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???

2 Answers
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(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...

Apr 21, 2018

Answer:

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