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For the question below, does anyone have a quick and easy method for finding the two numbers that when added together gives 11x, and when multiplied together gives -180? I know the numbers are 20 and -9 but it takes me forever to find them? Is there is quick way?
Hello
I don't now if this will help, but I do it this way. It doesn't take ages.
You know you have either:
1,15 or 3,5
and your looking for a difference of 11
Write out:
(12x )(x )
(6x )(2x )
(4x )(3x )
It doesn't tke long to work out the difference of the possible products and see that 4x5 - 3x3 = 11
So you have
(4x - 3)(3x + 5)
If you have a polynomialand the zero's
are known then the polynomial can be factored as:
(where
is the coefficient of the highest degree term).
Offcourse in this case you have to find the zero's first (by using the quadratic formula) ...
Hello, sparky!
Quote:
For the question below, does anyone have a quick and easy method for
finding the two numbers whose difference is 11 and whose product 180?
I know the numbers are 20 and -9, but it takes me forever to find them?
Is there is quick way?
I use a primitive approach.
It takes a bit of time, but it is direct.
Factor 180 into pairs of factors, and note the difference.
. .
How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.
When can we stop?
We can stop at the integer part of
We see that the pairhas a difference of 11.
(Of course, we can stop listing the moment we find it.)
We have: .
We want the middle term to be +11x, so we will use -9x and +20x.
So we have: .
Factor "by grouping": .
Wow, I love this! Why didn't I think of this? This was so obvious! ThanksQuote:
Originally Posted by Soroban
Hello, sparky!
I use a primitive approach.
It takes a bit of time, but it is direct.
Factor 180 into pairs of factors, and note the difference.
. .
How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.
When can we stop?
We can stop at the integer part of
We see that the pair
(Of course, we can stop listing the moment we find it.)
We have: .
We want the middle term to be +11x, so we will use -9x and +20x.
So we have: .
Factor "by grouping": .
^this. is beautiful.Quote:
Originally Posted by SorobanHello, sparky!
I use a primitive approach.
It takes a bit of time, but it is direct.
Factor 180 into pairs of factors, and note the difference.
. .
How do we find these pairs?
Divide 180 by 1, 2, 3, . . .
Some of the divisions do not "come out even", of course.
When can we stop?
We can stop at the integer part of
We see that the pair
(Of course, we can stop listing the moment we find it.)
We have: .
We want the middle term to be +11x, so we will use -9x and +20x.
So we have: .
Factor "by grouping": .