Question

I don't want a direct answer, I want someone to help me complete the problem by guiding me through it. And yes, medals will be given.

Determine two different values of “b” in x^2 + bx + 30 so that the expression can be factored into the product of two binomials. Explain how you determined those values, and show each factorization. Explain how your process would change if the expression was 2x^2 + bx + 30.

Answer 1

@triciaal @mathmale @mathstudent55 @zepdrix

Answer 2

In order for the expression to be factorable, the discriminant (\(\color{#000000 }{ \displaystyle \Delta }\)) has the be a perfect square.

You know that
\(\color{#000000 }{ \displaystyle \Delta=b^2-4ac }\) (referring to \(\color{#000000 }{ \displaystyle ax^2+bx+c }\))

So, you are given
\(\color{#000000 }{ \displaystyle a=2 }\)
\(\color{#000000 }{ \displaystyle c=30 }\)

And thus, you have:
\(\color{#000000 }{ \displaystyle \Delta=b^2-4\cdot 2\cdot 30=b^2-240}\)

So, for what values of b, will \(\color{#000000 }{ \displaystyle \Delta}\) be a perfect square?

Answer 3

That is the first thing that comes to my mind...

Answer 4

@SolomonZelman I lost you after the Δ

Answer 5

What does Δ mean?

Answer 6

Oh, there:
256-240=16

Answer 7

wut

Answer 8

@zepdrix Can you be a translator or are you as lost as I am?

Answer 9

For the first expression you were given,
Think about the ways you can break up the 30,\[\large\rm x^2 + bx + \color{orangered}{30}\]\[\large\rm x^2 + bx + \color{orangered}{5\cdot6}\]\[\large\rm x^2 + bx + \color{orangered}{3\cdot10}\]\[\large\rm x^2 + bx + \color{orangered}{15\cdot2}\]

Answer 10

Alright, I get that

Answer 11

If you choose 5*6, then your b is 5+6,
and so on.

Answer 12

@SolomonZelman One question, if I choose 5*6 why does it turn into 5 + 6??

Answer 13

\[\large\rm x^2 + (5+6)x + \color{orangered}{5\cdot6}\quad = x^2+11x+30\]\[\large\rm x^2 + (3+10)x + \color{orangered}{3\cdot10}\quad=x^2+13x+30\]\[\large\rm x^2 + (15+2)x + \color{orangered}{15\cdot2}\quad=x^2+17x+30\]
These are a bunch of values of b that will make the trinomial factorable :)
Lot of fun options because 30 has a lot of factors.

Answer 14

If you look at the last problem you did with Tricia,\[\large\rm x^2+9x+20\]What you were really doing was, turning the 20 into 5*4,\[\large\rm x^2+9x+5\cdot4\]Because you knew that 9 could be rewritten as 5+4.

Answer 15

\[\large\rm x^2+5x+4x+5\cdot4\]

Answer 16

I kinda get it

Answer 17

Ish

Answer 18

haha XD

Answer 19

Lemme just jam through that last problem you did, really fast,
with this notation.

Answer 20

Hooray! My smartscore is 50! Oh wait, your smartscores are like 99. :(

Answer 21

\[\large\rm x^2+9x+20\]\[\large\rm x^2+4x+5x+4\cdot5\]\[\large\rm (x^2+4x)+(5x+4\cdot5)\]\[\large\rm x(x+4)+5(x+4)\]\[\large\rm (x+5)(x+4)\]50? Noiceee

Answer 22

Alright, I kinda get it @zepdrix

Answer 23

Still in the same spot? XD Ok hmm

Answer 24

So basically, all I have to do is find factors of 30, and those factors can equal b? @SolomonZelman @zepdrix

Answer 25

The `sum of those factors` can equal b, yes.

Answer 26

Then I would factor by grouping?

Answer 27

Let's try one of them just to see how it works.
If we choose 3 and 10 for our factors of 30,
then b should be 3+10.\[\large\rm x^2+13x+30\]\[\large\rm x^2+3x+10x+30\]And then yes, grouping from there :)
Maybe I won't steal the fun on this one.

Answer 28

@zepdrix brb

Answer 29

@zepdrix Alright I'm back