6p^{2}+5pq-q

2x^{3}+2x^{2}y-12xy^{2}

6p^{2}+5pq-q

2x^{3}+2x^{2}y-12xy^{2}

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6p^{2}+5pq-q^{2}

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

Attn: I assumed -q^{2} as the last term.

2x^{3}+2x^{2}y-12xy^{2}

= 2x(x^{2}+xy-6y^{2})

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

(1.) ** 6p**^{2}** + 5pq - q**^{2}** ** ==> ( )( )

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 q^{2} 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) = 6p^{2} + 6pq - pq -q^{2} = 6p^{2} + 5pq - q^{2}

(2.) **2x**^{3}** + 2x**^{2}**y - 12xy**^{2}

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 ( x^{2} + xy - 6y^{2}) = 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.

Hello Karla,

The first problem 6p^{2 }+ 5pq - q^{2}

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

6p^{2} + 5pq - q^{2}

= 6p^{2} + 6pq - pq - q^{2 } (since 5pq = +6pq - pq)

Group first two terms and last two terms

= (6p^{2} + 6pq)+ (- pq - q^{2})

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

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

Second problem

2x^{3}+2x^{2}y-12xy^{2}

GCF of 2x^{3}, 2x^{2}y, 12xy^{2} is 2x. So, 2x is common in each term.

= 2x(x^{2} + xy - 6y^{2})

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(x^{2} - 2xy + 3xy - 6y^{2}) (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.