9-1 Adding and Subtracting Polynomials
Polynomials have some special names which we will want to know:
Monomial – One Term
Binomial – Two Terms
Trinomial – Three Terms
The degree of a monomial (or degree of the term) is the sum of the exponents on the variables
Example: 3x2y has a degree of 3 since the exponents on the variables add up to 3
The degree of a polynomial is the highest degree of a term. So to find the degree, begin by finding the degree of each term, whichever term has the highest degree, that is also the degree of the polynomial.
Example: 2x2y5 + 3x4y2 – 6x5y3 + 7x3y
This has 4 terms, the degree of the terms would be as follows:
2x2y5 + 3x4y2 – 6x5y3 + 7x3y
7 6 8 4
Thus the degree of the polynomial would be 8 since that is the highest degree we see in the polynomial.
When adding and subtracting polynomials, you can only combine like terms – or terms with the EXACT SAME VARIABLE including the exponent!
Example: 6x – 5 – (4x + 9) distribute the negative through to get 6x – 5 – 4x – 9 then combine the like terms to get 2x – 14
Sometimes polynomials are stacked vertically, but you can do them horizontally if it is easier to look at distributing when needed.
9-2 Multiplying and Factoring
When multiplying terms, we are really looking at the distributive property – watch your signs!!
When factoring polynomials, we are looking at “un-distributing” by pulling out the biggest common factor, or the GCF.
Example: 6a3b + 24a2b2 – 18ab3
Begin by factoring each term: 6a3b + 24a2b2 – 18ab3
6a3b + 24a2b2 – 18ab3
2*3*a*a*a*b + 2*2*2*3*a*a*b*b - 2*3*3*a*b*b*b
Look to see what they have in common – the GCF
2*3*a*a*a*b + 2*2*2*3*a*a*b*b - 2*3*3*a*b*b*b
The GCF is 2*3*a*b = 6ab so this is what gets pulled out front in factoring, the leftovers stay inside the parenthesis.
6ab(a2 + 4ab – 3b2)
9-3 Multiplying Binomials
When multiplying binomials, you will be distributing for each piece in the first set of parenthesis.
Again, watch the signs!
Multiplying Binomials is sometimes called the FOIL method:
Example: (3x + 2)(5x – 2)
Multiply the First Terms of each binomial (3x + 2)(5x – 2) = 15x2
Multiply the Outside Terms of each (3x + 2)(5x – 2) = - 6x
Multiply the Inside Terms of each (3x + 2)(5x – 2) = 10x
Multiply the Last Terms of each (3x + 2)(5x – 2) = -4
Then add them all together combining like terms where possible:
15x2 – 6x + 10x – 4 = 15x2 + 4x – 4
The FOIL method ONLY works with two binomials, but the process of distributing will work with
any two polynomials being multiplied together. As a quick check, before you combine like terms, you should have the number of terms that would be the product of the number you started with.
Example: (2x + 7) (3x2 – 4x + 8) has 2 x 3 terms, so it should have 6 terms BEFORE you simplify.
Let’s distribute to see:
9-4 Multiplying Special Cases
A Binomial squared can be done the same as two separate binomials, or with a shortcut,
“square, (product) double, square”
Example: (6a + 2b)2 can either be treated as (6a + 2b)(6a + 2b) and FOIL-ed out or you can use the shortcut:
Square the first term (36a2)
Double the product of the two terms (24ab)
Square the last term (4b2).
Put them together to get 36a2 + 24ab + 4b2
When multiplying two binomials that are the same except for the signs, the middle terms will be identical but have opposite signs, which means they will cancel out when you simplify:
(a + b)(a – b) = a2 – ab + ab – b2 = a2 – b2 the result is the difference of squares – notice a2and b2 are both perfect squares
This concept comes in quite handy when factoring – since we know what creates this difference of two perfect squares, we can “undo” the process by taking the square root of each perfect square and making them terms in 2 binomials, with opposite signs.
Example: 49y2 – 16
Both of these terms are perfect squares and we have a difference (subtraction) so we can take the square root of each and create two binomials, identical but with opposite signs.
49y2 has a square root of 7y, and 16 has a square root of 4 so the two binomials look like:
(7y + 4)(7y – 4)
These can always be checked by multiplying the binomials back out.
9-5 Factoring Trinomials With Leading Coefficient of One : x2 + bx + c
When you think about the process of multiplying to get a trinomial, what terms are needed to
get the first term, second term or third term? The key to factoring trinomials is to read the polynomial backwards. It will tell you what to do.
In the form x2 + bx + c, we are looking for factors of c that add (or subtract if the c is negative) to get b. The sign of the middle term is always going to go with the larger factor.
Example:
X2 – 16x + 28
We need factors of 28 that add to get 16.
(that would be 2 & 14 since 2*14 = 28 and 2 + 14 = 16)
The larger factor is 14, so it will be negative – in order to get a positive 28, the 2 will also have to be negative since a negative times a negative is a positive.
Your result looks like: (x – 2)(x – 14) You can always check it by multiplying it back out.
Let’s try another:
X2 + 3x – 54
9-6 Factoring Trinomials With Leading Coefficient Not of One : ax2 + bx + c
This can become quite complex to get the leading coefficient through factors, because the leading coefficient factor will ultimately affect the middle term, which as we know is also affected by the factors of the last term. As a general rule, you should always check to see if there is a GCF you can factor out of a polynomial before you begin your parenthesis factoring.
Example: 12x2 + 14 x - 48
You could start by looking at how all the factors of 12 and 48 could be combined together to get 14, which it won’t take long to discover that could be quite a feat! Instead, begin by pulling out the GCF. In this case, 2.
2(6x2 + 7x – 24)
Now you can look at the factors of 6 and 24 and see what might combine and subtract to give you 7, but that is still quite the ordeal:
Factors of 6: 1, 2, 3, 6
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Instead, try my shortcut:
Multiply the leading coefficient by the last term, change the leading coefficient to 1
Factor as before.
Put the leading coefficient back into BOTH of the binomials
Divide by the GCF of each binomial
Let’s walk through that:
2(6x2 + 7x – 24)
Step one: 2( x2 + 7x – 144) (Multiplied 6 by 24)
Step Two: 2( x + 16)( x - 9) (Factors of 144 that subtract to get 7 – follow sign rules)
Step Three: 2(6x + 16)(6x – 9) (Put leading coefficient back in BOTH binomials)
Step Four: 2(3x + 8)(2x – 3) (Divide by the GCF of each binomial, 2 in the 1st and 3 in the 2nd)
Let’s try another:
8x2 + 18 x + 9
9-7 Factoring Special Cases
A perfect square Trinomial is the result of squaring a binomial. Factoring a PST is reversing the
“Square Double Square” shortcut of multiplying the binomial by itself. So to “undo” that process, we must ask ourselves, can I get the square root of each end? Is the middle twice the product of those roots? If so, it is a Perfect Square Trinomial. Of course you can just use the “regular” rules for factoring as well.
Example: 4x2 - 36x + 81
Both 4x2 and 81 are perfect squares, and 36 is twice the product of those roots. (Since 2x * 9 = 18 and 18 * 2 = 36)
So we write: (2x-9)2 You can always check it by multiplying it out or using the square double square shortcut.
The sign of the binomial will match the middle sign. Since squaring a term ALWAYS results in a positive number, a trinomial can not be a perfect square trinomial if the last term is negative!
Again we revisit the Difference of Squares as well. Recall that the difference of squares comes from multiplying two binomials that have the same terms but different signs. The middle terms cancel out leaving the difference of two perfect squares. Thus to factor a difference of squares, find the root of each term, put it into each of the parenthesis and put one sign positive, the other negative.
Example: 169x4 – 144y2
Both are perfect squares and we are taking the difference. Find each root (13x2 & 12y) put them in each set of parenthesis and use opposite signs.
(13x2 + 12y) (13x2 - 12y)
You can always check by multiplying it out!
Again, don’t forget your first rule of factoring, to look for a GCF before you begin!
9-8 Factoring by Grouping
If a polynomial has more than three terms, you may have to factor only part of it at a time. This will require you to group together terms that have a GCF, factor that part, then see if there is another GCF from among the pieces you grouped together.
Example: 4x3 + 8x2 – 5x – 10
If you group the first two terms and the last two terms you can pull out a GCF from each
(4x3 + 8x2)+( – 5x – 10)
Notice I put a + between them to avoid thinking I was multiplying – that could be dangerous!
4x2(x + 2) + (-5)(x + 2)
I pulled out the GCF of each – notice pulling out a negative changed the – to a positive! Now each of these has a parenthesis in common!
Pull the (x + 2) out and the leftovers go in the second set of parenthesis
(x + 2)(4x2 + -5) or simply (x + 2)(4x2 -5) Again, you can check it by multiplying it back out.
Let’s try another!
6y3 + 3y2 + 2y + 1
You might also have to factor by grouping three terms together, and factor them as a polynomial.
Sometimes I call this method of factoring “Sesame Street Factoring” because we often are looking for something that doesn’t belong.
Example: 4x2 + 4xy + y2 – 9
One of these things is not like the other, one of these things just doesn’t belong! In this case it is the -9 that doesn’t belong. Everything else fits together because of the x-y connection. So group them like this (4x2 + 4xy + y2) – 9 then factor the first three terms as a trinomial (This one happens to be a perfect square trinomial) so it factors as (2x + y)2 – 9 which happens to be the difference of two squares.
Thus we break each perfect square down and use opposite signs to get:
(2x + y + 3) and (2x + y - 3) Whew! It’s like the energizer bunny! Just keeps going and going and going and…
Polynomials have some special names which we will want to know:
Monomial – One Term
Binomial – Two Terms
Trinomial – Three Terms
The degree of a monomial (or degree of the term) is the sum of the exponents on the variables
Example: 3x2y has a degree of 3 since the exponents on the variables add up to 3
The degree of a polynomial is the highest degree of a term. So to find the degree, begin by finding the degree of each term, whichever term has the highest degree, that is also the degree of the polynomial.
Example: 2x2y5 + 3x4y2 – 6x5y3 + 7x3y
This has 4 terms, the degree of the terms would be as follows:
2x2y5 + 3x4y2 – 6x5y3 + 7x3y
7 6 8 4
Thus the degree of the polynomial would be 8 since that is the highest degree we see in the polynomial.
When adding and subtracting polynomials, you can only combine like terms – or terms with the EXACT SAME VARIABLE including the exponent!
Example: 6x – 5 – (4x + 9) distribute the negative through to get 6x – 5 – 4x – 9 then combine the like terms to get 2x – 14
Sometimes polynomials are stacked vertically, but you can do them horizontally if it is easier to look at distributing when needed.
9-2 Multiplying and Factoring
When multiplying terms, we are really looking at the distributive property – watch your signs!!
When factoring polynomials, we are looking at “un-distributing” by pulling out the biggest common factor, or the GCF.
Example: 6a3b + 24a2b2 – 18ab3
Begin by factoring each term: 6a3b + 24a2b2 – 18ab3
6a3b + 24a2b2 – 18ab3
2*3*a*a*a*b + 2*2*2*3*a*a*b*b - 2*3*3*a*b*b*b
Look to see what they have in common – the GCF
2*3*a*a*a*b + 2*2*2*3*a*a*b*b - 2*3*3*a*b*b*b
The GCF is 2*3*a*b = 6ab so this is what gets pulled out front in factoring, the leftovers stay inside the parenthesis.
6ab(a2 + 4ab – 3b2)
9-3 Multiplying Binomials
When multiplying binomials, you will be distributing for each piece in the first set of parenthesis.
Again, watch the signs!
Multiplying Binomials is sometimes called the FOIL method:
Example: (3x + 2)(5x – 2)
Multiply the First Terms of each binomial (3x + 2)(5x – 2) = 15x2
Multiply the Outside Terms of each (3x + 2)(5x – 2) = - 6x
Multiply the Inside Terms of each (3x + 2)(5x – 2) = 10x
Multiply the Last Terms of each (3x + 2)(5x – 2) = -4
Then add them all together combining like terms where possible:
15x2 – 6x + 10x – 4 = 15x2 + 4x – 4
The FOIL method ONLY works with two binomials, but the process of distributing will work with
any two polynomials being multiplied together. As a quick check, before you combine like terms, you should have the number of terms that would be the product of the number you started with.
Example: (2x + 7) (3x2 – 4x + 8) has 2 x 3 terms, so it should have 6 terms BEFORE you simplify.
Let’s distribute to see:
9-4 Multiplying Special Cases
A Binomial squared can be done the same as two separate binomials, or with a shortcut,
“square, (product) double, square”
Example: (6a + 2b)2 can either be treated as (6a + 2b)(6a + 2b) and FOIL-ed out or you can use the shortcut:
Square the first term (36a2)
Double the product of the two terms (24ab)
Square the last term (4b2).
Put them together to get 36a2 + 24ab + 4b2
When multiplying two binomials that are the same except for the signs, the middle terms will be identical but have opposite signs, which means they will cancel out when you simplify:
(a + b)(a – b) = a2 – ab + ab – b2 = a2 – b2 the result is the difference of squares – notice a2and b2 are both perfect squares
This concept comes in quite handy when factoring – since we know what creates this difference of two perfect squares, we can “undo” the process by taking the square root of each perfect square and making them terms in 2 binomials, with opposite signs.
Example: 49y2 – 16
Both of these terms are perfect squares and we have a difference (subtraction) so we can take the square root of each and create two binomials, identical but with opposite signs.
49y2 has a square root of 7y, and 16 has a square root of 4 so the two binomials look like:
(7y + 4)(7y – 4)
These can always be checked by multiplying the binomials back out.
9-5 Factoring Trinomials With Leading Coefficient of One : x2 + bx + c
When you think about the process of multiplying to get a trinomial, what terms are needed to
get the first term, second term or third term? The key to factoring trinomials is to read the polynomial backwards. It will tell you what to do.
In the form x2 + bx + c, we are looking for factors of c that add (or subtract if the c is negative) to get b. The sign of the middle term is always going to go with the larger factor.
Example:
X2 – 16x + 28
We need factors of 28 that add to get 16.
(that would be 2 & 14 since 2*14 = 28 and 2 + 14 = 16)
The larger factor is 14, so it will be negative – in order to get a positive 28, the 2 will also have to be negative since a negative times a negative is a positive.
Your result looks like: (x – 2)(x – 14) You can always check it by multiplying it back out.
Let’s try another:
X2 + 3x – 54
9-6 Factoring Trinomials With Leading Coefficient Not of One : ax2 + bx + c
This can become quite complex to get the leading coefficient through factors, because the leading coefficient factor will ultimately affect the middle term, which as we know is also affected by the factors of the last term. As a general rule, you should always check to see if there is a GCF you can factor out of a polynomial before you begin your parenthesis factoring.
Example: 12x2 + 14 x - 48
You could start by looking at how all the factors of 12 and 48 could be combined together to get 14, which it won’t take long to discover that could be quite a feat! Instead, begin by pulling out the GCF. In this case, 2.
2(6x2 + 7x – 24)
Now you can look at the factors of 6 and 24 and see what might combine and subtract to give you 7, but that is still quite the ordeal:
Factors of 6: 1, 2, 3, 6
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Instead, try my shortcut:
Multiply the leading coefficient by the last term, change the leading coefficient to 1
Factor as before.
Put the leading coefficient back into BOTH of the binomials
Divide by the GCF of each binomial
Let’s walk through that:
2(6x2 + 7x – 24)
Step one: 2( x2 + 7x – 144) (Multiplied 6 by 24)
Step Two: 2( x + 16)( x - 9) (Factors of 144 that subtract to get 7 – follow sign rules)
Step Three: 2(6x + 16)(6x – 9) (Put leading coefficient back in BOTH binomials)
Step Four: 2(3x + 8)(2x – 3) (Divide by the GCF of each binomial, 2 in the 1st and 3 in the 2nd)
Let’s try another:
8x2 + 18 x + 9
9-7 Factoring Special Cases
A perfect square Trinomial is the result of squaring a binomial. Factoring a PST is reversing the
“Square Double Square” shortcut of multiplying the binomial by itself. So to “undo” that process, we must ask ourselves, can I get the square root of each end? Is the middle twice the product of those roots? If so, it is a Perfect Square Trinomial. Of course you can just use the “regular” rules for factoring as well.
Example: 4x2 - 36x + 81
Both 4x2 and 81 are perfect squares, and 36 is twice the product of those roots. (Since 2x * 9 = 18 and 18 * 2 = 36)
So we write: (2x-9)2 You can always check it by multiplying it out or using the square double square shortcut.
The sign of the binomial will match the middle sign. Since squaring a term ALWAYS results in a positive number, a trinomial can not be a perfect square trinomial if the last term is negative!
Again we revisit the Difference of Squares as well. Recall that the difference of squares comes from multiplying two binomials that have the same terms but different signs. The middle terms cancel out leaving the difference of two perfect squares. Thus to factor a difference of squares, find the root of each term, put it into each of the parenthesis and put one sign positive, the other negative.
Example: 169x4 – 144y2
Both are perfect squares and we are taking the difference. Find each root (13x2 & 12y) put them in each set of parenthesis and use opposite signs.
(13x2 + 12y) (13x2 - 12y)
You can always check by multiplying it out!
Again, don’t forget your first rule of factoring, to look for a GCF before you begin!
9-8 Factoring by Grouping
If a polynomial has more than three terms, you may have to factor only part of it at a time. This will require you to group together terms that have a GCF, factor that part, then see if there is another GCF from among the pieces you grouped together.
Example: 4x3 + 8x2 – 5x – 10
If you group the first two terms and the last two terms you can pull out a GCF from each
(4x3 + 8x2)+( – 5x – 10)
Notice I put a + between them to avoid thinking I was multiplying – that could be dangerous!
4x2(x + 2) + (-5)(x + 2)
I pulled out the GCF of each – notice pulling out a negative changed the – to a positive! Now each of these has a parenthesis in common!
Pull the (x + 2) out and the leftovers go in the second set of parenthesis
(x + 2)(4x2 + -5) or simply (x + 2)(4x2 -5) Again, you can check it by multiplying it back out.
Let’s try another!
6y3 + 3y2 + 2y + 1
You might also have to factor by grouping three terms together, and factor them as a polynomial.
Sometimes I call this method of factoring “Sesame Street Factoring” because we often are looking for something that doesn’t belong.
Example: 4x2 + 4xy + y2 – 9
One of these things is not like the other, one of these things just doesn’t belong! In this case it is the -9 that doesn’t belong. Everything else fits together because of the x-y connection. So group them like this (4x2 + 4xy + y2) – 9 then factor the first three terms as a trinomial (This one happens to be a perfect square trinomial) so it factors as (2x + y)2 – 9 which happens to be the difference of two squares.
Thus we break each perfect square down and use opposite signs to get:
(2x + y + 3) and (2x + y - 3) Whew! It’s like the energizer bunny! Just keeps going and going and going and…