# Math Help - Binomial and Trinomial factoring

1. ## Binomial and Trinomial factoring

I need some help. at some point i either was never taught or missed this but binomial/trinomials seem so confusing to me. Is there a certain spot i should backtrack to to understand the concepts/formulas for it? the math book we use does really explain it much and neither does the professor he assumes we should know it all already.

for example. Solve each equation 2q-9 = q(q+3) + q

the 9 has a exponent of 2. and the last Q on the right has an exponent of 2.

im someone that cant learn something with multiple steps just by watching a teach do it on the board. because i just dont remember the steps.

could anyone break this problem down for me or show me a site that can break the steps down in answering it?

yesterday in class when the teacher was talking about see if its squared, then cubed, then something else. blah i hate math

2. ## Correct Equation?

$(2q-9)^2=q(q+3)+q^2$

3. Originally Posted by Knowledge
$(2q-9)^2=q(q+3)+q^2$
If so, then to solve for q:

Expand (multiply) to:

$4q^2-36q+81=q^2+3q+q^2$

Collect like terms:

$4q^2-36q+81=2q^2+3q$

Isolate the variable to one side of the equation:

$2q^2-39q+81=0$

Now, you have a quadratic equation in the standard form of $ax^2+bx+c=0$

Is this enough information to get you started?

4. jtown82, part of your problem may be that many equation can't be factored in terms of factors with integer coefficients. In particular, the polynomial equation, $2q^2- 39q+ 81= 0$, to which Knowledge reduced your equation, has no rational roots and so cannot be "factored". It can be solved by completing the square or using the quadratic formula.

5. Originally Posted by Knowledge
If so, then to solve for q:

Expand (multiply) to:

$4q^2-36q+81=q^2+3q+q^2$

Collect like terms:

$4q^2-36q+81=2q^2+3q$

Isolate the variable to one side of the equation:

$2q^2-39q+81=0$

Now, you have a quadratic equation in the standard form of $ax^2+bx+c=0$

Is this enough information to get you started?
see this is where im lost. how do you come up with
$4q^2-36q+81=2q^2+3q$ ? what is the process to do this?

6. Originally Posted by jtown82
see this is where im lost. how do you come up with
$4q^2-36q+81=2q^2+3q$ ? what is the process to do this?
In one word - multiplication. Take the left side:

$(2q-9)^2$

$(2q-9)*(2q-9)$

$(2q*2q)+(2q*-9)+(-9*2q)+(-9*-9)$

$4q^2+(-18q)+(-18q)+81$

$4q^2-36q+81$

Now, you give the right side a try...