How do you factor 5y^2-12y+4=0?

Aug 10, 2016

$\left(5 x - 2\right) \left(x - 2\right) = 0$

Explanation:

$\textcolor{t u r q u o i s e}{5} {y}^{2} \textcolor{\mathmr{and} a n \ge}{-} \textcolor{m a \ge n t a}{12} y \textcolor{red}{+} \textcolor{t u r q u o i s e}{4} = 0$

We need to find factors of $\textcolor{t u r q u o i s e}{5 \mathmr{and} 4}$

The factors of 5 and 4 must $\textcolor{red}{\text{(ADD)}}$ together to give $\textcolor{m a \ge n t a}{12}$

The signs in the brackets will be $\textcolor{red}{\text{THE SAME}}$
They will BOTH be $\textcolor{\mathmr{and} a n \ge}{\text{(NEGATIVE)}}$"

Use factors of 5 and 4 and test the cross-products:

$\textcolor{w h i t e}{\ldots . .} 5 \text{ "2 " } \Rightarrow 1 \times 2 = 2$
$\textcolor{w h i t e}{\ldots . .} 1 \text{ "2" " rArr 5xx2 = 10" } 10 + 2 = \textcolor{m a \ge n t a}{12}$

The top row gives the factors in one bracket.
The second gives the factors in the other bracket.

$\left(5 x - 2\right) \left(x - 2\right) = 0$

Solving would lead to $x = \frac{2}{5} \mathmr{and} x = 2$

Aug 10, 2016

(5x - 2)(x - 2)

Explanation:

Use the new AC Method (Socratic Search)
$f \left(y\right) = 5 {y}^{2} - 12 y + 4 = 5 \left(x + p\right) \left(x + q\right)$
Converted trinomial $f ' \left(y\right) ' = {y}^{2} - 12 y + 20 =$ (x + p')(x + q').
Method. Find p' and q' then divide them by a to get p and q.
p' and q' have same sign because ac > 0.
Factor pairs of (ac = 20) --> (2, 10)(-2, -10). This last sum is -12 = b.
Then, p' = -2 and q' = -10
Back to f(y) --> $p = \frac{p '}{a} = - \frac{2}{5}$. and $q = \frac{q '}{a} = - \frac{10}{5} = - 2$.
Factor form of f(y):
$f \left(y\right) = 5 \left(x - \frac{2}{5}\right) \left(x - 2\right) = \left(5 x - 2\right) \left(x - 2\right)$