SOLUTION: How do you factor 5x^2+21x+4=0? The AC method has been slightly puzzling me, and I cannot quite figure it out.

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Question 1031275: How do you factor 5x^2+21x+4=0? The AC method has been slightly puzzling me, and I cannot quite figure it out.
Found 2 solutions by Alan3354, MathTherapy:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
How do you factor 5x^2+21x+4=0? The AC method has been slightly puzzling me, and I cannot quite figure it out.
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Just by observation, 5*4 + 1*1 = 21
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--> (x + 4)*(5x + 1)
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

How do you factor 5x^2+21x+4=0? The AC method has been slightly puzzling me, and I cannot quite figure it out.
ac = + 5 * + 4 = + 20

Looking for a PAIR of factors that multiply to give: ac, or + 20, and SUM to "b", or + 21 (value of b, since the general form of a quadratic is: ).

These 2 factors are + 20 and + 1



 ------- Replacing middle term: + 21x with + 20x and + x

5x2 + 20x + x + 4 -------- Separating polynomials in order to factor each

5x(x + 4) + 1(x + 4)



Continue from here, setting each binomial equal to 0, and then solving for x 

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