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

we have

$\frac{3x^{2}+8x-3}{x+3}$

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

$x+\frac{4}{3} =(+/-)\frac{5}{3}$

$x=-\frac{4}{3}(+/-)\frac{5}{3}$

$x=-\frac{4}{3}+\frac{5}{3}=\frac{1}{3}$

$x=-\frac{4}{3}-\frac{5}{3}=-3$

so

$3x^{2}+8x-3=3(x-\frac{1}{3})(x+3)=(3x-1)(x+3)$

substitute

$\frac{3x^{2}+8x-3}{x+3}=\frac{(3x-1)(x+3)}{x+3}=(3x-1)$

therefore

the answer is the option B

$(3x-1)$

Answer 2

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

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

$Q=(3x^2+8x-3)\div(x+3)~~~~~~~~~~~~~~~~~~~(i)$

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

$Q\\\\\\=(3x^2+8x-3)\div(x+3)\\\\\\=\dfrac{3x^2+9x-x-3}{x+3}\\\\\\=\dfrac{3x(x+3)-1(x+3)}{x+3}\\\\\\=\dfrac{(x+3)(3x-1)}{(x+3)}\\\\\\=3x-1.$

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

Option (B) is correct.