SOLUTION: how can I factor 4x^2+16x+13?

Algebra.Com
Question 270291: how can I factor 4x^2+16x+13?
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
Solved by pluggable solver: Factoring using the AC method (Factor by Grouping)


Looking at the expression , we can see that the first coefficient is , the second coefficient is , and the last term is .



Now multiply the first coefficient by the last term to get .



Now the question is: what two whole numbers multiply to (the previous product) and add to the second coefficient ?



To find these two numbers, we need to list all of the factors of (the previous product).



Factors of :

1,2,4,13,26,52

-1,-2,-4,-13,-26,-52



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to .

1*52 = 52
2*26 = 52
4*13 = 52
(-1)*(-52) = 52
(-2)*(-26) = 52
(-4)*(-13) = 52


Now let's add up each pair of factors to see if one pair adds to the middle coefficient :



First NumberSecond NumberSum
1521+52=53
2262+26=28
4134+13=17
-1-52-1+(-52)=-53
-2-26-2+(-26)=-28
-4-13-4+(-13)=-17




From the table, we can see that there are no pairs of numbers which add to . So cannot be factored.



===============================================================





Answer:



So doesn't factor at all (over the rational numbers).



So is prime.
RELATED QUESTIONS

Factor... (answered by jim_thompson5910)
Factor: 4x^2 - 16x -... (answered by tommyt3rd)
How do I Factor Completely factory... (answered by edjones)
how to factor completely: 16x^2 + 4x - 2 =... (answered by richwmiller)
Can you please show me how to factor these two trinomials? 4x^2 + 16x - 8 x^2 -... (answered by MathLover1)
Factor. Check by multiplying... (answered by Fombitz)
How do I factor this? 4x^3y^5z^2 - 12x^4y^6z^4 + 16x^2y^5z^3 (answered by solver91311)
Can someone review this problem for me? I was supposed to factor 4x^2-7xy+16x-28y... (answered by venugopalramana)
How do I factor this trinomoial:... (answered by jim_thompson5910,stanbon)