Question

what are your objectives when factoring the sum of two squares

Answer 1

like what are you trying to do

Answer 2

Using complete sentences, compare and contrast the ways to factor the "sum of two squares" and the "difference of two squares." Provide an explanation and example of each, including any similarities and differences.

Answer 3

this the actual question

Answer 4

The sum of squares factors into complex conjugates:
\[\large a^2+b^2\qquad=\qquad (a+b i)(a-b i)\]

The difference of squares factors into conjugates:\[\large a^2-b^2\qquad=\qquad (a-b)(a+b)\]

So both methods are fairly similar.
What are you trying to actually do when you factor the `sum of squares`?
Ummm I guess an example might help.

\[\large y=x^2+4\]To find `roots` of this polynomial we set it equal to zero, then solve for x.\[\large 0=x^2+4\]Solving for x gives us,\[\large x=\pm \sqrt{-4}\]Which gives us two complex roots, \(\large x=2i\) and \(\large x=-2i\).
Another way we could have gotten these roots is by using the formula for the sum of squares.\[\large 0=x^2+4 \qquad\to\qquad 0=x^2+2^2\]
Using our rule we can factor this as such,\[\large 0=(x+2i)(x-2i)\]And from there set each factor equal to zero and solve for x.\[\large x+2i=0 \qquad\to\qquad x=-2i\]\[\large x-2i=0 \qquad\to\qquad x=2i\]

Answer 5

when factoring a square i am trying to
Step 1: Check for a GCF.

Step 2: Take the square root of the first term.

Step 3: Take the square root of the last term.

Step 4: Check the sign for the middle term:

If the sign of the middle term is positive, the factored binomial follows the pattern (a + b)2
If the sign of the middle term is negative, the factored binomial follows the pattern (a – b)2
so an example would be x^2+8x+16
find the square root of x^2 and 16 so it would be (x+4)^2
take that and foil and you have the original equation but I just want to know what the purpose of doing this kind of math is

Answer 6

Oh you're talking about starting with something like \(\large x^2+8x+16\). Oh ok. :o

A parabola like \(\large y=x^2\) when we have it in standard form looks like:\[\large y=(x-\color{green}{h})^2+\color{blue}{k}\]Where the vertex of the parabola is located at \(\large (\color{green}{h},\color{blue}{k})\).
The standard form helps us to graph it much easier.

If we have a function \(\large y=x^2+8x+16\) it's a little bit difficult to graph.
We know it's a parabola because the largest degree of x is 2. But beyond that it's a little tricky.

If we factor it as you did,\[\large y=[x+4]^2 \qquad\to\qquad y=[x-\color{green}{(-4)}]^2\]we can see that the vertex has been shifted 4 to the left (since our h=-4).