Is it possible to factor $y=x^2+8x+14$ ? If so, what are the factors?

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Answer 1 · 2016-01-02T23:52:36

Yes, you have to solve the equation though.

$x^2+8x+14=(x+4-sqrt(2))(x+4+sqrt(2))$

By finding the roots $x_1$ and $x_2$ you can factor using the formula:

$ax^2+bx+c=a(x-x_1)(x-x_2)$

$x^2+8x+14=0$

$Δ=b^2-4*a*c=8^2-4*1*14=8$

$x=(-b+-sqrt(Δ))/(2*a)=(-8+-sqrt(8))/(2*1)=-8/2+-sqrt(4*2)/2=$
$=-4+-(sqrt4)*sqrt(2))/2=-4+-(2*sqrt(2))/2=-4+-sqrt(2)$

Now that the roots $x_1=-4+sqrt(2)$ and $x_2=-4-sqrt(2)$ are found the factoring can be done as follows:

$x^2+8x+14=1*(x-(-4+sqrt(2)))(x-(-4-sqrt(2)))=$
$=(x+4-sqrt(2))(x+4+sqrt(2))$

An easier example for a better understanding

Factor the following function:

$y=x^2+8x+12$

We solve the equation:

$x^2+8x+12=0$

$Δ=b^2-4*a*c=8^2-4*1*12=16$

$x=(-b+-sqrt(Δ))/(2*a)=(-8+-sqrt(16))/(2*1)=-8/2+-4/2=-4+-2$

$x_1=-4+2=-2$
$x_2=-4-2=-6$

Now we have:

$x^2+8x+12=1(x-(-2))(x-(-6))=(x+2)(x+6)$

Is it possible to factor y=x^2+8x+14 y=x^2+8x+14 ? If so, what are the factors?