(my way of factoring Trinomials.)
If you have had Algebra I or are taking it, some time in the second semester you
will be asked to factor a trinomial. They look like the following:
- X² + 4X – 12
- 3X² – 9X + 6
- 5X² + 8X – 4
When we were in elementary school we factored 12 into 2² x 3, but that was
prime factoring. Here we are factoring into two binomials. We have already
spent a weak or 2 multiplying 2 binomials into a trinomial;
(2X + 5) (X– 3) = 2X² – 6X + 5X – 15 = 2X² – X– 15
Now our job is to start with 2X² – X – 15 and factor it to (2X + 5) ( X – 3).
These are the important parts:
2X² is the first term, and the 2 is called the leading coefficient
– X is the Linear term and its coefficient is –1.
–15 is the constant because there is no variable.
The leading coefficient and the constant is where we start. If you have
read some of my blogs, one contained a game called
"I'm thinking of two numbers".
X² + 4X – 12 The leading coefficient is 1 and the constant is (–12).
Thus I am looking for 2 numbers whose product is
1 x (–12) = –12. The linear coefficient tells us what
sum will be. In this case +4.
∴ I am looking for 2 numbers whose
product is –12 and whose sum is +4.
Answer: 6 and –2, thus my solution is (X + 6)(X – 2).
X² – 9X + 20 Like the previous example my leading
coefficient is 1. Again I look at my constant
to help me with the product and the linear
coefficient to tell me what the sum will be.
∴ I am looking for 2 numbers whose
product is 20 and whose sum is –9
My options are 4 x 5, 2 x 10, –2 x –10, –4 x –5, 1 x 20, –1 x –20
Answer: –4 and –5, thus my solution is (X – 4)(X – 5)
It is easy when the leading coefficient is 1. However, before we move
on to a more difficult problem, lets do something that perhaps we have
already learned, but maybe not.
What if I was looking for 2 numbers whose product was –120 and
whose sum was –37.Perhaps a solution is not quickly at hand. It came
from a problem like:
8X² – 37X – 15 As above 8 ⋄ (–15) = –120 which is
the product we are looking for, and
the sum is –37.
Where do I start?
What do I do with the 8?
Both are good questions, and even though I understood my
Algebra I teacher's (Mr. Ward) method of solving this problem,
I looked for an easier method. Lets try:
Mr. W's 2 magic numbers!
Start with 8 ⋄ –15, we know that's wrong so adjust. Divide one
by a positive integer and multiply the other by
the same. –15 ÷ 3, and 8 x 3
24 ⋄ – 5, better but not far enough apart, so divide 24
by 8 = 3, and multiply the –5 by 8 = 40.
3 ⋄ –40, This is it 3 +(–40) = –37
Now I have my 2 magic numbers:
3 and (–40)
Set up two binomials: ( ) ( )
Our leading coefficient should divide evenly into one of these
2 numbers. Sometimes it does not and in that case we need to
be a bit clever, however in this case, it does.
Thus (8X ) ( X )
Now divide the 8 into the –40, it equals –5
∴ (8X )( X – 5) See the –40?
Now divide the –5 into the constant –15 equals 3
(8x + 3) (X – 5)
We will come back to doing more of these, but first we must
always remember, the first thing we look for is a common monomial
factor. Is there something in all the terms that I can factor out.
3X² – 9X + 6 There is a 3 in all the terms
thus: 3(X² – 3X + 2) Can I factor the trinomial?
I'm looking for 2 numbers
whose product is 2 and whose
sum is –3. They are –1 and –2
∴ 3(X – 1)(X – 2)
Now lets try 5X² – 8X – 4: I'm looking for 2 numbers
whose product is –20, and
whose sum is –8.
That's easy –10 and 2
5X² – 8X – 4
( ) ( ) –10 2
(5X ) (X ) 5X ⋄ X = 5X² and that's
is your only choice.
(5X ) (X – 2) 5 divided into –10 = –2
∴ (5X + 2) (X – 2) –2 divided into –4 = 2
We will return to this with X² – 9, the
"DIFFERENCE OF 2 SQUARES"
I'M BACK!
So what about X² – 9?
Well, sometimes the story is in what is not written.
This is really X² + 0X – 9
I'm looking for 2 numbers whose product
is –9 and whose sum is 0. They are 3 and –3
∴ (X + 3)(X – 3) is the solution
It is one of the easiest relationships in Math to recognize. We know that
X² is a perfect square; X⋄X and by Algebra we should know our perfect
squares to 625, which is 25 x 25.
FACTOR 4X² – 64
1st: 4 goes into both 4 and 64, thus
4(X² – 16)
2nd: We factor what is in the Parenthesis;
∴ 4 (X + 4)(X – 4)
There is one other "Special Case" to look at. It is a, ta–ta–ta–da.
"TRINOMIAL PERFECT SQUARE"
We know 100 is a perfect square and 10000 is a perfect square, but
how does a trinomial get to be a perfect square? When the binomials
that create the trinomial are equal.
(X + 5)² = (X + 5)(X + 5) = X² + 10X + 25
Just like any other perfect square, it comes from a number
being multiplied by itself. All of the following are trinomial
perfect squares.
X² + 6X + 9
4X² – 24X + 36
81X² + 18X + 1
49X² – 70X + 25
100X² + 40XY + 4Y²
How do I recognize a trinomial perfect square?
Step 1: The First term and the Constant must both be perfect squares.
Step 2: Find the Square Root of both of them.
Step 3: Multiply them together and then double the solution of the product.
Step 4: Check to see that that product is equal to the linear term.
LETS DO A COUPLE OF THE ABOVE
4X² – 24X + 36
1. √4X² = 2X √36 = 6
2. 2X 6
3. 2X ⋄ 6 = 12X
4. 12X doubled is 24X
4X² – 24X + 36 = (2X – 6)(2X – 6) = (2X – 6)²
100X² + 40XY + 4Y²
1. √100X² = 10X √4Y² = 2Y
2. 10X 2Y
3. 10X ⋄ 2Y = 20XY
4. 20XY doubled is 40XY
100X² + 40XY + 4Y² = (10X + 2Y)(10X + 2Y) = (10X + 2Y)²
So much of Math is just recognizing what is in front of you.
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