4x^2+14x+6

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Parviz F. | Mathematics professor at Community CollegesMathematics professor at Community Colle...

4X^2 +14X + 6

Let 's factor 2 first:

2 ( 2X^2 +7X + 3)

To factor 2X^2 + 7X + 3 / 2 numbers, whose Sum is 7 . and product is 2*3 = 6, are( 6 ,1)

2 ( 2X^2 + 6X + X + 3 ) / break 7x= 6X +X

2 ( 2X^2 +6X + X + 3 ) =

2 [ 2X ( X + 3 ) + ( X +3) ] =

2 ( X+3) ( 2X + 1)

Factor 4x^{2}+14x+6

The factors will be either (4x+m)(x+n) or (2x+m)(2x+n)

(4x+m)(x+n) = 4x^{2} + (4n+m)x + mn, so (4m+n)=14, mn=6

or

(2x+m)(2x+n) = 4x^{2} + (2m+2n) + mn, so (2m+2n)=14, mn=6

Start by looking at the factors of m*n = 6: 6*1, 3*2, 2*3, 1*6. Let's look at the first potential factors, (4x+m)(x+n). If we pick m=2 and n=3, then

4x^{2} + (4*3+2)x + (3*2) = 4x^{2} + 14x + 6 **which works!**

So our factors are:

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