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Transcript 
example 2
Combining Graphical and Algebraic Methods
Solve the equation x3  9 x 2  610 x  600  0 .
Chapter 6.4
Solve the equation x3  9 x 2  610 x  600  0 .
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
P(x)
P( x)  x 3  9 x 2  610 x  600
x
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
P(x)
P( x)  x 3  9 x 2  610 x  600
(1,0)
x
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
P(x)
P( x)  x 3  9 x 2  610 x  600
(1,0)
x
P(1)  1?  9 1  610 1  600  0
3
2
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
P(x)
P( x)  x 3  9 x 2  610 x  600
(1,0)
x
P(1)  1  9 1  610 1  600  0
3
2
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
P(x)
P( x)  x 3  9 x 2  610 x  600
(1,0)
x
P(1)  1  9 1  610 1  600  0
3
2
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
1. Arrange the coefficients in descending powers of x, with a 0 for any
missing power. Place a from x - a to the left of the coefficients.
1 1  9  610  600
1  10  600
1  10  600 
0
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
1. Arrange the coefficients in descending powers of x, with a 0 for any
missing power. Place a from x - a to the left of the coefficients.
1 1  9  610  600
1  10  600
1  10  600 
0
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
2. Bring down the first coefficient to the third line. Multiply the last number
in the third line by a and write the product in the second line under the
next term.
1 1  9  610  600
1  10  600
1  10  600 
0
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
2. Bring down the first coefficient to the third line. Multiply the last number
in the third line by a and write the product in the second line under the
next term.
1 1  9  610  600
1  10  600
Multiply
1  10  600 
0
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
3. Add the last number in the second line to the number above it in the first
line. Continue this process until all numbers in the first line are used.
1 1  9  610  600
1  10  600
1  10  600 
0
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
3. Add the last number in the second line to the number above it in the first
line. Continue this process until all numbers in the first line are used.
1 1  9  610  600
1  10  600
1  10  600 
0
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
3. Add the last number in the second line to the number above it in the first
line. Continue this process until all numbers in the first line are used.
1 1  9  610  600
1  10  600
1  10  600 
0
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
3. Add the last number in the second line to the number above it in the first
line. Continue this process until all numbers in the first line are used.
1 1  9  610  600
1  10  600
1  10  600 
0
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
3. Add the last number in the second line to the number above it in the first
line. Continue this process until all numbers in the first line are used.
1 1  9  610  600
1  10  600
1  10  600 
0
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
4. The third line represents the coefficients of the quotient, with the last
number the remainder. The quotient is a polynomial of degree one less
than the dividend.
1 1  9  610  600
1  10  600
1  10  600 
0
Remainder
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
4. If the remainder is 0, x – a is a factor of the polynomial, and the
polynomial can be written as the product of the divisor x - a and the
quotient.
x3  9 x 2  610 x  600   x  1  x 2  10 x  600 
1 1  9  610  600
1  10  600
1  10  600 
0
Remainder
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
4. If the remainder is 0, x – a is a factor of the polynomial, and the
polynomial can be written as the product of the divisor x - a and the
quotient.
Quotient
Divisor
x3  9 x 2  610 x  600   x  1  x 2  10 x  600 
1 1  9  610  600
1  10  600
1  10  600 
0
Remainder
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
x 3  9 x 2  610 x  600  0
 x  1  x 2  10 x  600   0
x  1  0 or
x 1
x 2  10 x  600  0
 x  30  x  20   0
x  30  0
or
x  30
x  20  0
x  20
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
x 3  9 x 2  610 x  600  0
 x  1  x 2  10 x  600   0
x  1  0 or
x 1
x 2  10 x  600  0
 x  30  x  20   0
x  30  0
or
x  30
x  20  0
x  20
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
x 3  9 x 2  610 x  600  0
 x  1  x 2  10 x  600   0
x  1  0 or
x 1
x 2  10 x  600  0
 x  30  x  20   0
x  30  0
or
x  30
x  20  0
x  20
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
x 3  9 x 2  610 x  600  0
 x  1  x 2  10 x  600   0
x  1  0 or
x 1
x 2  10 x  600  0
 x  30  x  20   0
x  30  0
or
x  30
x  20  0
x  20
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
x 3  9 x 2  610 x  600  0
 x  1  x 2  10 x  600   0
x  1  0 or
x 1
x 2  10 x  600  0
 x  30  x  20   0
x  30  0
or
x  30
x  20  0
x  20
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
x 3  9 x 2  610 x  600  0
 x  1  x 2  10 x  600   0
x  1  0 or
x 1
x 2  10 x  600  0
 x  30  x  20   0
x  30  0
or
x  30
x  20  0
x  20
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
x 3  9 x 2  610 x  600  0
 x  1  x 2  10 x  600   0
x  1  0 or
x 1
x 2  10 x  600  0
 x  30  x  20   0
x  30  0
or
x  30
x  20  0
x  20
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
P(x)
P( x)  x 3  9 x 2  610 x  600
x
2009 PBLPathways
Solve the equation x3  9 x 2  610 x  600  0 .
P(x)
(-30,0)
(1, 0)
P( x)  x 3  9 x 2  610 x  600
(20, 0)
x
2009 PBLPathways
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