SOLUTION: x^2-5x+4, a^4+a^3+a^2, 3b^2+b+4

Algebra ->  Distributive-associative-commutative-properties -> SOLUTION: x^2-5x+4, a^4+a^3+a^2, 3b^2+b+4      Log On


   

Question 78006: x^2-5x+4,
a^4+a^3+a^2,
3b^2+b+4
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If you want to factor these then:
Lets factor x%5E2-5x%2B4
Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor 1%2Ax%5E2%2B-5%2Ax%2B4, first we need to ask ourselves: What two numbers multiply to 4 and add to -5? Lets find out by listing all of the possible factors of 4


Factors:

1,2,4,

-1,-2,-4,List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 4.

1*4=4

2*2=4

(-1)*(-4)=4

(-2)*(-2)=4

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to -5? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to -5

||||
First Number|Second Number|Sum
1|4|1+4=5
2|2|2+2=4
-1|-4|-1+(-4)=-5
-2|-2|-2+(-2)=-4
We can see from the table that -1 and -4 add to -5.So the two numbers that multiply to 4 and add to -5 are: -1 and -4 Now we substitute these numbers into a and b of the general equation of a product of linear factors which is: %28x%2Ba%29%28x%2Bb%29substitute a=-1 and b=-4 So the equation becomes: (x-1)(x-4) Notice that if we foil (x-1)(x-4) we get the quadratic 1%2Ax%5E2%2B-5%2Ax%2B4 again





Now lets factor a%5E4%2Ba%5E3%2Ba%5E2

a%5E2%28a%5E2%2Ba%2B1%29 Factor out a%5E2 This is the simplified form



Now lets factor 3b%5E2%2Bb%2B4 (we're going to use x instead of b)

Solved by pluggable solver: Factoring Quadratics with a leading coefficient of 1 (a=1)
In order to factor 3%2Ax%5E2%2B1%2Ax%2B4, first we need to ask ourselves: What two numbers multiply to 4 and add to 1? Lets find out by listing all of the possible factors of 4


Factors:

1,2,4,

-1,-2,-4,List the negative factors as well. This will allow us to find all possible combinations

These factors pair up to multiply to 4.

1*4=4

2*2=4

(-1)*(-4)=4

(-2)*(-2)=4

note: remember two negative numbers multiplied together make a positive number

Now which of these pairs add to 1? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 1

||||
First Number|Second Number|Sum
1|4|1+4=5
2|2|2+2=4
-1|-4|-1+(-4)=-5
-2|-2|-2+(-2)=-4
substitute a=-1 and b=-4 So the equation becomes: (x-1)(x-4) Notice that if we foil (x-1)(x-4) we get the quadratic 3%2Ax%5E2%2B1%2Ax%2B4 again None of these factors add to 1. So this quadratic cannot be factored. In order to solve for x, we need to use the quadratic formula.



So this means we cannot factor 3b%5E2%2Bb%2B4