SOLUTION: Factor the trinomial using the AC method. 4x^2-7x+5

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Question 718705: Factor the trinomial using the AC method.
4x^2-7x+5

Answer by MathLover1(20850) About Me  (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 4x%5E2-7x%2B5, we can see that the first coefficient is 4, the second coefficient is -7, and the last term is 5.



Now multiply the first coefficient 4 by the last term 5 to get %284%29%285%29=20.



Now the question is: what two whole numbers multiply to 20 (the previous product) and add to the second coefficient -7?



To find these two numbers, we need to list all of the factors of 20 (the previous product).



Factors of 20:

1,2,4,5,10,20

-1,-2,-4,-5,-10,-20



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



These factors pair up and multiply to 20.

1*20 = 20
2*10 = 20
4*5 = 20
(-1)*(-20) = 20
(-2)*(-10) = 20
(-4)*(-5) = 20


Now let's add up each pair of factors to see if one pair adds to the middle coefficient -7:



First NumberSecond NumberSum
1201+20=21
2102+10=12
454+5=9
-1-20-1+(-20)=-21
-2-10-2+(-10)=-12
-4-5-4+(-5)=-9




From the table, we can see that there are no pairs of numbers which add to -7. So 4x%5E2-7x%2B5 cannot be factored.



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Answer:



So 4%2Ax%5E2-7%2Ax%2B5 doesn't factor at all (over the rational numbers).



So 4%2Ax%5E2-7%2Ax%2B5 is prime.