Question

12x^(2) (x + 3) + 7x (x + 3) - 45 (x + 3) factor completely. I got as far as factoring out the (x + 3). where do you go from there?

Answer 1

\[(x+3)(12x^2+7x-45)\]

Answer 2

What are you left with after you factor out (x+3)? Or, see above :)

Answer 3

going to need to use quadratic or maybe you can split up the middle

i'd pick quadratic since your a and c are pretty big.

Answer 4

\[x=\frac{-b\pm \sqrt {b^2-4ac}}{2} \]

for
\[0=ax^2+bx+c\]

Answer 5

\[\frac{-7\pm \sqrt{49-4(12)(-45)}}{24}\]

Answer 6

the denominator of the formulae shoudl be 2a not just 2

Answer 7

I have no idea what that is outkast.

Answer 8

\[\frac{-7\pm \sqrt{2209}}{24}=\frac{-7\pm 47}{24}\]

Answer 9

quadratic formulae is a way of finding zeroes/roots. Usually it's used in a sense where you have a complex or rational zeroes/roots.

Answer 10

Im sorry. Im not that smart when it comes to math, so I don't really understand anything that you're explaining.

Answer 11

when you factor something you are trying to basically find as many roots/zeroes as you can.

a zero/root is when you equate a multiple of equation to 0, the whole equation will become 0. For example, before you did anything , you pulled out a (x+3). -3 is a root because if you let x=-3, (x+3)=0, which then would make

\[(x+3)(12x^2+7x-45)=0\]
\[0(12x^2+7x-45)=0\]
\[0=0\]

Answer 12

what the quadratic formulae does is find the zeroes/roots just like x=-3 so that you can factor it completely. Say if you use the quadratic formula above and you get x=5, this would mean that

\[x+c=0\] where c is just a number. Since we know x=5 we now get

\[5+c=0\]
subtracting 5 would get

\[c=-5\] so now we know that

\[x-5=0\] when x=5, and we can factor (x-5) out now

Answer 13

if this is new to you we'll just use the other method

so take

\[12x^2+7x-45=0\]
where \[ax^2+bx-c=0\]

multiply a*c

\[12*-45=-540\]

now find factors of -540 that add up to b, which equals 7