# How do you factor 10y^2 - 29y + 10?

Aug 29, 2015

$10 {y}^{2} - 29 y + 10 = \left(5 y - 2\right) \left(2 y - 5\right)$

#### Explanation:

The matching coefficient of the ${y}^{2}$ term and the constant term indicate the symmetry we see in this factorisation, essentially leaving two possibilities with integer coefficients to try:

$\left(10 y - 1\right) \left(y - 10\right) = 10 {y}^{2} - 101 y + 10$

and

$\left(5 y - 2\right) \left(2 y - 5\right) = 10 {y}^{2} - 29 y + 10$

Aug 30, 2015

Factor: $f \left(y\right) = 10 {y}^{2} - 29 y + 10$

Ans: (5y - 2)(2y - 5)

#### Explanation:

$f \left(y\right) = 10 {y}^{2} - 29 y + 10 =$ 10(x + p)(x + q)
I use the new AC Method (Socratic search)
Converted trinomial: $f ' \left(y\right) = {y}^{2} - 29 y + 100 =$ (y + p')(y + q')
p' and q' have same sign. Factor pairs of 100 --> (2, 50)(4, 25). This sum is 29 = -b. Then p' = -4 ans q' = -25.
Back to original trinomial: $p = - \frac{4}{10} = - \frac{2}{5}$ and $q = - \frac{25}{10} = - \frac{5}{2}$.

Factoring form: $f \left(y\right) = \left(10\right) \left(y - \frac{2}{5}\right) \left(y - \frac{5}{2}\right) = \left(5 y - 2\right) \left(2 y - 5\right)$