factor completely

Question

6q²-13qv-63v²

Answer 1

Hey Ruby,

 

First, see if there are any GCF from all three terms.  There isn't one in this case.

 

Let's try 3 & 2   and   9 & 7.

(3      ) (2     )

 

Since middle term is negative, the bigger one wins the sign: 9 will be negative with 2q to multiply with 3 from other binomial.  If they are in same parenthesis, it will be written as one term and cannot be multiplied.

 

If 9v is with 3q, then this equation will have GCF for all three terms.  In this case, there wasn't.

 

9v had to be negative because sum of middle terms had to become negative.  To achieve that, bigger number (between 9 & 7) had to take that sign.

 

(3q + 7v) (2q - 9v)

 

If you still not sure, you can always try quadratic formula.  In this case, a = 6, b = -13, c = -63.

Answer 2

First, know that factoring (like most math) takes practice, LOT and LOTS of practice.

 

1. is there anything that goes into all the terms?

        in this case: no, 6 and 63 are both divisible by 3, but 13 is not and there is not a variable that occurs in all three terms.

 

2.trinomial often factor into two binomials (aq+bv)(cq+dv). when we Multiply this out (don't forget to distribute) we need to get what we started with

6q²- 13qv- 63v²= aqcq+aqdv+bvcq+bvdv

                        =acq²+adqv+bcqv+bdv²

6q² -13qv -63v²=acq²+(ad+bc)qv+bdv²

 

I have color-coded the like terms on each side of the equation. we get

6q²=acq²          -63v²=bdv^{2                -13qv= (ad+bc)qv}   

6=ac                 -63=bd                    -13=(ad+bc)

 

For integer solutions we get options, a and c could b3 (1 and 6) or (2 and 3) notice these are the Factors  of 6

b and d could be (9 and -7) or (-9 and 7) or (3 and -21) or (-3 and 21)

 

you need to figure out what combination satisfies -13=(ad+bc)