Download Multiplying and Factoring Polynomials Part 1 Students should feel

Multiplying and Factoring Polynomials Part 1
Students should feel fairly comfortable performing the following operations on polynomials.
• Adding and Subtracting: (4a2 + 2b) + (5a2 − 3b) = 9a2 − b
• Multiplying: (Using the distributive property)
– 3x(4x + 9) = 3x × 4x + 3x × 9 = 12x2 + 27x
– (x + 9)(2x − 6) = 2x2 − 6x + 18x − 54 = 2x2 + 12x − 54
In the second example I have used the FOIL method, which is really just a convenient
way to distribute. I multiply the following pairs in this order.
First (x × 2x) + Outer (x × −6) + Inner (9 × 2x) + Last (9 × −6)
• Factoring: Remember a factor of a number is an integer that evenly divides that number. For
example 5 is a factor of 15. We can say 15 = 5 × 3. When working with polynomials we use
the same concept. A factor of a polynomial is a number or polynomial that evenly divides
it. For example consider the polynomial 2x2 + 5x. If I am asked to factor this problem I need
to consider what numbers/polynomials will multiply together to produce this one? This can
be challenging at first but once you learn some tricks and start to see patterns it gets easier.
In our example I notice that I can write the polynomial given as a product in this manner:
2x2 + 5x = x(2x + 5). Therefore x and 2x + 5 are factors of 2x2 + 5x.
Being able to factor polynomials of second degree (x2 ), which are also called quadratics, is
essential for success on this exam. Students should be familiar with how to factor the following
types of quadratics and the procedures for factoring them.
– Simple Trinomials: These are of the form x2 + bx + c, where b is an integer coefficients
and c is an integer.
– Complex Trinomials with a leading coefficient: These are of the form ax2 + bx + c,
where a and b are integer coefficients and c is an integer.
– Perfect Square Trinomials: These are of the form (x + y)2 = (x + y)(x + y) =
x2 + 2xy + y 2 ,
– Difference of Squares: These are polynomials of the form x2 − y 2 = (x + y)(x − y)
The following three exam questions utilize a students knowledge of multiplying and factoring polynomials.
1. Expand and Simplify: (x − 4)3
Expand and Simplify simply means Multiply and gather like terms together through addition
or subtraction. For this problem we first need to note that (x − 4)3 = (x − 4)(x − 4)(x − 4)
Now we simply use the FOIL method for multiplication twice to find our answer.
(x − 4)(x − 4)(x − 4) = (x − 4)(x2 − 4x − 4x + 16)
Notice I multiplied the final two monomials together.
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(x − 4)(x2 − 4x − 4x + 16) = (x − 4)(x2 − 8x + 16)
Next I combined the like terms.
My next step will be to multiply these two polynomials. It will be slightly more complicated
because one of them has three parts rather then just two. What I will do is multiply x by
each part of the second polynomial and then multiply -4 by each part. I’ll put brackets
around the results so you can differentiate them.
(x − 4)(x2 − 8x + 16) = [x3 − 8x2 + 16x] + [−4x2 + 32x − 64]
Now I simply group like terms together - that is the Simplify part!
[x3 − 8x2 + 16x] + [−4x2 + 32x − 64] =
x3 − 8x2 − 4x2 + 16x + 32x − 64 =
x3 − 12x2 + 48x − 64
2. Factor y 2 − 81
You may notice that this is a special kind of quadratic, one called a difference of squares. As
the name states they are simply the difference of two perfect squares. These are polynomials
that factor in the following pattern: x2 − y 2 = (x + y)(x − y). The trick with identifying these
polynomials is to check that each element is a perfect square, there are only two elements,
and they are being subtracted. It is possible to factor this problem without knowing this
definition, but if you know it it will speed up your progress.
x2 − y 2 = (x + y)(x − y)
y 2 − 81 = (y + 9)(y − 9)
3. Given that the area of a rectangle below is 2x2 +9x−5, determine its length if the width is x+5.
length
Rectangle
x+5
In solving this problem we need to remember that to find the area of a rectangle we multiply
length times width.
area of a rectangle = length × width
2x2 + 9x − 5 = (length) × (x + 5)
Essentially this problem is asking us to factor 2x2 + 9x − 5 and giving us one of the factors.
2x2 + 9x − 5 = (?x−!)(x + 5)
I can work backwards using the FOIL method to figure out what ? and ! are.
I know ?x × x = 2x, so ? = 2
2x2 + 9x − 5 = (2x−!)(x + 5)
Using this information I know that 2x × 5 plus −! × x equals 9x
10x−!x = 9x, therefore ! = 1
2x2 + 9x − 5 = (2x − 1)(x + 5)
The length of my rectangle is (2x − 1).
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In order to make sure my calculations are correct I can expand (2x − 1)(x + 5). It should
equal 2x2 + 9x − 5
(2x − 1)(x + 5) = 2x2 + 10x − x − 5 = 2x2 + 9x − 5
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