How do find the quotient of $\frac{6 x^{3} + 5 x^{2} + 2 x + 1}{2 x + 3}$ $(6x^3+5x^2+2x+1)/ (2x+3)$ ?

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How do find the quotient of $\frac{6 x^{3} + 5 x^{2} + 2 x + 1}{2 x + 3}$ $(6x^3+5x^2+2x+1)/ (2x+3)$ ?

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Answer 1 · 2016-05-12T17:04:10

$6 x^{3} + 5 x^{2} + 2 x + 1 = (2 x + 3) (3 x^{2} - 2 x + 4) - 11$ $6x^3+5x^2+2x+1 = (2x+3)(3x^2-2x+4)-11$

Explanation:

Suppose that $q (x)$ $q(x)$ is the quocient and $n (x) = 6 x^{3} + 5 x^{2} + 2 x + 1, d (x) = 2 x + 3, r (x) = c_{0}$ $n(x) = 6x^3+5x^2+2x+1, d(x)=2x+3,r(x)=c_0$ . Then $q (x) d (x) + r (x) = n (x)$ $q(x)d(x)+r(x)=n(x)$ Putting $q (x) = a x^{2} + b x + c$ $q(x) = ax^2+bx+c$ (Here $q (x)$ $q(x)$ degree must be $3 - 1$ $3-1$ because $d (x)$ $d(x)$ degree is $1$ $1$ . By the same reason $r (x)$ $r(x)$ degree is $0$ $0$ ).
After multiplication of $q (x) d (x) + r (x) - n (x) = 0$ $q(x)d(x)+r(x)-n(x)=0$ we get:
$(2 a - 6) x^{3} + (3 a + 2 b - 5) x^{2} + (3 b + 2 c - 2) x + 3 c + c_{0} - 1 = 0$ $(2a-6)x^3+(3a+2b-5)x^2+(3b+2c-2)x+3c+c_0-1=0$ . This equation must be identically null for all $x$ $x$ . Solving for $a, b, c, c_{0}$ $a,b,c,c_0$ we obtain:
$q (x) = 3 x^{2} - 2 x + 4, r (x) = - 11$ $q(x)=3x^2-2x+4, r(x)=-11$

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How do find the quotient of $\frac{6 x^{3} + 5 x^{2} + 2 x + 1}{2 x + 3}$ $(6x^3+5x^2+2x+1)/ (2x+3)$ ?

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How do find the quotient of $\frac{6 x^{3} + 5 x^{2} + 2 x + 1}{2 x + 3}$ $(6x^3+5x^2+2x+1)/ (2x+3)$ ?