Question

Factor completely: 5x^2+30x+45

Answer 1

i think it is

2 numbers that produce 45 and add up to 30

a*b=45
a+b=30

Answer 2

5 (x+3)^2 well its actually this then

Answer 3

5(x^2 + 6x + 9)
5(x+3)^2

Answer 4

im sorry im still confused about this. Can you explain I really want to understand this

Answer 5

you start off by dividing 5 from the whole equation. it's so that you can factor easier.
after factoring out the 5 you get a less complex equation. With this you can easily find that 3 times 3 is 9 and 3+3 is 6. with that you can get the answer.
5(x^2 + 6x + 9) you repeat the 3 in both parentheses so you just have only one and put a ^2 next to it.
5(x+3)^2
sorry if it made you even more confused. best of luck to you.

Answer 6

@timo86m first post. This is false, you have to multiply the constant by 5 before you can use the grouping method. That's why it's best to simplify the equation first by taking out 5 as a common factor.

Answer 7

@Iamapineapple okay, so when im done multiplying i get 5x^2+30x+225 what's the next thing

Answer 8

@ineedyouubiebs, using that method probably isn't the best way to do it - factoring 225 in your head isn't easy. And when I said multiply the constant by 5, I didn't actually mean in the equation, just in your head :P

I'll explain the easy method:

You equation:
5x^2+30x+45

Answer 9

Take a Common Factor of 5 OUT of the equation, so that means I divide everything by 5 and take it outisde. You should ALWAYS take out a common factor if you can
5(x^2 + 6x + 9)

Now I can use the method everyone mentioned.
I have to find two number that multiply together to give 9 and add to give 6.

So

a x b = 9
a x b = 6

I know that the only things that multiply to give 9 are 1x9 or 3x3.
1+9 =10, therefore a and b must be 3

so after you find your numbers, you can write it like this
5[(x+3)(x+3)]
Because (x+3) is multiplied twice, I can use something called perfect squares. This means I can write (x+3)(x+3) as (x+3)^2

So your answer it 5(x+3)^2

Answer 10

OH MY GOD THIS MAKES SENCE NOW, THANK YOU SO MUCH. WOULD YOU MIND WORKING ON ANOTHER EXAMPLE IN ANOTHER POST PLEASE?

Answer 11

sure

Answer 12

yes so i think if you have the standard quadtratic equation

ax^2+bx+c then if possible factor out the a

first.