Question

Factor completely 3x^2 + x + 7.

(3x + 1)(x + 7)

(3x + 7)(x + 1)

Prime

(3x + 4)(x + 3)

Answer 1

@geerky42 can you help?

Answer 2

Try to FOIL A, B, and D. if none of them matches \(3x^2 + x + 7\), then answer is C.

Answer 3

This is a method for factoring trinomials of the form \(ax^2 + bx + c\), where \(a\ne 1\).

Answer 4

First, multiply a and c.
In your case, a = 3, and c = 7.
What is 3 * 7 = ?

Answer 5

21

Answer 6

@mathstudent55

Answer 7

Good.

Answer 8

Now we use 21, and we need two numbers that multiply to 21 and add to b.
In your case, b = 1.
So, once again, you need to numbers that multiply to 21 and add to 1.
Can you find two such numbers?

Answer 9

no, because it would only be 3x7 really

Answer 10

would it be prime @mathstudent55

Answer 11

Sorry. My computer went nuts.
I'm back now.

Answer 12

OKay, Its fine :)

Answer 13

You are correct.
The only factors of 21 are 3 * 7 and 1 * 21 (and the negative pairs -3 * (-7) and -1 * (-21)).

None of those choices give you a sum of 1. That means the trinomnial can't be factored, and it's prime.

Answer 14

Thank you For your help @mathstudent55 :D

Answer 15

You're welcome.