# 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

#### Answer:

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

#### Answer:

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