Factoring Help?

Question

I need someone to give me step by step instructions on how to solve a problem like this. Please, give the step by step instructions as if I had no clue how to do math at all. Thanks!

Factor: p^2 + p + 6

Please do not factor using the quadractic method.

Answer 1

Ok you need to find factors of 6 that has a difference of 1(look at the middle term(p). the only factors of 6 w/ a difference of 1 is 2 & 3.
I think you're last term is incorrect & should be (-6) not +6 because if both the 2 & 3 are positive when we use the foil method it would add up to 5 not be a difference of 1. I'm gonna assume that your last term is incorrect so my answer would be :
(p-2)(p+3)=p^2+p-6.

Answer 2

You use the quadratic formula, because there really is no other method to solve this, when this equation cannot be factored using the method where there must be two numbers that add up to one and multiply to 6 (there are no two numbers of the sort) -

x = [-b +/- sqrt(b - 4ac)] / 2a
(In this case, x would be the same as p)

First of all, in your equation: p^2 + p + 6
you must name the values that are a, b, and c.
a = the coefficient of the variable that is being squared - this would be 1 (when there is no number in front of a variable, it is automatically assumed that it is 1)
b = the coefficient of the single variable - this would be 1 as well
c = the single number with no variable - this would be 6

Now input these variables into the equation:
p = [-1 +/- sqrt(1 - 4(1)(6))] / 2(1)

And solve, like so:
p = [-1 +/- sqrt(1 - 24] / 2
p = [-1 +/- sqrt(-23)] / 2

Now because there is a negative number under the square root sign, there is no answer to this problem (you CANNOT take the square root of any negative number - it is impossible!). It is because of this non-existent or "imaginary number", that the entire equation cannot be answered.

I hope this helped ^^;

Answer 3

Note
ax^2 + bx + c = 0 is the generic equation of a quadratic

p^2 + p + 6 = 0
I am assuming that it equals zero in order to find a solution

Three options
#1 - Factor -- but this equation does not factor easily (although it does factor!!)

#2 - quadratic equation

p = [ -b +/- sqrt(b^2 - 4ac) ] / (2a)
a = 1
b = 1
c = 6
so
p = [ -1 +/- sqrt(1^2 - 4*1*6 ] / (2*1)
p = [-1 +/- sqrt(1 - 24) ] / 2
p = [ -1 +/- sqrt(-23) ] /2
p = [ -1 +/- i sqrt(23) ] /2

Where i = sqrt(-1), the imaginary number (although not really imaginary, we call it that after DesCartes - but Einstein taught that it is very practical0

Answer 4

Your expression does not factorise.
Assume question is p² + 5p + 6
First step is to find two values of p such that p² is obtained.
These are p and p
(p-------).(p -------)
Now require two numbers that multiply to give + 6 and add/subtract to give 5p
These are 3 and 2:_
(p + 3).(p + 2) is then required answer.
Check
p.(p + 2) + 3.(p + 2)
= p² + 2p + 3p + 6
= p² + 5p + 6 (as required)

Answer 5

Looking at this, the only way to get that answer is to use i, the imaginary number that is equal to the square root of -1.
(p + 3 + i) (p - 2 + i)

Answer 6

Use the quadratic formula. Given ax^2 + bx + c = 0,
x = (-b +/- sqrt(b^2 - 4 * a * c))/(2 * a)

a = 1
b = 1
c = 6

p = (-1 +/- sqrt(1^2 - 4 * 1 * 6))/(2 * 1) =
(-1 +/- sqrt(1 - 24))/2 =
(-1 +/- sqrt(-23))/2
= -0.5 +/- sqrt(-23)

Answer 7

it isnt factorable over the intergers, but anyways if it was:
set it up like this:

(p+___)(p+___)
so for the two blanks, you need two numbers that multiply into 6 and add up to 1.
none of these are integers but thats the idea of factoring.