1. Factoring trinomial

$\displaystyle x^2&+5x&+4$
$\displaystyle (x+4)(x+1)$

2. Re: Factoring trinomial

I guess the second degree polynomial has to be: $\displaystyle x^2+5x+4$, in that case your answer is correct.

3. Re: Factoring trinomial

Originally Posted by Siron
I guess the second degree polynomial has to be: $\displaystyle x^2+5x+4$, in that case your answer is correct.
Sorry for that You are right

No problem!

5. Re: Factoring trinomial

Originally Posted by Siron
No problem!
I'm sorry to bother you but I'm know on my second set , and the instructions are (Factor the trinomial. If the trinomial cannot be factored say so.).

The Trinomial is$\displaystyle 2x^2+7x+3$

I tried $\displaystyle (2X+1)(X+3)$ But obviously there wasn't even a point to try that.

6. Re: Factoring trinomial

You're not bothering me . And your answer is correct again. You can check it by yourself by expanding: $\displaystyle (2x+1)(x+3)=2x^2+x+6x+3=2x^2+7x+3$ which is the given polynomial.

7. Re: Factoring trinomial

Originally Posted by Siron
You're not bothering me . And your answer is correct again. You can check it by yourself by expanding: $\displaystyle (2x+1)(x+3)=2x^2+x+6x+3=2x^2+7x+3$ which is the given polynomial.
Sorry for my lack of understanding and communicating in english but therefore it cannot be factored since 3+1 is 4?

8. Re: Factoring trinomial

You don't have to excuse and 3+1 is indeed 4 but in this case it's 3x1 which is 3.

9. Re: Factoring trinomial

Originally Posted by Siron
You don't have to excuse and 3+1 is indeed 4 but in this case it's 3x1 which is 3.
But doesn't it also have to be equal to 7 when you add it?
or since there is a coefficient on the $\displaystyle x^2$ the rules change?