Re: Factorization Question 2

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)

Re: Factorization Question 2

If you have a polynomial and 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) ...

Re: Factorization Question 2

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 pair has 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": .

Re: Factorization Question 2

Wow, I love this! Why didn't I think of this? This was so obvious! Thanks

Re: Factorization Question 2