# How do you factor y^4+6y^2+25?

Jul 23, 2016

$\left({y}^{4} + 6 {y}^{2} + 25\right)$

= $\left(y - 1 - 2 i\right) \left(y + 1 + 2 i\right) \left(y + 1 - 2 i\right) \left(y - 1 + 2 i\right)$

#### Explanation:

If we use ${y}^{2} = x$, the polynomial ${y}^{4} + 6 {y}^{2} + 25$ is equivalent to ${x}^{2} + 6 x + 25$, a quadratic polynomial. In this the discriminant $\left({b}^{2} - 4 a c\right)$ for $\left(a {x}^{2} + b x + c\right)$ is ${6}^{2} - 4 \times 1 \times 25 = 36 - 100 = - 64$ and hence ${x}^{2} + 6 x + 25$ does not have real zeros or factors.

Its zeros are given by quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$ are $\frac{- 6 \pm \sqrt{{6}^{2} - 4 \times 1 \times 25}}{2 \times 1} = \frac{- 6 \pm \sqrt{- 64}}{2}$ or

$\frac{- 6 \pm 8 i}{2} = - 3 \pm 4 i$

Hence ${x}^{2} + 6 x + 25 = \left(x + 3 - 4 i\right) \left(x + 3 + 4 i\right)$ and

${y}^{4} + 6 {y}^{2} + 25 = \left({y}^{2} + 3 - 4 i\right) \left({y}^{2} + 3 + 4 i\right)$

Now to factorize further, let us find zeros of $\left({y}^{2} + 3 - 4 i\right)$ and $\left({y}^{2} + 3 + 4 i\right)$

For $\left({y}^{2} + 3 - 4 i\right)$, quadratic formula gives its zeros as $\frac{\pm \sqrt{- 4 \times 1 \times \left(3 - 4 i\right)}}{2}$ or $\left(\pm \sqrt{- 3 + 4 i}\right)$ and assuming
$- 3 + 4 i = {\left(a + b i\right)}^{2}$, then

${a}^{2} - {b}^{2} + 2 a b i = - 3 + 4 i$ which gives us

${a}^{2} - {b}^{2} = - 3$ and $2 a b = 4$. As such ${a}^{2} + {b}^{2} = \sqrt{{\left({a}^{2} - {b}^{2}\right)}^{2} + 4 {a}^{2} {b}^{2}} = \sqrt{9 + 16} = 5$.

Hence ${a}^{2} = 1$ and ${b}^{2} = 4$, and possible solutions for $\left(a , b\right)$ are $\left(1 , 2\right)$ and $\left(- 1 , - 2\right)$ (as $a b$ is positive).

Hence zeros of $\left({y}^{2} + 3 - 4 i\right)$ are $1 + 2 i$ and $- 1 - 2 i$ and hence

$\left({y}^{2} + 3 - 4 i\right) = \left(y - 1 - 2 i\right) \left(y + 1 + 2 i\right)$.

Similarly $\left({y}^{2} + 3 + 4 i\right) = \left(y + 1 - 2 i\right) \left(y - 1 + 2 i\right)$ and hence

$\left({y}^{4} + 6 {y}^{2} + 25\right) = \left(y - 1 - 2 i\right) \left(y + 1 + 2 i\right) \left(y + 1 - 2 i\right) \left(y - 1 + 2 i\right)$