Question

Is this a sum of squares? Can someone factor this: g(x) = x^2+92

Answer 1

Is it \(92\) or \(9^2\)?
Depending on what it is, the answer is either no or yes.

Answer 2

\(a^2 - b^2 = (a + b)(a - b)\)

\(a^2 + b^2 = a^2 - (-b^2) = (a + bi)(a - bi)\)

Answer 3

it's 9^2 sorry @mathstudent55

Answer 4

but actually could you explain to me Function: g(x) = x^2 + 2^2 can't be factored? I think it's something with prime numbers but I can't remember

Answer 5

@mathstudent55 you there?

Answer 6

@FortyTheRapper could you help explain for me

Answer 7

Is that x^2+4 basically?

Answer 8

i think so, because 2 squared is 4

Answer 9

It's pretty much because of that + sign. In order to factor these types, they would need to be a minus sign

Answer 10

How would I put that in writing as to why it can't be factored. Would I say because the equation is prime?

Answer 11

A sum of squares cannot be factored.
A difference of squares can be factored.

Answer 12

^ I like that one

Answer 13

A sum of squares can be factored, if you are willing to use complex numbers.

\[x^2+9^2 = x^2-(-1)(9^2) = x^2-(i^2)(9^2) = (x+3i)(x-3i)\]where \(i=\sqrt{-1}\)

Answer 14

oh okay!

Answer 15

sorry, that should be \[(x+9i)(x-9i)\]at the end...

Answer 16

what would you say the method is to factor a difference of squares?

Answer 17

recognition of a pattern. Anytime I see <something>^2 - <something else>, the little bell goes off and I try to figure out how I can express <something else> as a square.

Answer 18

what do you mean?

Answer 19

\[x^2-9\]well, obviously \(x^2\) is something squared...so now I look at the \(9\) and say "how do I write \(9\) as some other quantity squared?"

well, \[9=a^2\]\[\sqrt{9} = a\]\[3=a\]so \[x^2-9 = x^2-a^2 = x^2-(3)^2\]so I can immediately write that factored as \[x^2-9 = (x+3)(x-3)\]

Answer 20

try factoring this:
\[16x^2-25\]

Answer 21

Do I start looking for the GCF?

Answer 22

there is no GCF in this case...just look at it and tell me if it is a difference of squares.

Answer 23

No, I think

Answer 24

Oh wait, I think it is

Answer 25

(4x - 5) (4x +5)

Answer 26

Or (4x + 5) (4x - 5) does the matter the order in which the + or - is placed?

Answer 27

nope, doesn't matter. multiplication is commutative.

\[(4x+5)(4x-5) = (4x-5)(4x+5)\]

Here's a slightly trickier one:
\[9x^2y^6-100z^4\]

Answer 28

Oh wow, um, (3xy^3 - 10z^2) (3xy^3 - 10z^2) ?

Answer 29

Could you help me understand a perfect square trinomial?

Answer 30

yes, you got it it right, well, almost right...doesn't one of those terms need a + instead of a -?

Answer 31

Oh Right, I forgot that part! It needs it to make the other question have the -100

Answer 32

a perfect square trinomial is one obtained by squaring a binomial.

\[(x+3)^2 = (x+3)(x+3) = x*x + x*3 + 3*x + 3*3 = x^2 + 6x + 9\]That is a perfect square trinomial. We can recognize a PST by a couple of traits:

1) 1st term is a positive perfect square
2) 3rd term is a positive perfect square
3) 2nd term is twice the product of the square root of the first term and square root of the third term

Let's check our prospective PST:

1) 1st term is positive perfect square. \(x^2\) is positive and a perfect square.
2) 3rd term is positive perfect square. \(9\) is positive and a perfect square.
3) 2nd term is twice product of square root of 1st term and square root of 3rd term.\[6x = 2\sqrt{x^2}\sqrt{9} = 2*x*3 = 6x\checkmark\]

Answer 33

\[x^2-6x+9\]Is that a PST?

Answer 34

No because the 1st and 3rd term does not result in the 2nd term right?

Answer 35

uh, well, I must have misstated the rule :-)
\[(x-3)^2=(x-3)(x-3) = x^2 - 3x -3x +9 = x^2 -6x + 9\]
So it IS a PST.

2nd term (ignoring sign) must be equal to 2*sqrt(1st term)*sqrt(3rd term)

Answer 36

Oh okay, so in other words: The function g(x) is a perfect square trinomial.
Function: g(x)=(x-3)^2
Factored: g(x)=x^2-6x+9

That's how my school teaches it

Answer 37

Also do you know what it means if a function could only have a GCF factored out of it? Would it almost be like the difference of squares?

Answer 38

uh, factored would be \[g(x) = (x-3)(x-3)\]

\[x^3+3x^2+7x\]can only have a GCF of \(x\) factored out of it:
\[x(x^2+3x+7)\]but that doesn't imply anything about what's left...

Answer 39

So it needs the gcf to be factored out in order to get a result?

Answer 40

if you can only factor out a GCF from the function, that's all the factoring you can do...

Answer 41

Oh! Okay, could you explain the rest of what would be implied about what's left?

Answer 42

That simply means that what is left is not factorable. For example, you can't factor \[x^2+3x+7\]if you stick to rational numbers. I can multiply that polynomial by anything I want, and that will give me another polynomial where I can factor out as a GCF whatever I multiplied by, but what's left is not factorable.

I could also multiply a polynomial that IS factorable by a GCF:

\[x^4-9x^2\]

Can you factor that?

Answer 43

Oh, that makes sense I think. And, x^2(x^2 - 9) ?

Answer 44

factor it all the way out...

Answer 45

There's more to that? Um (x^2-3) (x^1+9) I'm not doing this right am I ahah?

Answer 46

\[x^2(x^2-9) = x*x*(x-3)(x+3)\]

Answer 47

\[x^2-9\]is a difference of squares, right?

Answer 48

no, but what could it be?

Answer 49

what do you mean, no?

\[(x+3)(x-3) = x*x -3x+3x-9 = x^2-9\]Is that not a difference of squares?

Answer 50

wouldn't it have to be x^4 - 9 in the end?

Answer 51

I mean x^2 - 9 is a difference of squares but that doesn't go back to x^4 - 9. Am I even making sense right now? Am I right or wrong

Answer 52

\(x^4 - 9\) can be rewritten as \((x^2)^2 - 3^2\) where you see clearly it is the difference of two squares. It is the square of \(x^2\) minus the square of \(3\).

\(x^4 - 9 = (x^2 + 3)(x^2 - 3) \)

Answer 53

\[x^4-9x^2\]can be factored as follows
both terms have a GCF of \(x^2\):
\[x^4-9x^2=x^2*x^2-9*x^2=x^2(x^2-9)\]the stuff in parentheses is a difference of squares\[x^2(x^-9)=x^(x^2-3^2)=x^2(x-3)(x+3)\]

\[x^4-9\]is also a difference of squares:\[(x^2)^2-3^2=(x^2-3)(x^2+3)\]if we had gone with a different exampl,\[x^4-81 = (x^2)^2-9^2=(x^2-9)(x^2+9)\]but \(x^2-9\) is a difference of squares so that factors further into\[x^4-81=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9)\]