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

Jan 2, 2016

Yes, you have to solve the equation though.

${x}^{2} + 8 x + 14 = \left(x + 4 - \sqrt{2}\right) \left(x + 4 + \sqrt{2}\right)$

Explanation:

By finding the roots ${x}_{1}$ and ${x}_{2}$ you can factor using the formula:

$a {x}^{2} + b x + c = a \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$

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

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:

${x}^{2} + 8 x + 14 = 1 \cdot \left(x - \left(- 4 + \sqrt{2}\right)\right) \left(x - \left(- 4 - \sqrt{2}\right)\right) =$
$= \left(x + 4 - \sqrt{2}\right) \left(x + 4 + \sqrt{2}\right)$

An easier example for a better understanding

Factor the following function:

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

We solve the equation:

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

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:

${x}^{2} + 8 x + 12 = 1 \left(x - \left(- 2\right)\right) \left(x - \left(- 6\right)\right) = \left(x + 2\right) \left(x + 6\right)$