Question

find the factors of x square + X upon 6 minus 1 upon 6​

Answer 1

Step-by-step explanation:

x^2+(x/6)-(1/6) = 0 .

multiply both the sides with 6.

6[x^2+(x/6)-(1/6)] =6[ 0 ]

=》6x2 + x - 1 =0.

Factoring  6x2 + x - 1 

The first term is,  6x2  its coefficient is  6 .

The middle term is,  +x  its coefficient is  1 .

The last term, "the constant", is  -1 

Step-1 : Multiply the coefficient of the first term by the constant   6 • -1 = -6 

Step-2 : Find two factors of  -6  whose sum equals the coefficient of the middle term, which is   1 .

     -6   +   1   =   -5

     -3   +   2   =   -1

     -2   +   3   =   1   That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -2  and  3 

                     6x2 - 2x + 3x - 1

Step-4 : Add up the first 2 terms, pulling out like factors :

                    2x • (3x-1)

              Add up the last 2 terms, pulling out common factors :

                     1 • (3x-1)

Step-5 : Add up the four terms of step 4 :

                    (2x+1)  •  (3x-1)

             Which is the desired factorization

2x+1 = 0 (or) 3x-1 = 0.

2x =-1 (or) 3x = 1

x=-1/2 (or) x=1/3.

therefore -1/2 & 1/3 are the factors of x^2+(x/6)-(1/6) = 0 .

Answer 2

Answer:

${x}^{2} + \frac{x}{6 } - \frac{1}{6} \\ = \frac{6 {x}^{2} + x - 1}{6} \\ = \frac{6 {x}^{2} + 3x - 2x - 1 }{6 } \\ = \frac{3x(2x + 1) - 1(2x + 1)}{6} \\ = \frac{(3x - 1)(2x + 1)}{6 } \\ hence \: factors \: are \\ \frac{1}{6} (3x - 1)(2x - 1)$

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