How do you factor 4y^2 + 4yz + z^2 - 1?

May 15, 2016

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

Explanation:

Focusing our attention on just the terms of degree $2$, notice that both $4 {y}^{2} = {\left(2 y\right)}^{2}$ and ${z}^{2}$ are perfect squares, and we find:

${\left(2 y + z\right)}^{2} = {\left(2 y\right)}^{2} + 2 \left(2 y\right) z + {z}^{2} = 4 {y}^{2} + 4 y z + {z}^{2}$

So we have:

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

Then since ${\left(2 y + z\right)}^{2}$ and $1 = {1}^{2}$ are both perfect squares, we can use the difference of squares identity:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

with $a = \left(2 y + z\right)$ and $b = 1$ as follows:

${\left(2 y + z\right)}^{2} - 1$

$= {\left(2 y + z\right)}^{2} - {1}^{2}$

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

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

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Bonus

One way of spotting that:

$4 {y}^{2} + 4 y z + {z}^{2} = {\left(2 y + z\right)}^{2}$

is to notice the pattern $4 , 4 , 1$ of coefficients on the left hand side.

You may know that $441 = {21}^{2}$ so notice that the pattern of coefficients on the right hand side is $2 , 1$.

When the number $21$ is squared to give $441$ there are no carries from one column to another, so we have:

$21 \times 21 = \left(20 + 1\right) \times \left(20 + 1\right) = \left(20 \times 20\right) + 2 \left(20 \times 1\right) + \left(1 \times 1\right)$

$= 400 + 40 + 1 = 441$

This is like putting $y = 10$ and $z = 1$ in $4 {y}^{2} + 4 y z + {z}^{2}$ and $\left(2 y + z\right)$.

This works for some other quadratics with small coefficients. For example:

${\left(x + 3\right)}^{2} = {x}^{2} + 6 x + 9$ like ${13}^{2} = 169$

$\left(2 x + 1\right) \left(x + 3\right) = 2 {x}^{2} + 7 x + 3$ like $21 \times 13 = 273$