Question

What is the quotient: (3x2 + 8x – 3) ÷ (x + 3) ?. Answer. A) 3x + 5, r = 1. B) 3x – 1. C) 3x + 1. D) 3x – 5.

Answer 1

Answer: The correct option is (B)
`(3x-1).`

Step-by-step explanation: We are given to find the quotient in the following division:

`Q=(3x^2+8x-3)/(x+3)~~~~~~~~~~~~~~~~~~~(i)`

To do so, we need to factorize the numerator and cancel one of the the factor with the denominator if possible. The remaining factor will be the required quotient.

From equation (i), we have

`Q\\\\\\=(3x^2+8x-3)/(x+3)\\\\\\=(3x^2+9x-x-3)/(x+3)\\\\\\=(3x(x+3)-1(x+3))/(x+3)\\\\\\=((x+3)(3x-1))/((x+3))\\\\\\=3x-1.`

Thus, the required quotient is
`(3x-1).`

Option (B) is correct.

Answer 2

we have

`(3x^(2)+8x-3)/(x+3)`

Simplify the numerator------> complete the square

equate the numerator to zero

`3x^(2)+8x-3=0`

Group terms that contain the same variable, and move the constant to the opposite side of the equation

`3x^(2)+8x=3`

Factor the leading coefficient

`3(x^(2)+(8x/3))=3`

Complete the square. Remember to balance the equation by adding the same constants to each side.

`3(x^(2)+(8x/3)+(16/9))=3+(16/3)`

`3(x^(2)+(8x/3)+(16/9))=(25/3)`

`(x^(2)+(8x/3)+(16/9))=(25/9)`

Rewrite as perfect squares

`(x+(4/3))^(2)=(25/9)`

Square root both sides

`x+(4)/(3) =(+/-)(5)/(3)`

`x=-(4)/(3)(+/-)(5)/(3)`

`x=-(4)/(3)+(5)/(3)=(1)/(3)`

`x=-(4)/(3)-(5)/(3)=-3`

so

`3x^(2)+8x-3=3(x-(1)/(3))(x+3)=(3x-1)(x+3)`

substitute

`(3x^(2)+8x-3)/(x+3)=((3x-1)(x+3))/(x+3)=(3x-1)`

therefore

the answer is the option B

`(3x-1)`