Question

Factoring 2 of 11

Answer 1

Restarting my Computer. Hang tight

Answer 2

\[2x^2-21x+49\] a=2
b=-21
c=49
ac=98

\[(7)(14)\] are my products to get B
\[(-7)+(-14)=-21\]
\[-7*-14=49\]
\[2x^2-(-7-14)x+49\]
\[2x^2+7x+14x+49\]
\[x(2x+7)+7(2x+7)\]
\[(x+7)(2x+7)\]

Answer 3

when you replace the 21 it must be replaced with An equivalent expression. You replaced +21 with -21

Answer 4

Do you ssee this?

Answer 5

It’s a negative 21 though?

Answer 6

I'll help you understand better:

This is what you were given:
\(2x^2-21x+49\)

If you put the 21 in parentheses we have:
\(2x^2-(21)x+49\)

Notice the 21 in parentheses is positive, so that means when you replace it, whatever the expression is must also be positive. So the positive expression for 21 in this case is \(7 + 14\)

Answer 7

Then I had to distribute the negative to inside the parentheses?

Answer 8

And now when we directly replace the \((21)\) with \((7 + 14)\) we will have:
\(x^2 - (7 + 14)x + 49\)

Answer 9

And then after that is when you distribute the negative.

Answer 10

The point is:
\(x^2 -(21)x + 49 = x^2 - (7 + 14)x + 49\)

I would advise setting it up this way going forward just to make sure you don't confuse yourself anymore on that step.

Answer 11

Okay. I gotta go catch the bus, will you be here in a hour?

Answer 12

Sure

Answer 13

I'll color code it for you just to make it more obvious:
\(x^2 -(\color\green{21})x + 49 = x^2 - (\color\green{7 + 14})x + 49\)

because \(21 = 7 + 14\)

Answer 14

Thanks 👍

Answer 15

You're welcome

Answer 16

See you in an hour

Answer 17

So my answer is wrong, correct?

Answer 18

You should start with the correct approach.

Answer 19

x2−(7+14)x+49 is my equation right?

Answer 20

Your expression. An equation has an equal sign with two expressions on either side of it.

Answer 21

Sorry ;-;

Answer 22

\[x^2-(7+14)x+49)\]

Answer 23

Wouldn't I distribute the negative to the 7 and 14 as well as the x which would be -7x and -14x

Answer 24

Yes, the only mistake you made earlier was not substituting the correct expression. Otherwise all your steps were correct from what I saw. So go ahead and do the complete factorization

Answer 25

\[x^2-(7+14)x+49\]

Answer 26

\[x^2-7x-14x+49\]

Answer 27

I lose a 2 in thee beginning, oops x'D

Answer 28

\[2x2-7x-14x+49\]

Answer 29

Ah, I didn't even see that. Hang on

Answer 30

okay, yes, it was me that did not include the 2. My bad.

Answer 31

\[x(2x+7) + 7 (2x+7)\]

Answer 32

\[(x+7)(2x+7)\]

Answer 33

My advice to you would be to never skip steps. Let's start back with the correct expression. This time, do what you did above. Solve the whole thing in one response:

\(2x^2 -(21)x + 49\)

Answer 34

And never assume you know what the end result will be.

Answer 35

I myself don't even assume I know the end result.

Answer 36

\[2x^2-(21)x+49\]
\[2x^2-(7+14)x+49\]
\[2x^2-7x-14x+49\]
\[(2x^2-7x)-(-14x+49)\]
\[x(2x-7)- 7(2+7)\]
I messed up e.e

Answer 37

Ughhhhh.... I can't do this... Everything I think I make progress I am doing something else wrong..

Answer 38

You're making huge progress. Just please do as I suggested and do not be discouraged.

Answer 39

You're speaking math, and math is like me speaking Latin to most of the people on this website, it makes no sense.

Answer 40

Please try again. The mistakes you're making are very tiny. It's just sign errors for the most part. You pretty much have the overall method in tact.

Answer 41

I see though were the confusion is coming in. Let me just post the correct steps. The confusion comes in when you factor out the negative you forget to flipflop

Answer 42

\(2x^2-(21)x+49\) =\(2x^2 - (7 + 14)x + 49\)
=\(2x^2 - 7x - 14x + 49\)
=\(x(2x - 7) - 7(2x - 7)\)
=\((2x - 7)(x - 7)\)

Answer 43

I dunno... I'm lost now...

Answer 44

Why? The fact is, when you start off, I should have emphasized not worrying about negatives to begin with. So when you are trying to figure out the factors focus only on the positive for the b value. In this case, find two numbers that multiply to get 98 yet add to get 21. I think that's what's throwing you off. When finding the two numbers, the b should always be positive. You don't ignore the negative. Sorry for confusing you. We'll just have to keep practicing with these. Don't be discouraged. Everything you're experiencing is normal.

Answer 45

There are plenty more we can practice with. You should have the frame of mind that you'll understand it after doing ten to fifteen practice problems, not after doing just a few. The only thing you're confused on as far as what I see is factoring and distributing negatives.

Answer 46

Can we please practice more?

Answer 47

I'll post another.