Factoring trinomials

**linds12** · October 18th 2012, 07:02 PM

Hello, I have a test tomorrow and 150 math problems due tomorrow morning.
I had 4 people explain this to me in 4 different ways, but I always get stuck when I work on problems by myself.

Can someone please explain to me step by step how to factor:

3x³+4xy+y²6x²-7xy-5y²15x²-31xy+10y²
^{Thank you}

**topsquark** · October 18th 2012, 07:29 PM

Originally Posted by linds12

Hello, I have a test tomorrow and 150 math problems due tomorrow morning.
I had 4 people explain this to me in 4 different ways, but I always get stuck when I work on problems by myself.

Can someone please explain to me step by step how to factor:

3x³+4xy+y²6x²-7xy-5y²15x²-31xy+10y²
^{Thank you}

This sounds awfully like you are giving us problems with your *graded* homework problems. Do NOT do this again or you will be banned.

-Dan

(For the record, the first problem cannot be factored. Typo?)

**querti09** · October 18th 2012, 07:57 PM

I'll give an explanation for the second problem: $6x^{2}-7xy-5y^{2}$
1) This is a trinomial of the form $ax^{2}+bx+c$
because we can identify an $x^{2}$
in the first term and an $x$
in the second one and the coefficient isn't the number $1$
2) We convert this polynomial to the form $x^{2}+bx+c$
because we then can solve (we know the procedure for polynomials of this another form)
We do this by multiplying the whole polynomial by the coefficient that has the $x^{2}$ , in this case $6$ , in the following way:
a)We will multiply the first term by $6$ , so the first term of the new polynomial is going to be: $36x^{2}$ , because $6\cdot6x^{2}=36x^{2}$
b)For the second term, we will write the multiplication but we won't actually do it. We only show the multiplication: So we get: $-7y(6x)$
This can be confusing but it's just reorganizing the letters and the quantities in a way that seems like the polynomial $x^{2}+bx+c$
We put into parenthesis the $(6x)$
because we know that $(6x)^{2}=36x^{2}$ , and we are looking for a pattern that is like $x^{2}+bx+c$
So $36x^{2}$
here is the same as $x^{2}$ , the first term in the polynomial that we take as a model: $x^{2}+bx+c$
And $6x$
here is the same as the $x$ in $bx$ , the second term in the polynomial that we take as a model: $x^{2}+bx+c$
c)For the last term, we just multiply by $6$
, as we have done with in a) and we get $30y^{2}$
, because $6\cdot-5y^{2}=-30y^{2}$
Now, we have our new polynomial: $36x^{2}-7y(6x)-30y^{2}$
We have to solve this as if it were $x^{2}+bx+c$
We create two sets of parenthesis and we plug the square root of the first term in both, we consider $36x^{2}$
as one quantity squared.
$(6x )(6x )$
The sign for these factors are as follow: The second one is the sign of the second term in the polynomial that you are factoring now, in this case minus.
The sign of the second factor is the result of multiplying the sign of the second term by the sign of the last term. This rule works always the same. For this case it's positive, because negative times negative is positive.
$(6x - )(6x+ )$
and now we look into $36x^{2}-7y(6x)-30y^{2}$
what 2 quantities you substract to get $7y$ AND you multiply to get $30y^{2}$
Those quantities are $10y$ and $3y$
You plug those quantities, and then you get $(6x-10y)(6x+3y)$
All you have to do now is to divide this factorization by $6$
, the quantity by which we multiplied everything at the beginning.
we know that $2\cdot3=6$
so instead of this: $\frac{(6x-10y)(6x+3y)}{6}$
we have this: $\frac{(6x-10y)(6x+3y)}{2\cdot3}$
And we can divide $\frac{(6x-10y)}{2}$
and $\frac{(6x+3y)}{3}$
to get our final answer: $(3x-5y)(2x+y)$

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