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by alan
Fri Jan 04, 2013 2:05 am
Forum: Intermediate Algebra
Topic: Factoring trinomials: 2a^2 + 3a + 1
Replies: 3
Views: 7653

Re: Factoring trinomials: 2a^2 + 3a + 1

When the coefficient of the x squared term is not 1 you need to use the method of decomposition to factor the trinomial. The first part you have done correctly. Find two numbers whose product is a*c and whose sum is b. In this case a*c=2 and b=3 so the two numbers are 2 and 1. Now, since the x squar...
by alan
Fri Jan 04, 2013 1:34 am
Forum: Intermediate Algebra
Topic: Is y=(2x+5)(3x-7) considered factored form of a quadratic?
Replies: 1
Views: 1718

Is y=(2x+5)(3x-7) considered factored form of a quadratic?

Factored form is stated as y=a(x-s)(x-t) so y=(2x+5)(3x-7) should be written as y=6(x+5/2)(x-7/3) to be in “proper” factored form. I'm trying to determine if y=(2x+5)(3x-7) would be considered to be in factored form. There may not be a definitive answer to this so feel free to offer your opinion. If...
by alan
Fri Jan 04, 2013 12:28 am
Forum: Intermediate Algebra
Topic: x^2-bx+18=0
Replies: 3
Views: 2370

Re: x^2-bx+18=0

The above post is correct, add bx to both sides and multiply by 1/x to get x+18/x=b Plug in any value of x to get a valid value for b. For example when x=1 then b=19 so x^2 -19x+18=0 is a solution. This solution is factorable as (x-1)(x-18). Another solution is x=-1 which gives b=-19 and the quadrat...

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