# How do you write  y=1.2(x+10)^2+6 into factored form?

Jun 2, 2015

${\left(x + 10\right)}^{2} \ge 0$, so $y \ge 6$ for all $x \in \mathbb{R}$.

Since there are no real values of $x$ for which $y = 0$,
$y = \frac{6}{5} \left({\left(x + 10\right)}^{2} - 5\right)$ has no linear factors with real coefficients.

Is there an error in the question?

How about $y = 1.2 {\left(x + 10\right)}^{2} - 6$ ?

Then $y = \frac{6}{5} \left({\left(x + 10\right)}^{2} - 5\right)$ when $x + 10 = \pm \sqrt{5}$

That is $y = 0$ when $x = - 10 \pm \sqrt{5}$

Hence

$y = 1.2 {\left(x + 10\right)}^{2} - 6$

$= \frac{6}{5} \left(x + 10 - \sqrt{5}\right) \left(x + 10 + \sqrt{5}\right)$