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Question 474068: Factor completely Remember to look first for a common factor. Check by multiplying. If a polynomial is prime, state this -4x^2-40x-100
Found 3 solutions by ewatrrr, stanbon, bucky:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi,
-4x^2-40x-100 = -4(x^2 + 10x + 25) = -4(x+5)(x+5) = -4(x+5)^2
F First terms x^2
O Outside terms 5x
I Inside terms 5x
L Last terms 25
Answer by stanbon(75887)  (Show Source):
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website! Given to factor completely: -4x^2 - 40x - 100
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The problem statement gives you the hint to look first for a common factor. Looking at the squared term, you can see that it's coefficient (multiplier) is -4. So the first thing to do is to see if every coefficient of the three terms is divisible by -4. The answer is yes, so you can factor out -4 from every term and the result is:
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-4*(x^2 + 10x + 25)
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This can be checked by distributive multiplication. Multiply -4 times each of the three terms inside the parentheses and you will see that it results in the original quadratic that you were given in the problem. Therefore, it is correct.
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The next step is to look at the three terms inside the parentheses and see if they can be factored. With a little experience you will recognize this particular quadratic is a perfect square. Other than that, you can try trial and error. Note that the constant term +25 in the quadratic has two sets of factors, either +25 and +1 or (-25 and -1) or +5 and +5 (or -5 and -5). The squared term has as its factors x and x (or -x and -x). Therefore, possible combination of pairs of factors are:
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(x + 25) and (x + 1) or
(x - 25) and (x - 1)
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(x + 5) and (x + 5) or
(x - 5) and (x - 5)
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(-x + 25) and (-x + 1) or
(-x - 25) and (-x - 1)
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(-x + 5) and (-x + 5) or
(-x - 5) and (-x - 5)
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These can each be multiplied out (using the FOIL pattern) to see if one or more of the eight possible sets doesn't multiply out to the quadratic in the parentheses.
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With a little more experience, you can recognize that middle term (the x term) in the original quadratic has a plus sign. Therefore, any FOIL multiplication of the Outside and Inside terms that results in two negatives can be dropped from the list of possible sets because they will add to give a negative sign on the x term instead of the +10x that is needed. Following that rule, the following sets of possible factors can be eliminated:
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(x - 25) and (x - 1)
(x - 5) and (x - 5)
(-x + 25) and (-x + 1) and
(-x + 5) and (-x + 5)
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This leaves as possible sets of factors:
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(x + 25) and (x + 1)
(x + 5) and (x + 5)
(-x - 25) and (-x - 1) and
(-x - 5) and (-x - 5)
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If you think about it, in those sets of factors that have as the constant parts 25 and 1 they are multiplied by x to get 25x and 1x. They cannot combine to give you the 10x term in the original quadratic. Therefore, these two sets of factors can also be eliminated and the only possibilities left are:
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(x + 5) and (x + 5) and the second set
(-x -5) and (-x -5)
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But -1 can be factored from each of the factors in the second set to give:
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-1*(x + 5) and -1*(x + 5)
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If you multiply the two -1 coefficients together they become +1 and this leaves:
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(x + 5) and (x + 5)
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which is exactly the same as the other possible set. So by the process of elimination you can finally say that the factors of the quadratic given in the problem are:
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-4*(x + 5)*(x + 5)
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That's the answer to the problem.
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You could also use the quadratic formula to find the factors, but I wasn't sure that you were that far along, so I didn't discuss that possibility.
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This is probably a little confusing, but hopefully it gives you a little insight into the process and with a little more practice you can understand what is going on in solving factoring problems such as this one.
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