How do you factor the trinomial $2x^2+10x+25$ ?

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Answer 1 · 2017-12-17T01:34:41

$x=(-5+5i)/2,$ $(-5-5i)/2$

$2x^2+10x+25$ is a quadratic equation in standard form:

$ax^2+bx+c$ ,

where:

$a=2$ , $b=10$ , and $c=25$

Set the equation to $0$ and solve for $x$ .

$0=2x^2+10x+25$

$x=(-b+-sqrt(b^2-4ac))/(2a)$

Plug in the known values.

$x=(-10+-sqrt(10^2-4*2*25))/(2*2)$

Simplify.

$x=(-10+-sqrt(-100))/4$

Prime factorize the square root.

$x=(-10+-sqrt(-2xx2xx5xx5))/4$

Simplify.

$x=(-10+-10i)/4$

Factor out the common $2$ in the numerator.

$x=(2(-5+-5i))/4$ .

Divide $4$ in the denominator by the $2$ in the numerator.

$x=(-5+-5i)/2$

Solutions for $x$ .

$x=(-5+5i)/2,$ $(-5-5i)/2$

How do you factor the trinomial 2x^2+10x+252x^2+10x+25?