The document discusses using synthetic division to evaluate polynomials at specific values and factor polynomials. It provides examples of using synthetic division to:
1) Evaluate polynomials like f(x) = x^2 - x + 5 at specific values such as f(-2).
2) Factor polynomials when one factor is known, such as factoring x^3 - 3x^2 - 13x + 15 after determining (x + 3) is a factor using synthetic division.
3) Determine if a binomial is a factor of a polynomial, such as showing (x - 3) is a factor of x^3 + 4x^2 - 15x - 18.
2. Simplify Polynomials using Synthetic Substitution
• Recall Synthetic Division can be used to divide a
polynomial by a binomial.
3. Simplify Polynomials using Synthetic Substitution
• Recall Synthetic Division can be used to divide a
polynomial by a binomial.
• Synthetic Division can also be used to evaluate a
polynomial at a specific value.
4. Simplify Polynomials using Synthetic Substitution
• Recall Synthetic Division can be used to divide a
polynomial by a binomial.
• Synthetic Division can also be used to evaluate a
polynomial at a specific value.
• Such as given f ( x ) = x − x + 5 , find f ( −2 )
2
5. Simplify Polynomials using Synthetic Substitution
• Recall Synthetic Division can be used to divide a
polynomial by a binomial.
• Synthetic Division can also be used to evaluate a
polynomial at a specific value.
• Such as given f ( x ) = x − x + 5 , find f ( −2 )
2
• The x value goes in the box and the coefficients of the
polynomial make up the first line. Proceed as usual.
6. Simplify Polynomials using Synthetic Substitution
• Recall Synthetic Division can be used to divide a
polynomial by a binomial.
• Synthetic Division can also be used to evaluate a
polynomial at a specific value.
• Such as given f ( x ) = x − x + 5 , find f ( −2 )
2
• The x value goes in the box and the coefficients of the
polynomial make up the first line. Proceed as usual.
−2 1 −1 5
−2 6
1 −3 11
7. Simplify Polynomials using Synthetic Substitution
• Recall Synthetic Division can be used to divide a
polynomial by a binomial.
• Synthetic Division can also be used to evaluate a
polynomial at a specific value.
• Such as given f ( x ) = x − x + 5 , find f ( −2 )
2
• The x value goes in the box and the coefficients of the
polynomial make up the first line. Proceed as usual.
• The remainder is the value. −2 1 −1 5
−2 6
1 −3 11
8. Simplify Polynomials using Synthetic Substitution
• Recall Synthetic Division can be used to divide a
polynomial by a binomial.
• Synthetic Division can also be used to evaluate a
polynomial at a specific value.
• Such as given f ( x ) = x − x + 5 , find f ( −2 )
2
• The x value goes in the box and the coefficients of the
polynomial make up the first line. Proceed as usual.
• The remainder is the value. −2 1 −1 5
• So −2 6
f ( −2 ) = 11 1 −3 11
9. Try it: Simplify using Synthetic Substitution
Find g ( 5 ) for g ( x ) = −x + x − 7x + 8
3 2
10. Try it: Simplify using Synthetic Substitution
Find g ( 5 ) for g ( x ) = −x + x − 7x + 8
3 2
5 −1 1 −7 8
−5 −20 −135
−1 −4 −27 −127
11. Try it: Simplify using Synthetic Substitution
Find g ( 5 ) for g ( x ) = −x + x − 7x + 8
3 2
5 −1 1 −7 8
−5 −20 −135
−1 −4 −27 −127
Therefore, g ( 5 ) = −127.
12. Try it: Simplify using Synthetic Substitution
Find f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6
4 3 2
13. Try it: Simplify using Synthetic Substitution
Find f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6
4 3 2
−3 1 −2 −3 5 −6
−3 15 −36 93
1 −5 12 −31 87
14. Try it: Simplify using Synthetic Substitution
Find f ( −3) for g ( x ) = x − 2x − 3x + 5x − 6
4 3 2
−3 1 −2 −3 5 −6
−3 15 −36 93
1 −5 12 −31 87
Therefore, f ( −3) = 87.
15. Try it: Simplify using Synthetic Substitution
Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1
3 2
Don’t be intimidated by the 2b. Use the same
process as before.
16. Try it: Simplify using Synthetic Substitution
Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1
3 2
Don’t be intimidated by the 2b. Use the same
process as before.
2b 1 3 −2 1
2b 4b 2 + 6b 8b 3 + 12b 2 − 4b
2 3 2
1 2b + 3 4b + 6b − 2 8b + 12b − 4b + 1
17. Try it: Simplify using Synthetic Substitution
Find h ( 2b ) for h ( x ) = x + 3x − 2x + 1
3 2
Don’t be intimidated by the 2b. Use the same
process as before.
2b 1 3 −2 1
2b 4b 2 + 6b 8b 3 + 12b 2 − 4b
2 3 2
1 2b + 3 4b + 6b − 2 8b + 12b − 4b + 1
Therefore, h ( 2b ) = 8b 3 + 12b 2 − 4b + 1.
18. Find Factors given a Polynomial & one factor
• Synthetic Division can be useful in factoring
polynomials that cannot be factored using traditional
methods.
19. Find Factors given a Polynomial & one factor
• Synthetic Division can be useful in factoring
polynomials that cannot be factored using traditional
methods.
• It’s a great way to ‘break down’ the big, nasty looking
polynomials.
20. Find Factors given a Polynomial & one factor
• Synthetic Division can be useful in factoring
polynomials that cannot be factored using traditional
methods.
• It’s a great way to ‘break down’ the big, nasty looking
polynomials.
• Care to factor this polynomial?
3 2
x − 3x − 13x + 15
21. Factoring when know 1 factor
3 2
x − 3x − 13x + 15
• Say we know one factor of this nasty polynomial is
(x + 3) or the zero of the function is x = -3. We can use
synthetic division to help us find the remaining factors.
22. Factoring when know 1 factor
3 2
x − 3x − 13x + 15
• Say we know one factor of this nasty polynomial is
(x + 3) or the zero of the function is x = -3. We can use
synthetic division to help us find the remaining factors.
• First step is to divide by the factor you know.
23. Factoring when know 1 factor
3 2
x − 3x − 13x + 15
• Say we know one factor of this nasty polynomial is
(x + 3) or the zero of the function is x = -3. We can use
synthetic division to help us find the remaining factors.
• First step is to divide by the factor you know.
−3 1 −3 −13 15
−3 18 −15
1 −6 5 0
24. Factoring when know 1 factor
3 2
x − 3x − 13x + 15
• Say we know one factor of this nasty polynomial is
(x + 3) or the zero of the function is x = -3. We can use
synthetic division to help us find the remaining factors.
• First step is to divide by the factor you know.
−3 1 −3 −13 15
−3 18 −15
1 −6 5 0
• Remainder is zero so this indicates that (x + 3) is
indeed a zero of the polynomial.
25. Continued...
−3 1 −3 −13 15
−3 18 −15
1 −6 5 0
• Write the new polynomial from your synthetic
division.
26. Continued...
−3 1 −3 −13 15
−3 18 −15
1 −6 5 0
• Write the new polynomial from your synthetic
division. 2
x − 6x + 5
27. Continued...
−3 1 −3 −13 15
−3 18 −15
1 −6 5 0
• Write the new polynomial from your synthetic
division. 2
x − 6x + 5
• Look, we now have a nice quadratic. Use any
method you’ve learned to factor the quadratic.
28. Continued...
−3 1 −3 −13 15
−3 18 −15
1 −6 5 0
• Write the new polynomial from your synthetic
division. 2
x − 6x + 5
• Look, we now have a nice quadratic. Use any
method you’ve learned to factor the quadratic.
( x − 5 ) ( x − 1)
29. Continued...
3 2
• So the factorization of x − 3x − 13x + 15
is ( x + 3) ( x − 5 ) ( x − 1)
30. Continued...
3 2
• So the factorization of x − 3x − 13x + 15
is ( x + 3) ( x − 5 ) ( x − 1)
• When writing the factorization, be sure to include the
first factor you were given because without it, you do
not have the original polynomial!
31. Continued...
3 2
Now solve x − 3x − 13x + 15 = 0.
32. Continued...
3 2
Now solve x − 3x − 13x + 15 = 0.
Remember the factors are ...
( x + 3) ( x − 5 ) ( x − 1)
33. Continued...
3 2
Now solve x − 3x − 13x + 15 = 0.
Remember the factors are ...
( x + 3) ( x − 5 ) ( x − 1)
Set each factor to 0 and solve.
34. Continued...
3 2
Now solve x − 3x − 13x + 15 = 0.
Remember the factors are ...
( x + 3) ( x − 5 ) ( x − 1)
Set each factor to 0 and solve.
x+3= 0 x−5=0 x −1= 0
35. Continued...
3 2
Now solve x − 3x − 13x + 15 = 0.
Remember the factors are ...
( x + 3) ( x − 5 ) ( x − 1)
Set each factor to 0 and solve.
x+3= 0 x−5=0 x −1= 0
x = −3
36. Continued...
3 2
Now solve x − 3x − 13x + 15 = 0.
Remember the factors are ...
( x + 3) ( x − 5 ) ( x − 1)
Set each factor to 0 and solve.
x+3= 0 x−5=0 x −1= 0
x = −3 x=5
37. Continued...
3 2
Now solve x − 3x − 13x + 15 = 0.
Remember the factors are ...
( x + 3) ( x − 5 ) ( x − 1)
Set each factor to 0 and solve.
x+3= 0 x−5=0 x −1= 0
x = −3 x=5 x =1
38. Continued...
3 2
Now solve x − 3x − 13x + 15 = 0.
Remember the factors are ...
( x + 3) ( x − 5 ) ( x − 1)
Set each factor to 0 and solve.
x+3= 0 x−5=0 x −1= 0
x = −3 x=5 x =1
The zeros are {-3, 1, 5}.
40. Your turn...
Show ( x − 3) is a factor of x + 4x − 15x − 18
3 2
3 1 4 −15 −18
3 21 18
1 7 6 0
Remainder is 0, therefor x - 3 is a factor.
41. Now take the last problem a little further...
3 2
Factor x + 4x − 15x − 18
Remember you were given a factor.
42. Now take the last problem a little further...
3 2
Factor x + 4x − 15x − 18
Remember you were given a factor.
3 1 4 −15 −18
3 21 18
1 7 6 0
43. Now take the last problem a little further...
3 2
Factor x + 4x − 15x − 18
Remember you were given a factor.
3 1 4 −15 −18
3 21 18
1 7 6 0
Write the new polynomial.
44. Now take the last problem a little further...
3 2
Factor x + 4x − 15x − 18
Remember you were given a factor.
3 1 4 −15 −18
3 21 18
1 7 6 0
2
Write the new polynomial. x + 7x + 6
45. Now take the last problem a little further...
3 2
Factor x + 4x − 15x − 18
Remember you were given a factor.
3 1 4 −15 −18
3 21 18
1 7 6 0
2
Write the new polynomial. x + 7x + 6
It’s a quadratic. Factor using
any method you like.
46. Now take the last problem a little further...
3 2
Factor x + 4x − 15x − 18
Remember you were given a factor.
3 1 4 −15 −18
3 21 18
1 7 6 0
2
Write the new polynomial. x + 7x + 6
It’s a quadratic. Factor using
any method you like. ( x + 6 ) ( x + 1)
47. Now take the last problem a little further...
3 2
Factor x + 4x − 15x − 18
Remember you were given a factor.
3 1 4 −15 −18
3 21 18
1 7 6 0
2
Write the new polynomial. x + 7x + 6
It’s a quadratic. Factor using
any method you like. ( x + 6 ) ( x + 1)
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
48. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
49. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
50. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set each factor to 0 and solve.
51. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set each factor to 0 and solve.
x−3= 0 x+6=0 x +1= 0
52. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set each factor to 0 and solve.
x−3= 0 x+6=0 x +1= 0
x=3
53. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set each factor to 0 and solve.
x−3= 0 x+6=0 x +1= 0
x=3 x = −6
54. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set each factor to 0 and solve.
x−3= 0 x+6=0 x +1= 0
x=3 x = −6 x = −1
55. And a little further...
3 2
Find the zeros of x + 4x − 15x − 18
Remember ...
x + 4x − 15x − 18 = ( x − 3) ( x + 6 ) ( x + 1)
3 2
Set each factor to 0 and solve.
x−3= 0 x+6=0 x +1= 0
x=3 x = −6 x = −1
The zeros are {-6, -1, 3}.
56. Practice some more...
• Follow this link to practice using synthetic division to
find the zeros of the polynomials.
