Question

Factoring expressions completley..... 10x^3+80

Answer 1

@Hero can you please help

Answer 2

Start by factoring out any common factors.

Answer 3

10(x^3+8) @whpalmer4

Answer 4

Okay, now look at what you have — the sum of two cubes, x*x*x and 2*2*2

Answer 5

Let's try an experiment:

\[(x+1)^3 = (x+1)(x+1)(x+1) = (x^2+2x+1)(x+1) = x^3+3x^2+3x+1\]
Hmm, that doesn't get us quite what we want, does it?

Answer 6

No it does not @whpalmer4

Answer 7

But what if we multiply (x+1) instead by something else? Let's try
\[(x+1)(x^2-x+1) = x^3-x^2+x+x^2-x+1 = x^3+1\]Look at that, we made the other terms go away. More generally, \[(x+a)(x+a)(x+a) = x^3 + 3ax^2 + 3a^2x + a^3\]and if we divide that by \((x+a)\) we get \(x^2-ax+a^2\). So, a sum of two cubes, \(x^3+a^3\) can always be factored to \[(x+a)(x^2-ax+a^2)\]

What does that mean for our problem? What are the factors?

Answer 8

x and ax? @whpalmer4

Answer 9

What is the value of a, if our problem is \(x^3+8\)?

Answer 10

Hint:\[x^3+a^3=x^3+8\]

Answer 11

would it be (x+2)(x^2-2x+4)

Answer 12

@whpalmer4

Answer 13

Yes, that's the factoring.

\[(x+2)(x^2-2x+4) = x^3-2x^2+4x+2x^2-4x+8 = x^3+8\]

Answer 14

So the whole thing would be
\[10(x+2)(x^2-2x+4)\]