# 20n squared -47n +21 3x squared +10x +4

20n squared -47n +21
3x squared +10x +4
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Kitamiliseakekw
$20{n}^{2}-47n+21$ factors into (4x-7)(5x-3)
$3{x}^{2}+10x+4$ cannot be factored. You will need to usethe binomial theorem to find the roots.

comAttitRize8
$20{n}^{2}-47n+21$
Factors of 20 : 1, 20 ; 2, 10; 4,5
Factors of 21: 1, 21 ; 3, 7
You can do this by trial and error and just try differentvaraitions, but if you cant get that you can do the "ACmethod" but thats a lot longer in my opinion, esp withthe A term and C term being large.
Trial and error. We want to try variations of the factors tillwe find one that works.
Try 4 and 5 for 20 and 7 and 3 for 21
(4n-7)(5n-3)
Multiplying the terms to check to see if we get 47n for themiddle term 12n + 35n = 47n
Since the last sign is positive, I know both signs have to bepositve or both negative.
Since the middle sign is negative, bothsign are negative.
(4n - 7)(5n -3)
$3{x}^{2}+10x+4$
Factors of 3 are 3 and 1
Factors of 4 are 1 and 4, 2 and 2
Since everything is positive are signs are positive.
(3x + 4)(x+1) 3x + 4x = 7x; No,does not equal 10x
(3x + 1) (x + 4) 12x + x = 13x; No
(3x + 2) (x + 2) 6x + 2x= 8x; No
Attempted all variations of the factors and none of themwork.
$3{x}^{2}+10x+4$ is a prime polynomial.