Question

Completely factor the polynomial below.
p^2+19p+60
A. (p+4)(p+15)
B. (p+4)(p-15)
C. (p-4)(p+15)
D. (p-4)(p+5)

Answer 1

Hi @Mal87 how far have you gotten with factoring this one?

Answer 2

I haven't started it because I don't know how to do it

Answer 3

I see. So you are given a polynomial form \(ax^2 + bx + c\) and asked to factor it to a form \((ax + b)(cx + d)\), product of two binomials

In order to do this, the first step would be to find two numbers \(m\) and \(n\) such that
\(m + n = |b|\)
\(m \times n = (a)(c)\)

In this case, \(a = 1, b = 19, \text{and } c = 60\)
So:
\(m + n = |19|\)
\(m \times n = (1)(60)\)

Are you with me so far?

Answer 4

Not really I am still confused

Answer 5

At what point are you confused?

Answer 6

The part where you said m+n=absolute value of b

Answer 7

Since b can be either positive or negative, we need to ensure that b is positive for the method we will use to factor this. There is no need to get hung up on that. In this case, since b is positive we can simply say
\(m + n = b\)
\(m \times n = (a)(c)\)

And then

\(m + n = 19\)
\(m \times n = 60\)

Are you okay with that so far?

Answer 8

@Mal87

Answer 9

Yes I was processing it all

Answer 10

Okay, good now after we find \(m\) and \(n\) we will replace \(b\) with \(m + n\) in the given polynomial. In other words, since \(19 = m + n\) we will perform the following step:

\(p^2 + (19)p + 60 \rightarrow p^2 + (m + n)p + 60\)

Answer 11

Nevertheless in order to finish factoring we'd have to find the values of \(m\) and \(n\).

Answer 12

Seems familiar e.e

Answer 13

How do we find the values of \(m\) and \(n\)? We must factor 60 several times until we find the two numbers.

Answer 14

First we begin with
(1)(60) but clearly this does not work because \(m + n = 60 + 1 \ne 19\)

Answer 15

Next we try:
(2)(30) but of course this does not work either because \(m + n = 2 + 30 = 32 \ne 19\)

Answer 16

Okay

Answer 17

Again we try
(3)(20) but still this does not work because \(m + n = 3 + 20 = 23 \ne 19\)

Answer 18

Alas we try
(4)(15) and then \(m + n = 4 + 15 = 19\)

So looks like we've found our \(m\) and \(n\). We'll let \(m = 4 \text{ and } n = 19\)

Answer 19

Now we can substitute \(19\) with \(4 + 15\) in the quadratic expression:

\(p^2 + 19p + 60 \text{ becomes } p^2 + (4 + 15)p + 60\)

Answer 20

Next we use the distributive property \(a(b + c) = ab + ac\) to distribute the term \((4 + 15)p\)

Answer 21

Upon doing so we have:
\(p^2 + 4p + 15p + 60\)

Answer 22

Then next step is to factor by grouping. When you factor by grouping you factor the first two terms, then the last two terms. In other words we will factor \(p^2 + 4p\) first then factor \(15p + 60\) afterwards.

Answer 23

@Mal87 would you mind attempting to factor both expressions?

Answer 24

I will try to factor them yes

Answer 25

Awesome. And please post the results of each factorization here.

Answer 26

(p+4) for the first one and then for the second (p+15)

Answer 27

So for the first one you factored out the \(p\) correct and you ended up with \(p(p + 4)\) as the factorization of \(p^2 + 4p\) correct?

Answer 28

I got the answer A.

Answer 29

Yes, but do you feel you have learned how to factor these completely? If so, would you mind posting your steps here so that I know you know what you are doing? It's more about understanding the process rather than finding the correct answer.

Answer 30

For example, suppose you had to factor something like x2+14x+48. Would you be able to solve that right now using the method I showed you?

Answer 31

More than likely not I opened a docs and posted what you have written down and I am going to use that as an example.

Answer 32

Well here is the complete factorization of \(p^2 + 19p + 60\) just so you have all the steps:
\(p^2 + 19p + 60 = p^2 + 4p + 15p + 60\)
\(=p(p + 4) + 15(p + 4)\) (factor p from first two, factor 15 from last two terms)
\(=(p + 4)(p + 15)\) (notice that p + 4 is common to both factorizations, then factor that out

Answer 33

Okay thank you

Answer 34

You're welcome.

Answer 35

Would you like to try factoring \(x^2 + 14x + 48\) now?

Answer 36

Yes

Answer 37

Okay great. Go ahead and post your steps here when you are ready.

Answer 38

Okay so I got this far =x(x+1) And then I got confused

Answer 39

Did you find your \(m\) and \(n\) values?

Answer 40

If so, how did you go about finding them?

Answer 41

No i have not i got stuck

Answer 42

So first you have to make sure you identify your a, b, and c values. Do you know how to find a, b, and c?

Answer 43

No

Answer 44

So remember that \(x^2 + 14x + 48\) is a quadratic polynomial, the general form of which is \(ax^2 + bx + c\). So that means for the given polynomial, \(a = 1, b = 14, \text{ and } c = 48\).

Answer 45

Okay and then would you plug them in together? or no

Answer 46

The next step is to find two numbers \(m\) and \(n\) such that
\(m + n = |b|\)
\(m \times n = (a)(c)\)

Remember, in this case, \(a = 1, b = 14, \text{ and } c = 48\)

Answer 47

Do you have an idea of what to do next from here?

Answer 48

I think you would put the numbers with their letters like 1 would go with a And 48 would go with C.
So it would kind of look like this: m*n=(1)(48)

Answer 49

I see that you setup the second equation. What about the first equation. How do you set that up?

Answer 50

m+n=the absolute value of 14

Answer 51

Sorry I dont know how to do the absolute value bars on here

Answer 52

Absolute value just means whatever is in the bars, make it positive. Which in this case 14 is already positive. So you write m + n = 14

Answer 53

Do you think you might be able to factor 48 until we find two values that will work for \(m + n = 14\)?

Answer 54

8 and 6

Answer 55

Very good. Do you know the next step after this?

Answer 56

Would you factor out the two numbers? for the greatest common factor of both of them

Answer 57

Actually the next step is to take the 14 in \(x^2 + 14x + 48\) and replace it with 6 + 8

Answer 58

Oh okay. So then would it look like this: (m+6)(n+8)

Answer 59

Yes, you are getting away with taking that short cut right now because these polynomial expressions are very simple right now because \(a\) is always \(1\) for what you have been given so far. But what if I gave you something like \(4x^2 + 14x + 12\) and asked you to factor this as a product of two binomials?

Answer 60

I am still here I am looking for my notes

Answer 61

All my notes have only A as one

Answer 62

4x^2+4x+1
(4^2+2x)(2x+1)
2x(2x+1)(2x+1)
(2x+1)(2x+1)
(2x+1)^2
Did i do it right?? @Hero

Answer 63

The expression I gave you was \(4x^2 + 14x + 12\)

Answer 64

I did the wrong one I guess but would i have gotten the steps right?

Answer 65

There are some things missing in your steps. I don't recommend factoring that way.

Answer 66

Okay can you tell me what I was missing