Question 909252: Hello, I am having trouble with Factoring using algebra tile can you please help me.
1. x^2-7x+12
2. x^2+2x-8
3. x^2-6x+8
4. x^2-3x-15
Found 2 solutions by Alan3354, Edwin McCravy: Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! I am having trouble with Factoring using algebra tile can you please help me.
1. x^2-7x+12
Find a pair of factors, both negative, that add to 7.
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2. x^2+2x-8
Find a pair of factors of 8 that differ by 2.
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3. x^2-6x+8
Similar to #2
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4. x^2-3x-15
The the factors of 15
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Factoring is mostly trial and error.
There's nothing gained by having someone do it for you.
Answer by Edwin McCravy(20056) (Show Source):
You can put this solution on YOUR website!
1. x^2-7x+12
Look at it from right to left:
First look at the 12, then the +, then the 7 then the -.
A. Look at the 12. Start with this partial sentence:
"Think of two positive integers which have product 12, such that when...
B. Look at the sign + before the 12, which tells us that the next word is
"added". So now our partial sentence is:
"Think of two positive integers which have product 12, such that when
added...
C. Look at the 7, which completes our sentence:
"Think of two positive integers which have product 12, such that when
added gives 7".
D. It's easy to do that and come up with 4 and 3, because 4×3=12 and 4+3=7
Now write this:
(x 4)(x 3)
E. Now look at the sign before the 7. It is a -. So put a - before the
larger one, which is 4, so we have:
(x+4)(x 3)
F. Since the sign of 12 is +, we use the same sign for 3, so the answer is
(x+4)(x+3).
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2. x^2+2x-8
Look at it from right to left:
First look at the 8, then the -, then the 2 then the +.
A. Look at the 8. Start with this partial sentence:
"Think of two positive integers which have product 8, such that when...
B. Look at the sign - before the 8, which tells us that the next word is
"subtracted". So now our partial sentence is:
"Think of two positive integers which have product 8, such that when
subtracted...
C. Look at the 2, which completes our sentence:
"Think of two positive integers which have product 8, such that when
subtracted gives 2".
D. It's easy to do that and come up with 4 and 2, because 4×2=8 and 4-2=2
Now write this:
(x 4)(x 2)
E. Now look at the sign before the 2. It is a +. So put a + before the
larger one, which is 4, so we have:
(x+4)(x 2)
F. Since the sign of 12 is -, we use the opposite sign for 2, so the answer is
(x+4)(x-2).
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3. x^2-6x+8
Look at it from right to left:
First look at the 8, then the +, then the 6 then the -.
A. Look at the 8. Start with this partial sentence:
"Think of two positive integers which have product 8, such that when...
B. Look at the sign + before the 8, which tells us that the next word is
"added". So now our partial sentence is:
"Think of two positive integers which have product 8, such that when
added...
C. Look at the 6, which completes our sentence:
"Think of two positive integers which have product 8, such that when
added gives 6".
D. It's easy to do that and come up with 4 and 2, because 4×2=8 and 4+2=6
Now write this:
(x 4)(x 2)
E. Now look at the sign before the 6. It is a -. So put a - before the
larger one, which is 4, so we have:
(x+4)(x 2)
F. Since the sign of 8 is +, we use the same sign for 2, so the answer is
(x+4)(x+2).
-----------------------
4. x^2-3x-15
Look at it from right to left:
First look at the 15, then the -, then the 3 then the -.
A. Look at the 15. Start with this partial sentence:
"Think of two positive integers which have product 15, such that when...
B. Look at the sign - before the 15, which tells us that the next word is
"subtracted". So now our partial sentence is:
"Think of two positive integers which have product 15, such that when
subtracted...
C. Look at the 3, which completes our sentence:
"Think of two positive integers which have product 15, such that when
subtracted gives 3".
D. Oh! oh! There are NO SUCH POSITIVE INTEGERS!!! So we cannot factor
this, so we STOP and write " This trinomial is PRIME".
Edwin
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