Factoring binomials with multiple variables p, x, and q

[Hoophy]

Homework Statement

Question: "If p and q are constants and x²+px+12 is equivalent to (x+3)(x+q), what is the value of p?"

I was told that the answer was 7 but I am not sure how to reach this value myself.

2. Homework Equations
I am not quite sure if this is applicable.

The Attempt at a Solution

1) To start I read the problem and saw that it said the two expressions were equivalent. So I set them equal to each other: x²+px+12=(x+3)(x+q)

2A) At this point I am confused about what to do next, so I try multiplying out the two factors: x²+px+12=x²+xq+3x+3q. With this I do not know what to do next, I believe I may have made things harder to solve and that this step was not useful. I then decide to start over and attempt to factor the trinomial from the question.

2B) x²+px+12: I have attempted to factor out the trinomial in the way I remember factoring in the past... This proved to be even more challenging. I started by multiplying the coefficient of x² 1 by the constant 12, I got 12. Then I try to find the factors of 12, I find 1 and 12, 2 and 6, and 3 and 4. However obviously none of these factors can be added to equal the middle term.

I am not sure if my mistake was setting the two expressions equal to each other, multiplying the two factors, attempting to factor the trinomial, or something else entirely.

This is my first day posting homework on the Physics Forum and I am not sure if I am doing it right. I realize that i did not even provide an attempted answer so please let me know if this is unacceptable, also if I did anything else wrong in this post please inform me. Thanks for the help!

 

[fresh_42]

What happens, if you compare all coefficients at $x^2 \, , \, x^1$ and $x^0 = 1$?

(Caution: (This might lead to confusion!) If you have polynomials with rational numbers and only consider rational equations, you may always think of $x$ as $\pi$. Of course this changes if you deal with real numbers where you can do arithmetic on $\pi$ and change it. But among rationals they both behave the same way.)

 

[Hoophy]

What happens, if you compare all coefficients at $x^2 \, , \, x^1$ and $x^0 = 1$?

(Caution: (This might lead to confusion!) If you have polynomials with rational numbers and only consider rational equations, you may always think of $x$ as $\pi$. Of course this changes if you deal with real numbers where you can do arithmetic on $\pi$ and change it. But among rationals they both behave the same way.)

I do not understand what you mean by comparing the coefficients of x², x¹, and x⁰. They are all 1 right? But what good does that do to know?

Also the second portion of your post did lead to confusion.

 

[fresh_42]

You have $x^2+px+12=x^2+xq+3x+3q$ or $x^2+px^1+12 x^0= 1 \cdot x^2 + (q+3) \cdot x^1 + 3q \cdot x^0$.
Coefficients are the numbers at the powers of the variable $x$: $1,p,12$ and $1,(q+3),3q$.
In order to have both polynomials equal, all coefficients must be the same in both polynomials.

(The example with $\pi$ only meant: if you replace $x$ by $\pi$ then there is only one way both expressions can be equal, because $\pi$ cannot be changed by means available in $\mathbb{Q}$. Forget it.)

 

[OmneBonum]

Ack! I posted an answer before but then I realized I was making things too complicated.

We already know that p has a value. That means we have to kill off q. If it also had a value we wouldn't be able to solve for p, so we have advance warning that this can be done.

So, look for a value of x that will assassinate q and use that find p.

Let me know if that helps and don't hesitate to ask more questions.

 

[TomHart]

Hoophy, you multiplied out the two terms and then set the equations equal. You are on the right track.

 

[OmneBonum]

It is not actually necessary to do any factoring, but the right side is composed of two factors, and it might help to see them that way. I don't know if Fresh 42's approach would work, but it is way, way more complicated than what's called for here!

 

[TomHart]

Here's a sample problem:
ax² + bx + c = 9x² + 4x + 13
Solve for constants a, b, and c.

 

[OmneBonum]

Hoophy, you multiplied out the two terms and then set the equations equal. You are on the right track.

Multiplying out the factors is completely unnecessary. You can get there that way but I would call it 'the long way round' rather than 'the right track.'

 

[haruspex]

I do not understand what you mean by comparing the coefficients of x², x¹, and x⁰. They are all 1 right?

fresh_42 means the coefficients on those terms in the quadratic.

 

[fresh_42]

Multiplying out the factors is completely unnecessary. You can get there that way but I would call it 'the long way round' rather than 'the right track.'

The primary goal in the homework section is not to find a nice trick or to deliver a solution, but to learn something.
At least it has been this, that drove my answer. It includes:

• forget the obscure term of equivalence in the OP and substitute it by equality, as it actually has been used in the OP
• what are coefficients
• which domain are we talking about
• what makes two polynomials equal
• a concept that holds true even in more general cases
• what is the difference between the variable $x$ and the constants $p$ and $q$
• what means that $\pi$ is a transcendental number

I admit, the last point had been a little too ambitious, but I couldn't know from the start. As soon as I recognized it, I first simplified the explanation and then dropped this point.

And the most difficult part of all: Students should get the knowledge, that they found it by themselves, or at least the feeling they could have. Therefore in this section it is more important than anywhere else to build upon what they already know and / or wrote. The points 2 and 3 in the homework template are there for good reasons!