Polynomial Help Please!

[Jamie27]

6x^2+17x+12=___?



A) (6x+3)(x+4)
B) (6x+1)(x+12)
C) (2x+3)(3x-4)
D)(2x+3)(3x+4)

I came up with none of these answering the top equation? The closest I came was A?

Thank you!!!!
Jamie :roll:

 

[stapel]

I came up with none of these answering the top equation? The closest I came was A?

How did you arrive at your answer?

Please reply showing your work and reasoning.

Thank you.

Eliz.

 

[Denis]

6x^2+17x+12=___?
A) (6x+3)(x+4)
B) (6x+1)(x+12)
C) (2x+3)(3x-4)
D)(2x+3)(3x+4)
I came up with none of these answering the top equation? The closest I came was A?

What do you get when you try A) ?

 

[Jamie27]

I came up with :

6x^2+13x+7 for A
6x^2+19x+13 for B
6x^2-7 for C
6x^2+7 for D

I know I am missing a step here. I did foil but this is what I come up with?

 

[TchrQbic]

I came up with :

6x^2+13x+7 for A
6x^2+19x+13 for B
6x^2-7 for C
6x^2+7 for D

I know I am missing a step here. I did foil but this is what I come up with?

If you used the FOIL procedure, you didn't show all your steps. For each one, show

first = 6x * ___
outer = 6x * ___
inner = ___ * ___
last = ___ * ___

If you work them more carefully, you'll find which is correct.

 

[Denis]

I came up with :
6x^2+13x+7 for A

That is not the solution for (6x + 3)(x + 4)

Where were you when your teacher covered this?

 

[Euler]

(a+b)(c+d)=(a)(c)+(a)(d)+(b)(c)+(b)(d)

 

[Jamie27]

I take online courses and have absolutely no help in answering questions that is why I come here..

 

[stapel]

I take online courses and have absolutely no help....

So are you actually asking for lessons on this topic, so you can learn how to answer these questions...?

Thank you.

Eliz.

 

[Jared Williams]

Hi Jamie, I am working on the same lesson as you and trust me i strugled through it. Than I got help from my math teacher and now it is easier.

I think that the answer is Prime, you just can't do it.

 

[stapel]

I got help from my math teacher and now it is easier. I think that the answer is Prime....

I would suggest that you might want to work a bit more with your teacher, as the polynomial in question is not prime.

Eliz.

 

[Mrspi]

6x^2+17x+12=___?



A) (6x+3)(x+4)
B) (6x+1)(x+12)
C) (2x+3)(3x-4)
D)(2x+3)(3x+4)

I came up with none of these answering the top equation? The closest I came was A?

Thank you!!!!
Jamie :roll:

Here's a method you can use to factor expressions of the form
ax<SUP>2</SUP> + bx + c

Step 1:
Multiply a * c (that is, the coefficient of the x<SUP>2</SUP> term and the constant term). In your problem, we multiply 6*12 to get 72.

Step 2: Look for two factors which produce the product ac, and which ADD UP to the coefficient of the middle term, b. In your problem, we need two factors of +72 which add up to +17 (the coefficient of the middle term). 9*8 is 72, and 9 + 8 is 17, so the numbers we want are 9 and 8.

Step 3: Use the numbers you found in step 2 to rewrite the middle term. For your problem, we'll use 9x + 8x instead of the middle term 17x:

6x<SUP>2</SUP> + 9x + 8x + 12

Step 4: Factor by grouping. Group the first two terms together, and the last two terms together:

6x<SUP>2</SUP> + 9x + 8x + 12

Remove the greatest common factor from each pair of terms:
3x(2x + 3) + 4(2x + 3)

Now, (2x + 3) is a common factor from both terms. Remove it:
(2x + 3)(3x + 4)

There's the factorization. This process works on any factorable (over the integers) quadratic polynomial and does not involve any trial-and-error.

I hope this helps you.