# How do you factor y = 4 (x - 4)^2 + 4 I have been working on this for days and I can’t find the answer. Can anybody help me???

Apr 2, 2018

There are a number of possible ways, but I've shared the way I'd do it below. In the end I didn't get a solution, though...

#### Explanation:

It's a toughie, and (for me at least) factoring always involves some trial and error. In the end I worked through the process with one guess that didn't work out, but hopefully that sets you up well to find the solution.

Here's how I'd approach it:

First expand the square:

y=4((x−4)(x-4))+4

$= 4 \left({x}^{2} - 8 x + 16\right) + 4$

Now multiply through with $4$:

$= 4 {x}^{2} - 32 x + 64 + 4$

$y = 4 {x}^{2} - 32 x + 68$

A starting guess for the factors might be:

$\left(4 x - a\right) \left(x - b\right)$

Why do I use minus signs both terms? Because the middle term of the quadratic is negative by the final term is positive.

We know that:

$a \times b = 68$

The tricky bit is that factor of $4$ in the first bracket. It means that:

$a + 4 b = - 32$

(expand the brackets to see why this is the case)

Now we have two equations in two unknowns, so we use our toolkit for simultaneous equations.

$b = \frac{68}{a}$

Substituting this value into the other equation:

$a + \frac{68}{a} = - 32$

Multiply through by $a$:

${a}^{2} + 68 = - 32 a$

Rearranging:

${a}^{2} + 32 a + 68 = 0$

Which is a quadratic, so now we need our quadratic tools. I tend to use the quadratic formula:

$a = \frac{32 \pm \sqrt{{\left(32\right)}^{2} + 4 \times 1 \times 68}}{2 \times 1}$

$= \frac{- 32 \pm \sqrt{1024 + 272}}{2}$

$= \frac{- 32 \pm 36}{2}$

Therefore $a = - 34$ or $a = 2$

If we substitute both values back into $a + 4 b = 32$, which we can rearrange to make $b$ the subject as:

$b = \frac{32 - a}{4}$

We get $b = 16 \frac{1}{2}$ or $b = 7 \frac{1}{2}$

Hmm, these fractional factors suggest to me that perhaps that first guess should have been:

$y = \left(2 x - a\right) \left(2 x - b\right)$

Nope: when I tried that and ran it through I get the square root of a negative number. We can deal with 'unreal roots', but probably not here.

I don't have the time or space here to go through the whole process again, so I'll leave it to you to work through it...

Apr 21, 2018

For me it is easy. See reasoning below

#### Explanation:

$y = 4 {\left(x - 4\right)}^{2} + 4$. Let be $t = x - 4$

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

We know that ${t}^{2} + 1$ has two complex roots $\pm i$. Then

$y = 4 \left(t + i\right) \left(t - i\right) = 4 \left(x - 4 + i\right) \left(x - 4 - i\right)$

Hope this help