Question

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???

Answer 1

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(x^2-8x+16)+4`

Now multiply through with `4`:

`=4x^2-32x+64+4`

`y=4x^2-32x+68`

A starting guess for the factors might be:

`(4x-a)(x-b)`

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:

`axxb=68`

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

`a+4b=-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=68/a`

Substituting this value into the other equation:

`a+68/a=-32`

Multiply through by `a`:

`a^2+68=-32a`

Rearranging:

`a^2+32a+68=0`

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

`a=(32+-sqrt((32)^2+4xx1xx68))/(2xx1)`

`=(-32+-sqrt(1024+272))/2`

`=(-32+-36)/2`

Therefore `a=-34` or `a=2`

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

`b=(32-a)/4`

We get `b=16 1/2` or `b=7 1/2`

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

`y=(2x-a)(2x-b)`

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...

Answer 2

For me it is easy. See reasoning below

Explanation:

`y=4(x-4)^2+4`. Let be `t=x-4`

`y=4t^2+4=4(t^2+1)`

We know that `t^2+1` has two complex roots `+-i`. Then

`y=4(t+i)(t-i)=4(x-4+i)(x-4-i)`

Hope this help