Question

factor completely
x2 + 9x + 20

Answer 1

\( \color{Black}{\Rightarrow x^2 + 4x + 5x + 20 }\)

Does that help you?

Answer 2

Factor the first two and the last two terms.

Answer 3

that's it??

Answer 4

(x+4)*(x+5)

Answer 5

hmm okay. I knew the 4 and 5 but couldn't figure out how to write them out...

Answer 6

Basically you have to express b as the sum of two numbers \(\alpha\) and \(\beta \) such that \(\alpha \times \beta = ac\)

Answer 7

@ParthKohli I'm sure you know this, but factoring by grouping isn't necessary here?
(x + 4)(x + 5) = 0

Answer 8

\( \color{Black}{\Rightarrow x( x + 4) + 5(x + 4) }\)
\( \color{Black}{\Rightarrow (x + 5)(x + 4) }\)

Can you do the rest?

Answer 9

@Calcmathlete we have to factor this, rather than solving for x :)

Answer 10

wow now everyone is confusing me... its all written different

Answer 11

May I know where it is written?

Answer 12

It's just easy stuff haha you just gotta believe in yourself =)

Answer 13

its a problem in my text book I am trying to understand how to factor

Answer 14

Hmmm how much did you understand?

Answer 15

the first answer made sense and then the rest was so confusing haha

Answer 16

Hmm where did you get confused? I'd be happy to help you :)

Answer 17

if you could just start from the beginning and re-answer for me that would be fabulous!

Answer 18

am I done after this: x2+4x+5x+20 ?

Answer 19

Okay, so first I'll split the middle terms in two parts:

\( \color{Black}{\Rightarrow x^2 + 4x + 5x + 20 }\)

Notice how 4 * 5 = a * c

Answer 20

Now, you'll have to factor the first two and last two terms.

Answer 21

I'm not sure how to do that....

Answer 22

What is x^2 + 4x?

\( \color{Black}{\Rightarrow x^2 + 4x = x(x + 4)}\)

Answer 23

Now what is 5x + 20?

\( \color{Black}{\Rightarrow 5(x + 4) }\)

Answer 24

Now if you write that:

\( \color{Black}{\Rightarrow x(x + 4) + 5(x + 4) }\)

Now just group them(remember the distributive property?)

\( \color{Black}{\Rightarrow (x + 5)(x + 4) }\)

Answer 25

wait... what happened to the x and the 5 outside ( ) ?

Answer 26

Do you remember the distributive property? we're doing the reverse of that.

Answer 27

can you explain? I'm super rusty on all of this

Answer 28

\( \color{Black}{\Rightarrow a(x + n) = ax + an }\)

we're essentially reversing that.

Answer 29

I think I get it

Answer 30

Hmm ok

Answer 31

thank you very much

Answer 32

@ParthKohli lol. I was just saying before that you can essentially get the same result by finding factors of 20 that add to 9: 4 and 5. Then by pairing it up with an individual x value since the leading coefficient is found by multiplying the first terms in binomials.
(x + 4)(x + 5)
Factoring by grouping is only necessary if the leading term has a coefficient other than 1. :)

Answer 33

so the answer is (x + 4)(x + 5) ?

Answer 34

Yes.