• Apr 21st 2013, 11:43 AM
Mathnood768
Factor: 3(2z + 3)^2 - 9(2z + 3) - 30

My attempt:

3(2z + 3)^2 - 9(2z + 3) - 30

3x^2 - 9x - 30 Replace (2z +3)^2 and (2z +3) with a variable (I choose x)

3(x^2 - 3x - 10) Find 2 integers whose product is ac = -10 and sum equals b = -3 (+2 and - 5)

3(x^2 + 2x )(- 5x - 10)

x(x + 2) -5(x + 2)

(x - 5) ( x + 2) Plug (2z +3)^2 and (2z +3) back into the x's.

((2z + 3)^2) - 5 ((2z + 3) + 2)

(2z - 2)(2z + 5)

I got it wrong, the correct answer is 6(z - 1)(2z + 5)

Where did I go wrong?
• Apr 21st 2013, 11:51 AM
MarkFL
We are given to factor:

\$\displaystyle 3(2z + 3)^2-9(2z + 3)-30\$

I would first factor our the 3:

\$\displaystyle 3((2z + 3)^2-3(2z + 3)-10)\$

Now, we need two factors of -10 whose sum is -3, and they are 2 and -5, hence:

\$\displaystyle 3((2z+3)+2)((2z+3)-5)\$

Further simplification will lead to the result you desire...
• Apr 21st 2013, 12:02 PM
Mathnood768
Why do you need two factors whose sum is -2 and not - 3? The middle term is -3.
• Apr 21st 2013, 12:10 PM
MarkFL
Quote:

Originally Posted by Mathnood768
Why do you need two factors whose sum is -2 and not - 3? The middle term is -3.

Sorry, that was a typo. I have edited my post.
• Oct 7th 2013, 03:32 AM
mathssolutions
Quote:

Originally Posted by Mathnood768
Factor: 3(2z + 3)^2 - 9(2z + 3) - 30

My attempt:

3(2z + 3)^2 - 9(2z + 3) - 30

3x^2 - 9x - 30 Replace (2z +3)^2 and (2z +3) with a variable (I choose x)

3(x^2 - 3x - 10) Find 2 integers whose product is ac = -10 and sum equals b = -3 (+2 and - 5)

3(x^2 + 2x )(- 5x - 10)

x(x + 2) -5(x + 2)

(x - 5) ( x + 2) Plug (2z +3)^2 and (2z +3) back into the x's.

((2z + 3)^2) - 5 ((2z + 3) + 2)

(2z - 2)(2z + 5)

I got it wrong, the correct answer is 6(z - 1)(2z + 5)

Where did I go wrong?

3(2z + 3)^2 - 9(2z + 3) - 30

This is an expression of the form ax2+bx+c so let x = 2z+3 hence 3x2-9x-30

Multiply 3 by -30 (-90) and find a pair of its factors that sum to -9

The pair are 6 and -15 so replace -9x with +6x and -15x

Hence 3x2 +6x -15x -30 now bracket off and factorise

[(3x2 +6x) -(15x -30)]

[3x(x+2)-15(x-2)]

(3x-15)(x+2) = 3(x-5)(x+2)

Now replace x with 2z+3

3[(2z+3)-5][(2z+3)+2]

3(2z-2)(2z+5) = 6(z-1)(2z+5)
• Oct 7th 2013, 03:44 AM
Prove It
Quote:

Originally Posted by Mathnood768
Factor: 3(2z + 3)^2 - 9(2z + 3) - 30

My attempt:

3(2z + 3)^2 - 9(2z + 3) - 30

3x^2 - 9x - 30 Replace (2z +3)^2 and (2z +3) with a variable (I choose x)

3(x^2 - 3x - 10) Find 2 integers whose product is ac = -10 and sum equals b = -3 (+2 and - 5)

3(x^2 + 2x )(- 5x - 10) Should be 3(x^2 + 2x - 5x - 10)

= 3[ x(x + 2) -5(x + 2) ]

=3[ (x - 5) ( x + 2) ] Plug (2z +3)^2 and (2z +3) back into the x's.

No, you just put 2x+3 wherever you see x

See edits...