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

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Answer 1 · 2015-06-02T20:37:01

$(x+10)^2 >= 0$ , so $y >= 6$ for all $x in RR$ .

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

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How about $y = 1.2(x+10)^2 - 6$ ?

Then $y = 6/5((x+10)^2-5)$ when $x+10 = +-sqrt(5)$

That is $y = 0$ when $x = -10+-sqrt(5)$

Hence

$y = 1.2(x+10)^2-6$

$= 6/5(x+10-sqrt(5))(x+10+sqrt(5))$

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