#2 Factor Polynomial
- Thread starter Stine
- Start date
Quadratic Formula
16x^2 + 36x + 9=0
I am having a hard time trying to find two integers whose product is 9 and whos sum is 36.
36= 18*2
36= 3*12 ex. of what I am coming up with intergers that = 36.
Can soemone please give me some more help?
Thank you
16x^2 + 36x + 9=0
I am having a hard time trying to find two integers whose product is 9 and whos sum is 36.
36= 18*2
36= 3*12 ex. of what I am coming up with intergers that = 36.
Can soemone please give me some more help?
Thank you
This cannot be intuitively factored using whole numbers.
Learn how to solve a polynomial using the completing the square method. This is how the quadratic formula is derived. Learn how to complete the square, and WHY the quadratic formula works. It is IMPORTANT to understand besides just plugging and chugging with the formula.
These links should help.
http://www.purplemath.com/modules/sqrquad2.htm#formula
http://www.purplemath.com/modules/sqrquad.htm
Cheers,
John
Learn how to solve a polynomial using the completing the square method. This is how the quadratic formula is derived. Learn how to complete the square, and WHY the quadratic formula works. It is IMPORTANT to understand besides just plugging and chugging with the formula.
These links should help.
http://www.purplemath.com/modules/sqrquad2.htm#formula
http://www.purplemath.com/modules/sqrquad.htm
Cheers,
John
Count Iblis
Junior Member
- Joined
- Feb 13, 2007
- Messages
- 108
Stine said:16x^2 + 36x +9=
(8x+3) (12x+3)
I tried several different ways but I have not come up with the correct answer yet.Can someone help me? Thank you!
On both sides of the equation there is a constant term of 9, so they cancel. This means that x = 0 is a solution. Divide this away and you are left with a linear equation.
Denis-
(Are you too busy to look up the quadratic formula?)
U r not very nice obviously I am having an issue with this problem and am seeking more help. You don't have to help me. I am seeking help for school not attitude. I appreciate all time that is put in to this site and the help that I get in return.
THANKS
(Are you too busy to look up the quadratic formula?)
U r not very nice obviously I am having an issue with this problem and am seeking more help. You don't have to help me. I am seeking help for school not attitude. I appreciate all time that is put in to this site and the help that I get in return.
THANKS
Count Iblis
Junior Member
- Joined
- Feb 13, 2007
- Messages
- 108
jwpaine said:Count Iblis said:Stine said:16x^2 + 36x +9=
(8x+3) (12x+3)
I tried several different ways but I have not come up with the correct answer yet.
On both sides of the equation there is a constant term of 9, so they cancel. This means that x = 0 is a solution. Divide this away and you are left with a linear equation.
0 is NOT a zero for this polynomial!
Cheers,
John
Oh, you just wrote an attempted factorization on the right hand side!
Anyway, solving quadratic equations is easy, you just get rid of the linear term of the quadratic a x^2 + b x + c by substituting x = y - b/(2a) and then all you have to do is to exctract the square root.
I don't know WHY some responders assume that a factoring expression is necessarily associated with the solutions to equation.
As I read this problem, we're asked to find an expression that is equivalent to
16x<SUP>2</SUP> + 36x + 9
Do the three terms have a common factor? If they do, removing that common factor would be the first step in factoring the expression. But, there is no common factor in this case.
Next, one might try the "a*c" approach. Multiply the coefficient of the x<SUP>2</SUP> term by the constant term. 16*9 = 144
Look for two factors of 144 which add up to the coefficient of the middle term, + 36.
Possible factors of 144:
144*1
72*2
36*4
18*8
9*16
3*48
Since none of these factor pairs add up to 36, then this trinomial is "prime" over the set of integers.
As I read this problem, we're asked to find an expression that is equivalent to
16x<SUP>2</SUP> + 36x + 9
Do the three terms have a common factor? If they do, removing that common factor would be the first step in factoring the expression. But, there is no common factor in this case.
Next, one might try the "a*c" approach. Multiply the coefficient of the x<SUP>2</SUP> term by the constant term. 16*9 = 144
Look for two factors of 144 which add up to the coefficient of the middle term, + 36.
Possible factors of 144:
144*1
72*2
36*4
18*8
9*16
3*48
Since none of these factor pairs add up to 36, then this trinomial is "prime" over the set of integers.
I think you did the same as I did, initially:
his equation is \(\displaystyle \L 16x^{2} + 36x +9 -(8x+3) (12x+3) = 0\)
=
\(\displaystyle \L -80x^2 - 24x = 0\)
\(\displaystyle \L -80{x}^2-24{x}\) can be factored into -8 x (3 + 10 x) = 0
so -8x = 0 or (3x+10) = 0
x = 0 or x = -(10/3)
And those are the two x intercepts.
This is the answer, Stine: now that everyone has thrown you a different way of solving it.
his equation is \(\displaystyle \L 16x^{2} + 36x +9 -(8x+3) (12x+3) = 0\)
=
\(\displaystyle \L -80x^2 - 24x = 0\)
\(\displaystyle \L -80{x}^2-24{x}\) can be factored into -8 x (3 + 10 x) = 0
so -8x = 0 or (3x+10) = 0
x = 0 or x = -(10/3)
And those are the two x intercepts.
This is the answer, Stine: now that everyone has thrown you a different way of solving it.
jwpaine said:
UMMMMM....why do you ASSUME that the expression is equal to 0? The poster did not say that in his/her original post.
the originall post, as I saw it, was
16x^2 + 36x + 9 = ?
NO zero on the right side!
Overcomplicating problems tends to confuse the posters, in my opinion.
I didn't see the poster asking for x-intercepts or y-intercepts.
I may also be operating under erroneous assumptions here, but I think the poster was looking for a factorization of the expression.
jwpaine said:
thus: 16x^2 + 36x +9 - (8x+3)(12x+3) = 0
and then you get -80x^2 - 24x = 0
cheers,
sorry, John.....
16x<SUP>2</SUP> + 36x + 9 = ?
WHICH WAS THE ORIGINAL POST does not imply that the orginal expression is equal to 0.
When poster gave (8x + 3)(12x + 3) as a possible answer, this was the poster's attempt at factoring.
Assuming that the original expression equals 0, OR that the original expression is equal to (8x + 3)(12x + 3) is going to produce some very strange and incorrect results.
If you don't have an equation to begin with, you can't "magically" create one.
That is what I thought in the beginning, Count Iblis confused me with his answer, as he combined both as one equation.
If my initial understanding is correct, than I apologize for re-thinking Stine's post and assuming that his binomial factors were part of the equation.
I will restate my initial post to clarify what I initially thought. (if your equation is equal to zero) if not, than you just cant factor this type of problem.
I'm sorry for any confusion.
Now, Stine: if you look at your factors, and you FOIL your solution of (8x+3)*(12x+3) , you will see that it expands to: 96x^2+60x+9 this does not match your expression, does it?
now IF this is an equation (if you meant it to equal zero) cannot be factored to solve for the x-intercepts...you will have to complete the square to get your two x-intercepts. (that is if this is an equation and not simply an expression for which you are trying to factor)
Cheers,
John
If my initial understanding is correct, than I apologize for re-thinking Stine's post and assuming that his binomial factors were part of the equation.
I will restate my initial post to clarify what I initially thought. (if your equation is equal to zero) if not, than you just cant factor this type of problem.
I'm sorry for any confusion.
Now, Stine: if you look at your factors, and you FOIL your solution of (8x+3)*(12x+3) , you will see that it expands to: 96x^2+60x+9 this does not match your expression, does it?
now IF this is an equation (if you meant it to equal zero) cannot be factored to solve for the x-intercepts...you will have to complete the square to get your two x-intercepts. (that is if this is an equation and not simply an expression for which you are trying to factor)
Cheers,
John