please help me factor completely 6p2+5pq-q2

Question

6p²+5pq-q

 

2x³+2x²y-12xy²

Answer 1

(1.)     6p² + 5pq - q²      ==>     (            )(            )

     First, look at the variables in the first and last terms and notice that they are perfect squares. With that, put a 'p' in the left hand side of each set of parentheses and a 'q' in the right hand side of each set of parentheses:

                    (  p         q )(  p          q)

Since the q² is a negative, you know that there has to be a '-' sign in one set of parentheses and a '+' sign in the other set:

                    (  p   -   q )(  p    +    q )

The first term has a coefficient of 6, which means we have find factors of 6:   1 * 6   and   2 * 3

The middle term has a coefficient of 5, which means that one of the factors of 6 we found above have to equal 5 when subtracted from one another. The only set of factors that will do this is 1 and 6. So, place a 6 in front of the 'p' in the first set of parentheses and you can leave the other one as is since it's coeffiecient is 1:

                        ( 6p  -  q )( p  +  q )

We see that this is the answer when we check the factorization works:

 (6p  -  q)(p  +  q) =  6p(p) + 6p(q) - q(p) - q(q) = 6p² + 6pq - pq -q² = 6p² + 5pq - q²

(2.)     2x³ + 2x²y - 12xy² 

     Notice that there is a greatest common factor, that being 2x. So we first factor out a 2x from every term in the equation:

                2x ( x² + xy - 6y²)   =   2x (            )(            )

Now we only need to factor what's inside the parentheses. Place an 'x' on the left hand side of each set of parentheses and a 'y' on the right hand side of each set of parentheses. Since the last term is negative, also place a '+' in on set and a '-' in the other set:

              2x ( x  +   y )( x   -   y )

Since the coefficient of the last term is a 6 and the middle term has a coefficient of 1, we need to find factors of 6 that will subtract from one another to equal 1. Those factors of 6 are 2 and 3, so we place a 3 in front of the y in the first set of parentheses and a 2 in front of the y in the other set:

             2x ( x  +  3y )( x  -  2y )

You can again check that this is the answer as we did in the problem above. 

Answer 2

Hello Karla,

The first problem 6p² + 5pq - q²

 Multiply the coefficient of first term and the last term 6 * -1 = -6. Now list the factors of -6.

1 * -6 = -6

-1 * 6 = -6

2 * -3 = -6

-3 * 2 = -6

But we need the sum of factors as 5. Since our middle term in the given equation is 5. So, if we sum the factors ( -1 + 6 = 5).

So we'll split the middle term and write

6p² + 5pq - q²

= 6p² + 6pq - pq - q²        (since 5pq = +6pq - pq)

Group first two terms and last two terms

= (6p² + 6pq)+ (- pq - q²)

= 6p(p + q) -q(p + q)

= (p + q)(6p - q) ------> answer

 

Second problem

2x³+2x²y-12xy²

GCF of 2x³, 2x²y, 12xy² is 2x. So, 2x is common in each term.

= 2x(x² + xy - 6y²)

Same as above, list the factors of -6. This time the sum of factors should be 1 (coefficient of second term) and if you multiply the factors you should get -6(coeffiecient of third term). So, -2 + 3 = 1 and -2 * 3 = -6.

= 2x(x² - 2xy + 3xy - 6y²)             (since xy = -2xy + 3xy)

= 2x(x(x - 2y) + 3y(x - 2y))           (group first two terms and last two terms)

= 2x((x - 2y)(x + 3y))

= 2x (x - 2y) (x + 3y) ---------> answer

Hope this helps you.

Answer 3

6p²+5pq-q²

= (6p-q)(p+q)

Attn: I assumed -q² as the last term.

 

2x³+2x²y-12xy²

= 2x(x²+xy-6y²)

= 2x(x+3y)(x-2y)