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Section 6.5
Factoring Binomials
Review of factoring techniques:
1. ALWAYS look for a GCF first with any polynomial
you are trying to factor.
2. After that, count how many terms are in the
polynomial. If there are
– FOUR terms: Use factoring by grouping.
– THREE terms ( a trinomial):
• If the leading coefficient is 1 (if it looks like x2 + bx + c),
– Find factors (x + t)(x + v) such that t + v = b and t*v = c.
• If the leading coefficient is not 1 (if it looks like ax2 + bx + c),
– Use the British method, which starts with multiplying a*c,
and then finding factors of that number that add up to b.
Today we are going to learn techniques for
factoring two-term polynomials (binomials).
Case one:
Binomials in which both terms are perfect squares.
Previously, we discovered a formula for finding the
product of the sum and difference of two terms:
(a – b)(a + b) = a2 – b2
We can use the reverse of this equation to factor the
difference of 2 squares.
a2 – b2 = (a – b)(a + b)
Examples:
a2 – b2 = (a – b)(a + b)
Factor x2 – 16.
Since this polynomial can be written as (x)2 – (4)2,
x2 – 16 = (x – 4)(x + 4).
Factor 9x2 – 4.
Since this polynomial can be written as (3x)2 – (2)2,
9x2 – 4 = (3x – 2)(3x + 2).
Factor 16x2 – 9y2.
Since this polynomial can be written as (4x)2 – (3y)2,
16x2 – 9y2 = (4x – 3y)(4x + 3y).
Examples
a2 – b2 = (a – b)(a + b)
Factor x8 – y6.
Since this polynomial can be written as (x4)2 – (y3)2,
x8 – y6 = (x4 – y3)(x4 + y3).
Factor x2 + 4.
This one is the sum of two squares, not the
difference of squares, so it can’t be factored.
This polynomial is a prime polynomial.
Example:
a2 – b2 = (a – b)(a + b)
Factor 36x2 – 64.
Remember that you should always factor out any common
factors first, before you start any other technique.
Step 1: Factor out the GCF, which in this case is 4.
36x2 – 64 = 4(9x2 – 16)
Step 2: Factor the polynomial 9x2 – 16
The polynomial can be written as (3x)2 – (4)2,
so (9x2 – 16) = (3x – 4)(3x + 4).
Our final result is 36x2 – 64 = 4(3x – 4)(3x + 4).
↑
(Don’t forget to write in 4 (the GCF)
as part of your final answer!)
Example from the online homework:
Note that 4096 is a HUGE NUMBER. But if it is a square, then we’re
in business.
How do you tell if it’s a perfect square? Take the square root
(You can do this with your on-line calculator.)
Answer: 4096 = 642. So s4 – 4096 = (s2)2 – 642 = (s2 – 64)(s2 + 64)
Are we done yet?
No, because s2 – 64 factors further into (s + 8)(s – 8)
Final answer: (s + 8)(s – 8)(s2 + 64)
There are two additional types of binomials that can be
factored easily by remembering a formula.
We have not studied these special products previously, as
they involve cubes of terms, rather than just squares.
a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a – b)(a2 + ab + b2)
[You can prove that these formulas work by multiplying out the
two factors on the right side of each equation. If you do this,
you’ll get six products at the first stage, but you’ll find that
everything cancels out except the a3 and b3 terms.]
NOTE: These formulas are on the formula sheet that
you can use on each test and quiz, so you don’t have to
memorize them, but you do need to know how to apply
them.)
Formulas:
Sum of cubes: a3 + b3 = (a + b)(a2 – ab + b2)
Diff. of cubes: a3 – b3 = (a – b)(a2 + ab + b2)
Examples
HINT: It will help you a lot when doing cube problems if you
always start by writing blank parentheses, then figure out what
goes in them. Those will be the “a” and “b” in your formula.
1. Factor x3 + 1.
HINT: This one looks like ( )3 + ( )3.
Since this polynomial can be written as (x)3 + (1)3,
we can use the sum of cubes formula, with a = x and b = 1
Answer: x3 + 1 = (x + 1)(x2 – x + 1).
2. Factor y3 – 64.
HINT: This one looks like ( y )3 - ( 4 )3. What goes in the parentheses?
Therefore we can use the difference of cubes formula with a = y and b = 4.
Answer: y3 – 64 = (y – 4)(y2 + 4y + 16).
Examples:
a3 + b3 = (a + b)(a2 – ab + b2)
a3 – b3 = (a – b)(a2 + ab + b2)
3. Factor 8t3 + s6.
HINT: This one looks like ( )3 + ( )3. What goes in the parentheses?
This polynomial can be written as (2t)3 + (s2)3, so
8t3 + s6 = (2t + s2)((2t)2 – (2t)(s2) + (s2)2)
= (2t + s2)(4t2 – 2s2t + s4).
4. Factor x3y6 – 27z3.
HINT: This one looks like ( )3 - ( )3. What goes in the parentheses?
This polynomial can be written as (xy2)3 – (3z)3, so
x3y6 – 27z3 = (xy2 – 3z)((xy2)2 + (3z)(xy2) + (3z)2)
= (xy2 – 3z)(x2y4 + 3xy2z + 9z2).
Remember to ALWAYS check to see if you can factor
out any common factors before attempting to use any
other factoring techniques or formulas.
Example
Factor 375y6 – 24y3.
Step 1: Factor out the GCF.
(Tip: Since 375 is such a big number, start by factoring the smaller
number 24, then see if any of its factors will divide into 375. You will find
that the number 3 is a divisor of both 24 and 375.)
375y6 – 24y3 = 3y3(125y3 – 8)
Since the second part can be written as (5y)3 – 23,
125y3 – 8 = (5y – 2)((5y)2 + (5y)(2) + 22)
= (5y – 2)(25y2 + 10y + 4).
Final answer: 3y3(5y – 2)(25y2 + 10y + 4).
The assignment on this material (HW 6.5)
Is due at the start of the next class session.
You may now OPEN
your LAPTOPS
and begin working on the
homework assignment.