Question

factor completely.

4w^2+52w+48

Answer 1

Well the first thing I notice, is that everything is divisible by 4...so we can factor out a 4

\[\large 4(w^2 + 13w + 12)\]

Now, how would we factor what is in the parenthesis?

Answer 2

Hint: 12 = 12 * 1

Answer 3

I think that's where I got stuck I wanna say we divide by the GCF of both and factor out the w^2?

Answer 4

Not quite, we did divide out by the 4, but now we cant factor out the w^2 because the other 2 terms dont have a w^2 variable...

So what we need to do

\[\large 4(w^2 + 13w + 12)\]

Is find 2 numbers, that multiply to make 12, but at the same time ADD to make 13

*Hint, look at @robtobey hint above :)

Answer 5

lol yes 12 and 1 =)

Answer 6

Indeed ^_^

So now we just write those 2 factors

\[\large 4(w^2 + 13w + 12)\]

would now become *in factored form*

\[\large 4((w + 12)(w + 1))\]

Answer 7

I thought everything ended up in parenthesis when factoring? =o

Answer 8

Well it does! The second set of parenthesis isn't QUITE needed here, I just like to be thorough :P

However we can write this answer a few different ways

\[\large 4(w + 12)(w + 1)\]

or if you dont like the look of the 4 there on the outside, we can multiply it into one of the parenthesis

\[\large (4w + 48)(w + 1)\]

Answer 9

oh ok got it haha I gues the four on the outside was just throwing me off =) thank you!

Answer 10

Yeah there's different ways to write quadratics...sometimes too many >.< lol

But of course, anytime :)

Answer 11

lol well thanks for imparting your knowledge to me I appreciate it =)

Answer 12

^_^