1. ## Why is this incorrect (quadratics)?

I'm supposed to solve:
3X^2+17X+20=0

This is my working:
-> 3X^2-3X+20X+20=0
-> 3X(X-1)-20(X-1)=0
-> (X-1)(3X-20)=0
(X-1) -> X=1
(3X-20) -> 3X=20 -> X=20/3
X= 1 or 20/3

However, the anser to that question is
-4 or -5/3

What did I do wrong?

2. Originally Posted by juliak
I'm supposed to solve:
3X^2+17X+20=0

This is my working:
-> 3X^2-3X+20X+20=0
-> 3X(X-1)-20(X-1)=0 This is wrong. Should be 3x(x-1) + 20(x+1) which doesn't help.
-> (X-1)(3X-20)=0
(X-1) -> X=1
(3X-20) -> 3X=20 -> X=20/3
X= 1 or 20/3

However, the anser to that question is
-4 or -5/3

What did I do wrong?
In general, $ax^2 + bx + c$ can be factored by finding two numbers that multiply to get $ac$ and add up to $b$, call them $r_{1}, r_{2}$. Then you split your quadratic: $ax^2 + r_{1}x + r_{2}x + c$ and you should be able to factor from there (assuming it was factorable in the first place)

For your quadratic, you have to find 2 numbers that multiply to (3)(20) = 60 that add up to 17. 12 and 5 should come to mind. So:
$3x^2 + {\color{blue}17x} + 20 \ = \ 3x^2 + {\color{blue}12x + 5x} + 20 \ = \ 3x(x + 4) + 5(x+4) \ = \ \hdots$

3. What's with all the negative signs already?

$3x^2 + 17x + 20 = 0$

$3x^2 - 3x + 20x + 20 = 0$

$=3x(x-1) + 20 (x+1) = 0$

... and your factorisation falls over.

What I do in this situation when I can't find the numbers to factor by is use the quadratic formula. Then when I know what the answer is (or I've looked it up in the answers in the back of the book) I go back and use those numbers to factor it properly and I don't tell the teacher I cheated.

4. $3x^2 + 17x+20$

$\frac{-17 ^+_-\sqrt{(17)^2-4(3)(20)}}{2(3)} = 0$