1. ## Factoting Trinomials

Ok, other post already. How on earth do I plan to past my final. *stresss* Anyways I don't remember how to factor Trinomials in the form of :

Ax(squared) + bx + c

so how do I solve?

3x(squared) + 10x + 8

P.S. sorry but I don't know how to get the squared thing.

2. Originally Posted by dizzycarly
Ok, other post already. How on earth do I plan to past my final. *stresss* Anyways I don't remember how to factor Trinomials in the form of :

Ax(squared) + bx + c

so how do I solve?

3x(squared) + 10x + 8

P.S. sorry but I don't know how to get the squared thing.
Are you asking how to solve for $x$ or how to factorize it? Anyways, to solve for $x$, you have:

3x² + 10x + 8 where

a = 3;
b = 10;
c = 8
;

$x_1, x_2 = -b \pm \frac {\sqrt {b^2-4ac}}{2a}$

To factorize it, you can use the box method:

3x² + 10x + 8

What multiplied by what gives 24 (a*c or 8*3), and added up gives 10 (b)? Obviously, in this case, 6 and 4.

3x² + 6x + 4x + 8;
(3x²+6x) + (4x+8);
3x(x+2) + 4(x+2);
(3x+4)(x+2)

3. Originally Posted by dizzycarly
Ok, other post already. How on earth do I plan to past my final. *stresss* Anyways I don't remember how to factor Trinomials in the form of :

Ax(squared) + bx + c

so how do I solve?

3x(squared) + 10x + 8

P.S. sorry but I don't know how to get the squared thing.
Let's look at this

(rx+s)(x+t)

if we foil we get

$rx^2+rtx+sx+st$
$=rx^2+(rt+s)x+st$

So, when factoring a quadratic trinomial, we must look for all factors of the first an last term whose sums add up to be the coefficient of the linear term.

In your case we can analyze as follows:

$3x^2+10x+8$

3 only has two factors, this makes things easier. 8 has factors, $1\cdot8,2\cdot4$.

Our possiblities are then

$(3x+1)(x+8)$ NO

$(3x+8)(x+1)$ NO

$(3x+2)(x+4)$ NO

$(3x+4)(x+2)$ Yes $=3x^2+6x+4x+8=3x^2+10x+2$

This takes practice to get good at, so my advice to you is to hit the books!