 0.078774
UCL = 0 ( SET TO ZERO )
The second p chart diagram is upload below;
NOTE; the red circle symbol on 28 denotes that the point is not used in computing the control limits
