Re: Quadratics Factoring 2

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...

Re: Quadratics Factoring 2

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

Re: Quadratics Factoring 2

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.

Re: Quadratics Factoring 2

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 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)**

Re: Quadratics Factoring 2

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...