We are given to factor:
I would first factor our the 3:
Now, we need two factors of -10 whose sum is -3, and they are 2 and -5, hence:
Further simplification will lead to the result you desire...
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?
We are given to factor:
I would first factor our the 3:
Now, we need two factors of -10 whose sum is -3, and they are 2 and -5, hence:
Further simplification will lead to the result you desire...
3(2z + 3)^2 - 9(2z + 3) - 30
This is an expression of the form ax^{2}+bx+c so let x = 2z+3 hence 3x^{2}-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 3x^{2} +6x -15x -30 now bracket off and factorise
[(3x^{2} +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)