Transcript Factoring Higher Degree Poly 1

Factoring Higher
Degree Polynomials
And the Factor Theorem (7.6)
POD
Factor these polynomial functions by grouping. In
this case, there is no middle term to split– we
just group the first pair and last pair.
f ( x)  x 3  2 x 2  3 x  6
f ( x)  x 3  2 x 2  5 x  6
When you can’t do this…
…there are other strategies you can use.
You need to find one factor to start. Synthetic
division will help. We’ll start with this
equation—the leading coefficient is 1, so it is a
simple example.
f ( x)  x 3  2 x 2  5 x  6
What is the constant? What are the factors of
that constant? Be sure to include negative
factors as well.
Using synthetic division
By tables, working on the board, divide one of
those factors into the polynomial using
synthetic division. Which number(s) give a
remainder of 0?
f ( x)  x 3  2 x 2  5 x  6
Using synthetic division
By tables, divide one of those factors into the
polynomial using synthetic division. Which
number(s) give a remainder of 0?
f ( x)  x 3  2 x 2  5 x  6
Let’s use 1. If we have a remainder of 0 when we
divide 1 into the polynomial, what does that
mean about (x-1)? What is the quotient?
Using synthetic division
Can we factor the quotient? If so, what is it?
f ( x)  x3  2 x 2  5x  6  ( x  1)( x 2  x  6)
Using synthetic division
Once we factor that quadratic quotient, we have
factored the entire polynomial.
f ( x)  x3  2 x 2  5x  6  ( x  1)( x  3)( x  2)
And we did it by starting with one factor.
The Factor Theorem
Let’s go back to that first step, when we
found a factor because the remainder
was 0.
The factor is (x-1). What is f(1)?
The Factor Theorem
f(1) = 0
That’s the Factor Theorem:
The (x – a) is a factor of f(x), if and only if
f(a) = 0.
In our polynomial, f(1) = 0, so (x – 1) is a
factor.
Expanding the tool
What about when the leading coefficient is not 1?
Let’s look at another example. This time, the
leading coefficient is 2.
g ( x)  2 x 3  x 2  x  3
What are the factors of 2? Of 3? Be sure to
include positive and negative factors.
Expanding the tool
We use synthetic division again to find a
remainder of 0, but modify what we plug in to
divide.
The divisors are fractions:
factors of constant/factors of leading coefficient.
3 1 3 1
 , , ,
1 1 2 2
In this case, we’ll divide by
. Do it
again by tables, working on the board, to save
some time.
Expanding the tool
Wow, a fraction.
3
g ( x)  2 x 3  x 2  x  3  ( x  )( 2 x 2  2 x  2)
2
Let’s tidy it up some. We want integers.
3
g ( x)  2 x 3  x 2  x  3  ( x  )( 2 x 2  2 x  2)  (2 x  3)( x 2  x  1)
2
I’ll help out by saying that the quadratic factor
cannot be factored any more, so we’re done.
Expanding the tool
Now, graph the polynomial. Where does it cross
the x-axis?
How does this match our factoring?
g ( x)  2 x 3  x 2  x  3  (2 x  3)( x 2  x  1)
When you’re stuck, another tool in the tool box is
to graph and find an x-intercept. That can give
you a zero, which means having a factor to start
with.
A caution
If you divide by all the possible numbers, and
don’t have any remainder of 0, then there are
no rational factors– the polynomial cannot be
factored over the set of rational numbers.
t ( x)  x 3  4 x 2  3 x  2
Finally…
…in answer to a question last year, here are the
patterns for factoring sums and differences of
powers greater than 3. What patterns do you
see? Could you factor a9 + b9?
a 5  b 5  (a  b)( a 4  a 3b  a 2b 2  ab 3  b 4 )
a 5  b 5  (a  b)( a 4  a 3b  a 2b 2  ab 3  b 4 )
a 7  b 7  (a  b)( a 6  a 5b  a 4b 2  a 3b 3  a 2b 4  ab5  b 6 )
a 7  b 7  (a  b)( a 6  a 5b  a 4b 2  a 3b 3  a 2b 4  ab 5  b 6 )