Question

Show the work please

Answer 1

Answer:

1) (x + 3)(3x + 2)

2) x= +/-root6 - 1 by 5

Step-by-step explanation:

3x^2 + 11x + 6 = 0 (mid-term break)

using mid-term break

3x^2 + 9x + 2x + 6 = 0

factor out 3x from first pair and +2 from the second pair

3x(x + 3) + 2(x + 3)

factor out x+3

(x + 3)(3x + 2)

5x^2 + 2x = 1 (completing squares)

rearrange the equation

5x^2 + 2x - 1 = 0

divide both sides by 5 to cancel out the 5 of first term

5x^2/5 + 2x/5 - 1/5 = 0/5

x^2 + 2x/5 - 1/5 = 0

rearranging the equation to gain a+b=c form

x^2 + 2x/5 = 1/5

adding (1/5)^2 on both sides

x^2 + 2x/5 + (1/5)^2 = 1/5 + (1/5)^2

(x + 1/5)^2 = 1/5 + 1/25

(x + 1/5)^2 = 5 + 1 by 25

(x + 1/5)^2 = 6/25

taking square root on both sides

root(x + 1/5)^2 = +/- root(6/25)

x + 1/5 = +/- root6 /5

shifting 1/5 on the other side

x = +/- root6 /5 - 1/5

x = +/- root6 - 1 by 5

x = + root6 - 1 by 5 or x= - root6 - 1 by 5