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?

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- Sep 6th 2008, 11:52 PMjuliakWhy 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? - Sep 6th 2008, 11:58 PMo_O
In general, $\displaystyle ax^2 + bx + c$ can be factored by finding two numbers that multiply to get $\displaystyle ac$ and add up to $\displaystyle b$, call them $\displaystyle r_{1}, r_{2}$. Then you split your quadratic: $\displaystyle ax^2 + r_{1}x + r_{2}x + c$ 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:

$\displaystyle 3x^2 + {\color{blue}17x} + 20 \ = \ 3x^2 + {\color{blue}12x + 5x} + 20 \ = \ 3x(x + 4) + 5(x+4) \ = \ \hdots$ - Sep 6th 2008, 11:58 PMMatt Westwood
What's with all the negative signs already?

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

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

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

... 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. - Sep 7th 2008, 12:07 AM11rdc11
$\displaystyle 3x^2 + 17x+20$

Use the quad formula

$\displaystyle \frac{-17 ^+_-\sqrt{(17)^2-4(3)(20)}}{2(3)} = 0$ - Sep 7th 2008, 12:07 AMjuliak
Oh okay I get it now.

Heh, thank you :D.