7.4 the remainder and factor theorems
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7.4 The Remainder and Factor Theorems. Use Synthetic Substitution to find Remainders. A review of Synthetic Substitution. Synthetic Division a method of dividing polynomials using the coefficients of the divided polynomial and its divisor. Synthetic Division definition. - PowerPoint PPT PresentationTRANSCRIPT
7.4 The Remainder and Factor Theorems
Use Synthetic Substitution to find Remainders
A review of Synthetic Substitution
Synthetic Division a method of dividing polynomials using the coefficients of the divided polynomial and its divisor.
Synthetic Division definition
Synthetic Division a method of dividing polynomials using the coefficients of the divided polynomial and its divisor.
Since the divided polynomial is in descending order, we can just use the coefficient as they are written.
4642 23 xxxx
Synthetic Division definition
2 | 1 -4 6 -4 Inside the box is the zero of x - 2
Synthetic can only be used when the divisor has a degree of 1
Synthetic Division definition
2 | 1 -4 6 -4 Inside the box is 2 -4 4 the zero of x - 2
1 - 2 2 0
We drop the first number the then multiply by 2 and add it to the next number. Repeat till done.
Synthetic Division definition
2 | 1 -4 6 -4 Inside the box is 2 -4 4 the zero of x - 2
1 - 2 2 0
We drop the first number the then multiply by 2 and add it to the next number. Repeat till done.
Synthetic Division definition
2 | 1 -4 6 -4 Inside the box is 2 -4 4 the zero of x - 2
1 - 2 2 0
Since the Remainder is Zero, (x – 2) is a factor of 464 23 xxx
A new way to Evaluate polynomials
Find
Use Synthetic Substitution
223)4( 234 xxxf
A new way to Evaluate polynomials
Find
Use Synthetic Substitution
4 | 3 -2 1 0 - 2
12
3
223)4( 234 xxxf
A new way to Evaluate polynomials
Find
Use Synthetic Substitution
4 | 3 -2 1 0 - 2
12 40
3 10
223)4( 234 xxxf
A new way to Evaluate polynomials
Find
Use Synthetic Substitution
4 | 3 -2 1 0 - 2
12 40 164
3 10 41
223)4( 234 xxxf
A new way to Evaluate polynomials
Find
Use Synthetic Substitution
4 | 3 -2 1 0 - 2
12 40 164 656
3 10 41 164
223)4( 234 xxxf
A new way to Evaluate polynomials
Find
Use Synthetic Substitution
4 | 3 -2 1 0 - 2
12 40 164 656
3 10 41 164 654
223)4( 234 xxxf
A new way to Evaluate polynomials
Find
Use Synthetic Substitution
4 | 3 -2 1 0 - 2
12 40 164 656
3 10 41 164 654
Remainder is 654, so f(4) = 654
223)4( 234 xxxf
To show a binomial is a Factor the remainder equals zero
Show (x – 3) is a factor of
x3 + 4x2 – 15x – 18.
To show a binomial is a Factor the remainder equals zero
Show (x – 3) is a factor of
x3 + 4x2 – 15x – 18.
3 | 1 4 -15 -18
3 21 18
1 7 6 0
Since the remainder is zero, ( x – 3) is a factor
To show a binomial is a Factor the remainder equals zero
Show (x – 3) is a factor of
x3 + 4x2 – 15x – 18.
3 | 1 4 -15 -18
3 21 18
1 7 6 0
If you were going to find the other factors, you would use the depressed equation.
x2 + 7x + 6 = (x + 6)(x + 1)
equationdepressed
x3 + 4x2 – 15x – 18 factor
Using the synthetic substitution and factoring the depressed equation gives us the factors of the polynomial.
x3 + 4x2 – 15x – 18 = (x – 3)(x + 1)(x + 6)
Homework
Page 368 – 369
# 13 – 29 odd
Homework
Page 368 – 369
# 14 – 30 even